Sequences and Series

1. INTRODUCTION

Sequence
A succession of numbers formed according to some definite rule is called a sequence. For example 1, 3, 5, 7,9 ……. is a sequence, here each term of the sequence can be obtained by adding 2 to the preceding term.

Types of Sequences
There are two types of sequence.

i) Finite sequence ii) Infinite sequence

a sequence is said to be a finite or infinite according as it has finite or infinite number of terms.

Series
If {f_n} be a sequence then an expression of the form f₁ + f₂ + …… + f_n is called series. In other word a series is the sum of the terms of the sequence.

Progression
If the terms of a sequence are written under specific condition then the sequence is called progression. There are three types of progressions.

i) Arithmetic Progression

ii) Geometric Progression

iii) Harmonic Progression

2. ARITHMETIC PROGRESSION

A sequence of numbers $<t_n>$ is said to be in arithmetic progression (A.P.) when the difference $t_n–t_{n–1}$ is a constant for all n ∈ N. This constant is called the common difference of the A.P. and is usually denoted by the letter d.
If ‘a’ is the first term and ‘d’ the common difference, then an A.P. can be represented as a,a+d,a+2d,a+3d

Example : 2, 7, 12, 17, 22, …… is an A.P. whose first term is 2 and common difference 5. Algorithm to determine whether a sequence is an A.P. or not.
Step I: Obtain  (the n^th term of the sequence).
Step II: Replace n by n – 1 in  to get $a_{n–1}$.
Step III: Calculate $a_n–a_{n–1}$.

If $a_n–a_{n–1}$ is independent from n, the given sequence is an A.P. otherwise it is not an A.P. ∴ $t_n=An+B$ represents the n^th term of an A.P. with common difference A. 

3. GENERAL TERM OF A.P

(1) Let ‘a’ be the first term and ‘d’ be the common difference of an A.P. Then its n^th term is i.e., $T_n=a+(n–1)d$.
(2) p^th term of an A.P. from the end : Let ‘a’ be the first term and ‘d’ be the common difference of an A.P. having n terms. Then p^th term from the end is ${\left(n–p+1\right)}^{th}$ term from the beginning
i.e., p^th term from the end = $T_{(n–p+1)}=a+(n–p)d$.

If last term of an A.P. is l then p^th term from end =  $l–(p–1) d$

4. SELECTION OF TERMS IN A.P

When the sum is given, the following way is adopted in selecting certain number of terms :

Number of terms	Terms to be taken
3	a – d, a, a + d
4	a – 3d, a – d, a + d, a + 3d
5	a – 2d, a – d, a, a + d, a + 2d

In general, we take a – rd, a – (r – 1)d, ……., a – d, a, a + d, ……, a + (r – 1)d, a + rd, in case we have to take (2r + 1) terms (i.e. odd number of terms) in an A.P. And, a –(2r – 1)d, a –(2r – 3)d……., a – d, a + d, ……, a + (2r – 1)d in case we have to take 2r terms in an A.P.
When the sum is not given, then the following way is adopted in selection of terms.

Number of terms	Terms to be taken
3	a, a+d, a+2d
4	a, a+d, a+2d, a+3d
5	a, a+d, a+2d, a+3d, a+4d

5. SUM OF n TERMS OF AN A.P

The sum of n terms of the series

a+(a+d)+(a+2d)+…..+ {a+(n-1)d} is given by

$S_n=\frac{n}{2}[2a+(n–1\left)d\right]$

Also, $S_n=\frac{n}{2}\left(a+l\right)$, where l = last term = a+(n-1)d.

6. PROPERTIES OF A.P

(1) If $a_1,a_2,a_3.....$ are in A.P. whose common difference is d, then for fixed non-zero number k ∈ R.
(i) $a_1\pm k,a_2\pm k, a_3\pm k,\dots .$ will be in A.P., whose common difference will be d.
(ii) $k{a}_1, k{a}_2, k{a}_3\dots$ will be in A.P. with common difference = kd.
(iii) $\frac{a_1}{k},\frac{a_2}{k},\frac{a_3}{k}\dots \dots$ will be in A.P. with common difference = d/k.

(2) The sum of terms of an A.P. equidistant from the beginning and the end is constant and is equal to sum of first and last term. i.e. $a_1+a_n=a_2+a_{n–1}=a_3+a_{n–2}=\dots$

(3) If number of terms of any A.P. is odd, then sum of the terms is equal to product of middle term and number of terms.

(4) If number of terms of any A.P. is even then A.M. of middle two terms is A.M. of first and last term.

(5) If the number of terms of an A.P. is odd then its middle term is A.M. of first and last term.

(6) If $a_1,a_2,.....a_n$ and $b_1,b_2,.....b_n$ are the two A.P.’s. Then $a_1\pm b_1,a_2\pm b_2,\dots \dots a_n\pm b_n$ are also A.P.’s with common difference $d_1\ne d_2$, where $d_1$ and $d_2$ are the common difference of the given A.P.’s.

(7) Three numbers a, b, c are in A.P. iff 2b=a+c.

(8) If $T_n,T_{n+1}$ and $T_{n+2}$ are three consecutive terms of an A.P., then $2T_{n+1}=T_n+T_{n+2}$.

(9) If the terms of an A.P. are chosen at regular intervals, then they form an A.P.

ARITHMETIC MEAN

If a, A, b are in A.P., then A is called A.M. between a and b.

(1) If $a,A_1,A_2,A_3,....A_n$,b are in A.P., then $A_1,A_2,A_3,\dots \dots ,A_n$ are called n A.M.’s between a and b.

(2) Insertion of arithmetic means

(i) Single A.M. between a and b : If a and b are two real numbers then single A.M. between a and b $=\frac{a+b}{2}$
(ii) n A.M.’s between a and b : If A₁, A₂, A₃, ……., A_n are n A.M.’s between a and b, then ,

$A_1=a+d=a+\frac{b–a}{n+1}$,   $A_2=a+2d=a+2\frac{b–a}{n+1}$

$A_3=a+3d=a+3\frac{b–a}{n+1},\dots \dots ,A_n=a+nd=a+n\frac{b–a}{n+1}$.

Illustration -1

If the 1st and the 2nd terms of an A.P are 1 and –3 respectively, find the nth term and the sum of the 1st n terms.

Solution

1st term = a, 2nd term = a + d where a = 1, a + d = -3,

⇒ d = -4 (Common difference of A.P.)

we have a_n = a + (n –1)d

= 1 + (n – 1) (-4) = 5 – 4n

S_n = $\frac{n}{2}${a + a_n}   = $\frac{n}{2}${1 + 5 – 4n} = n (3 – 2n)

Illustration -2

If 6 arithmetic means are inserted between 1 and 9/2, find the 4^th arithmetic mean.

Solution

Let a₁, a₂, a₃, a₄, a₅, a₆ be six arithmetic means.

Then  1, a₁, a₂, …, a₆, $\frac{9}{2}$ will be in A.P.

Now, $\frac{9}{2}$ = 1 + 7d

⇒$\frac{7}{2}$ = 7d  ⇒ d = $\frac{1}{2}$

Hence    a₄ = 1 + 4$\left(\frac{1}{2}\right)$ = 3

Illustration -3

The interior angles of a polygon are in arithmetic progression. The smallest angle is 120^o and the common difference is 5^o. Find the number of sides of the polygon.

Solution

Let the number of sides of the polygon be n. The sum of the interior angles of  the  polygon = (n -2)$\mathrm{\pi }$ = (n-2).180^o

Also the first term  of the A.P.= a = 120^o

The common difference = d = 5^o

$\Rightarrow \frac{\mathrm{n}}{2}\left[2.120°+(\mathrm{n}–1)5°\right]=(\mathrm{n}–2).180°$

$\Rightarrow \mathrm{n}[48+(\mathrm{n}–1\left)\right]=72(\mathrm{n}–2), \Rightarrow {\mathrm{n}}^2–25\mathrm{n}+144=0$

$\Rightarrow (n–9)(n–16)=0\Rightarrow n=9\text{ or }16$

For n=16, the largest  angle = 120^o+15 x 5^o = 195 ^o. This is not possible as an interior angle of a polygon cannot be greater than 180^o.  Hence n = 9.

Illustration -4

Let S_n denote  the  sum upto n terms of an A.P..If S_n = n²P and S_m = m²P ,  where m, n  and p are positive integers and  m$\ne$n, then find S_p.

Solution

Let first term be a and common difference be d, then

S_n = $\frac{\mathrm{n}}{2}$(2a + (n –1)d) = n²P

S_m = $\frac{\mathrm{m}}{2}$(2a + (m –1)d) = m²P

⇒ (n –m)d = (2n –2m)P ⇒ d = 2P and a = p

⇒ S_p = $\frac{\mathrm{p}}{2}$[2a + (p –1)d] = $\frac{\mathrm{p}}{2}$[2p + (p –1)2p] ⇒ S_p = p³.

Illustration -5

Find the number of terms in the series 20, $19\frac{1}{3},18\frac{2}{3}$, ….. of which  the sum is 300.

Solution

Clearly here a = 20, d = $–\frac{2}{3}$ and S_n = 300.

∴  $\left(\frac{\mathrm{n}}{2}\right)\left(2\times 20+(\mathrm{n}–1)\left(–\frac{2}{3}\right)\right)=300$.

Simplifying, n² – 61n + 900 = 0  n = 25 or 36.

Since common difference is negative and S₂₅ = S₃₆ = 300, it shows that the sum of the eleven terms i.e., T₂₆, T₂₇ , ….., T₃₆ is zero.

Illustration -6

If the sum of n terms of an A. P. is (pn + qn²), where p and q are constants, find the common difference.

Solution

$S_{\mathrm{n}}=\frac{\mathrm{n}}{2}(2\mathrm{a}+(\mathrm{n}–1\left)\mathrm{d}\right)$

$=\mathrm{n}\left(\mathrm{a}–\frac{\mathrm{d}}{2}\right)+{\mathrm{n}}^2\frac{\mathrm{d}}{2}$

$\mathrm{a}–\frac{\mathrm{d}}{2}=\mathrm{p}\text{ and }\mathrm{q}=\frac{\mathrm{d}}{2}$

$\Rightarrow \mathrm{a}=\mathrm{p}+\mathrm{q} \& \mathrm{d}=2\mathrm{q}$

on comparing S_n with given sum

Illustration -7

If the p^th term of an A.P. be q and q^th term be p, then its r^th term will be

(A) p+q+r          (B) p+q-r          (C) p+r-q          (D) p-q-r

Solution

(B)

Given that,  $T_p=a+(p–1)d=q$    …..(i)

and  $T_q=a+(q–1)d=p$  ….. (ii)

From (i) and (ii), we get $d=–\frac{(p–q)}{(p–q)}=–1$

Putting value of  in equation (i), then   a = p+q-1

Now, r^th term is given by A.P.

$T_r=a+(r–1)d$ = (p+q-1)+(r-1)(-1) = p+q-r

Illustration -8

If $2x, x+8, 3x+1$ are in A.P., then the value of x will be

(A) 3          (B) 7          (C) 5                             (D) – 2

Solution

(C)

$2x, x+8, 3x+1$are in A.P.

Therefore  $(x+8)=\frac{\left(2x\right)+(3x+1)}{2}=\frac{5x+1}{2}$ $\Rightarrow 2x+16=5x+1$ $\Rightarrow 3x=15\Rightarrow x=5$.

Illustration -9

If the 9^th term of an A.P. is 35  and 19^th is 75, then its 20^th terms will be

(A) 78                (B) 79            (C) 80           (D) 81

Solution

(B)

$T_9=a+8d=35$ and $T_{19}=a+18d=75$

Solving the equations, we get  d=4 and a=3

Hence 20^th term of A.P. a+19d = 3+19⨯4 = 79.  

Illustration -10

The 9^th term of the series $27+9+5\frac{2}{5}+3\frac{6}{7}+.......$ will be

(A) $1\frac{10}{17}$          (B)  $\frac{10}{17}$             (C)  $\frac{16}{27}$            (D) $\frac{17}{27}$

Solution

(A)

Given series $27+9+5·\frac{2}{5}+3·\frac{6}{7}+\dots \dots$ = $27+\frac{27}{3}+\frac{27}{5}+\frac{27}{7}+\dots ..+\frac{27}{2n–1}+\dots \dots$

Hence n^th term of given series $T_n=\frac{27}{2n–1}$

So, $T_9=\frac{27}{2\times 9–1}=\frac{27}{17}=1\frac{10}{17}$.

Illustration -11

If the p^th, q^thand r^th term of an arithmetic sequence are a , b and  respectively, then the value of  $\left[\mathrm{a}\right(\mathrm{q}–\mathrm{r})+\mathrm{b}(\mathrm{r}–\mathrm{p})+\mathrm{c}(\mathrm{p}–\mathrm{q}\left)\right]$ =

(A) 1                 (B) -1                            (C) 0                             (D) 1/2

Solution

(C)

Suppose that first term and common difference of A.P.’s are and D respectively.

Now, p^th term = A+(p-1)D = a         …..(i)

q^th term = A+(q-1)D = b   ……(ii)

and  r^th  term = A+(r-1)D = c        …..(iii)

So, a(q-r)+b(r-p)+c(p-q)

$=a\left\{\frac{b–c}{D}\right\}+b\left\{\frac{c–a}{D}\right\}+c\left\{\frac{a–b}{D}\right\}$ 

= $\frac{1}{D}(ab–ac+bc–ab+ca–bc)=0$.

Illustration -12

If p times the p^th term of an A.P. is equal to q times the q^th term of an A.P., then (p+q)^th term is

(A) 0              (B) 1           (C) 2                 (D) 3

Solution

(A)

$p\{a+(p–1\left)d\right\}=q\{a+(q–1\left)d\right\}$

$\Rightarrow a(p–q)+\left(p^2–q^2\right)d+(q–p)d=0$

$\Rightarrow (p–q)\{a+(p+q–1\left)d\right\}=0$

$\Rightarrow a+(p+q–1)d=0\Rightarrow T_{p+q}=0, \{\because p\ne q\}$.

Illustration -13

The sums of n terms of two arithmetic series are in the ratio 2n+3 : 6n+5, then the ratio of their 13^th terms is

(A) 53 : 155              (B) 27 : 77          (C) 29 : 83          (D) 31 : 89  

Solution

(A)

We have $\frac{S_{n_1}}{S_{n_2}}=\frac{2n+3}{6n+5}$

$\Rightarrow \frac{\frac{n}{2}\left[2a_1+(n–1)d_1\right]}{\frac{n}{2}\left[2a_2+(n–1)d_2\right]}=\frac{2n+3}{6n+5}$

$\Rightarrow \frac{2\left[a_1+\left(\frac{n–1}{2}\right)d_1\right]}{2\left[a_2+\left(\frac{n–1}{2}\right)d_2\right]}=\frac{2n+3}{6n+5}$

$\Rightarrow \frac{a_1+\left(\frac{n–1}{2}\right)d_1}{a_2+\left(\frac{n–1}{2}\right)d_2}=\frac{2n+3}{6n+5}$

Put, n=25 then $\frac{a_1+12d_1}{a_2+12d_2}=\frac{2\left(25\right)+3}{6\left(25\right)+3}$$\Rightarrow \frac{T_{{13}_1}}{T_{{13}_2}}=\frac{53}{155}$.

Illustration -14

Let $T_r$ be the r^th term of an A.P. for r = 1,2,3,…… If for some positive integers m,n we have $T_m=\frac{1}{n}$ and  $T_n=\frac{1}{m}$, then$T_{mn}$  equals

(A)  $\frac{1}{mn}$         (B) $\frac{1}{m}+\frac{1}{n}$                (C) 1                   (D) 0

Solution

(C)

$T_m=a+(m–1)d=\frac{1}{n}$ 

and $T_n=a+(n–1)d=\frac{1}{m}$

On solving   $a=\frac{1}{mn}\text{ and }d=\frac{1}{mn}$

$\therefore T_{mn}=a+(mn–1)d=\frac{1}{mn}+(mn–1)\frac{1}{mn}=1$

Illustration -15

The sum of the series $\frac{1}{2}+\frac{1}{3}+\frac{1}{6}+.....$ to 9  terms is

(A)  $–\frac{5}{6}$                         (B) $–\frac{1}{2}$               (C)1                (D) $–\frac{3}{2}$

Solution

(D)

Given series $\frac{1}{2}+\frac{1}{3}+\frac{1}{6}+.....$

Here $a=\frac{1}{2}$, common difference $d=–\frac{1}{6}$ and n=9.

$\Rightarrow S_9=\frac{9}{2}\left[2\times \frac{1}{2}+(9–1)\left(–\frac{1}{6}\right)\right]=–\frac{3}{2}$.

Illustration -16

The first term of an A.P. is 2 and common difference is 4. The sum of its 40 terms will be

(A) 3200            (B) 1600            (C) 200              (D) 2800

Solution

(A)

First term a=2 and common difference d=4 and n=40.

Then $S_n=\frac{n}{2}[2a+(n–1\left)d\right]$.

= $20[4+39\times 4]=20[4+156]=160\times 20=3200$

Illustration -17

If the sum of the series 2+5+8+11…….. is 60100, then the number of terms are

(A) 100              (B) 200             (C) 150            (D) 250

Solution

(B)

Series,2+5+8+11+……

a=2, d=3 and let number of terms is n

then sum of A.P. = $\frac{n}{2}\{2a+(n–1\left)d\right\}$

$\Rightarrow 60100=\frac{n}{2}\{2\times 2+(n–1\left)3\right\}\Rightarrow 120200=n(3n+1)$

$\Rightarrow 3n^2+n–120200=0\Rightarrow (n–200)(3n+601)=0$

Hence n=200.

Illustration -18

If the first term of an A.P. be 10, last term is 50 and the sum of all the terms is 300, then the number of terms are

(A) 5              (B) 8                   (C) 10                (D) 15

Solution

(C)

Given that first term a=10, last term l=50 and sum S=300.

$\therefore S=\frac{n}{2}(a+l)\Rightarrow 300=\frac{n}{2}(10+50)\Rightarrow n=10$.

Illustration -19

The ratio of sum of m and n terms  of an A.P. is $m^2 : n^2$, then the ratio of $m^{th}$ and $n^{th}$ term will be

(A) $\frac{m–1}{n–1}$                (B) $\frac{n–1}{m–1}$           (C) $\frac{2m–1}{2n–1}$         (D) $\frac{2n–1}{2m–1}$

Solution

(C)

Given that   $\frac{\frac{m}{2}[2a+(m–1\left)d\right]}{\frac{n}{2}[2a+(n–1\left)d\right]}=\frac{m^2}{n^2}$

$\Rightarrow \frac{2a+(m–1)d}{2a+(n–1)d}=\frac{m}{n}$$\Rightarrow \frac{a+\frac{1}{2}(m–1)d}{a+\frac{1}{2}(n–1)d}=\frac{m}{n}$

$\Rightarrow an+\frac{1}{2}(m–1)nd=am+\frac{1}{2}(n–1)md$

$\Rightarrow a(n–m)+\frac{d}{2}[mn–n–mn+m]=0$

$\Rightarrow a(n–m)+\frac{d}{2}(m–n)=0\Rightarrow a=\frac{d}{2}\text{ or }d=2a$

So, required ratio, $\frac{T_m}{T_n}=\frac{a+(m–1)d}{a+(n–1)d}=\frac{a+(m–1)2a}{a+(n–1)2a}$ $=\frac{1+2m–2}{1+2n–2}=\frac{2m–1}{2n–1}$.

Trick : Replace m by 2m-1 and n by 2n-1. Obviously if $S_m$ is of degree 2, then $T_m$ is of 1  linear.

Illustration -20

Four numbers are in arithmetic progression. The sum of first and last term is 8 and the product of both middle terms is 15. The least number of the series is

(A) 4                     (B) 3               (C) 2            (D) 1

Solution

(D)

Let A₁, A₂, A_3……. and A₄ are four numbers in A.P.

$A_1+A_4=8$…..(i)    and   $A_2.A_3=15$…..(ii)

The sum of terms equidistant from the beginning and end is constant and is equal to sum of first and last terms.

Hence, $A_2+A_3=A_1+A_4=8$        …..(iii)

From (ii) and (iii),

$A_2+\frac{15}{A_2}=8\Rightarrow A_2^2–8A_2+15=0$

$A_2=3\text{ or }5\text{ and }{A}_3=5\text{ or 3}$

As we know,  $A_2=\frac{A_1+A_3}{2}\Rightarrow A_1=2A_2–A_3$

$\Rightarrow A_1=2\times 3–5=1\text{ and }{A}_4=8–A_1=7$

Hence the series is, 1, 3, 5, 7.

So that least number of series is 1.

8. GEOMETIRC PROGRESSION

A progression is called a G.P. if the ratio of its each term to its previous term is always constant. This constant ratio is called its common ratio and it is generally denoted by r.

Example: The sequence 4, 12, 36, 108, ….. is a G.P., because $\frac{12}{4}=\frac{36}{12}=\frac{108}{36}=\dots ..=3$, which is constant.

Clearly, this sequence is a G.P. with first term 4 and common ratio 3.

The sequence $\frac{1}{3},–\frac{1}{2},\frac{3}{4},–\frac{9}{8},.....$ is a G.P. with first term $\frac{1}{3}$ and common ratio $\left(–\frac{1}{2}\right)/\left(\frac{1}{3}\right)=–\frac{3}{2}$. 

9. GENERAL TERMS OF G.P

(1) We know that, $a,ar,a{r}^2,a{r}^3,\dots ..a{r}^{n–1}$ is a sequence of G.P.

Here, the first term is ‘a’ and the common ratio is ‘r’.

The general term or n^th term of a G.P. is $T_n=a{r}^{n–1}$.

It should be noted that, $r=\frac{T_2}{T_1}=\frac{T_3}{T_2}=\dots \dots$.

(2) p^th term from the end of a finite G.P. : If G.P. consists of ‘n’ terms, p^th term from the end $=(n–p+1{)}^{th}$ term from the beginning = $a{r}^{n–p}$.

Also, the p^th term from the end of a G.P. with last term l and common ratio r is $l{\left(\frac{1}{r}\right)}^{n–1}$.

10. SELECTION OF TERMS IN G.P

(1) When the product is given, the following way is adopted in selecting certain number of terms:

Number of terms	Terms to be taken
3	$\frac{a}{r},a,ar$
4	$\frac{a}{r^3},\frac{a}{r},ar,a{r}^3$
5	$\frac{\mathrm{a}}{{\mathrm{r}}^2},\frac{\mathrm{a}}{\mathrm{r}},\mathrm{a},\mathrm{ar},{\mathrm{ar}}^2$

(2) When the product is not given, then the following way is adopted in selection of terms  

Number of terms	Terms to be taken
3	a, ar, ar2
4	a, ar, ar2, ar3
5	a, ar, ar2, ar3, ar4

11. SUM OF n TERMS OF AN G.P

If a be the first term, r the common ratio, then sum $S_n$ of first n terms of a G.P. is given by

$S_n=\frac{a\left(1–r^n\right)}{1–r}\text{ and }{S}_n=\frac{a–lr}{1–r}$,     (when |r|< 1)

$S_n=\frac{a\left(r^n–1\right)}{r–1}\text{ and }{S}_n=\frac{lr–a}{r–1}$,     (when |r|> 1)

$S_n=na$,   (when r = 1)

12. SUM OF INFINITE TERMS OF A G.P.

(1) When |r|< 1,    (or -1 < r < 1); $S_{\infty }=\frac{a}{1–r}$.

(2) If r $\ge$ 1, then $S_{\infty }$ doesn’t exist.

13. GEOMETRIC MEAN

If a, G, b are in G.P., then G is called G.M. between a and b.

(1) If $a,G_1,G_2,G_3,\dots .G_n,b$ are in G.P. then $G_1,G_2,G_3,\dots .G_n$ are called n G.M.’s between a and b.

(2) Insertion of geometric means : (i) Single G.M. between a and b : If a and b are two real numbers then single G.M. between a and b = $\sqrt{ab}$.

(ii) n G.M.’s between a and b : If $G_1,G_2,G_3,\dots .G_n$ are n G.M.’s between a and b, then

$G_1=ar=a{\left(\frac{b}{a}\right)}^{\frac{1}{n+1}}, G_2=a{r}^2=a{\left(\frac{b}{a}\right)}^{\frac{2}{n+1}}$,

$G_3=a{r}^3=a{\left(\frac{b}{a}\right)}^{\frac{3}{n+1}}$, ……………….., $G_n=a{r}^n=a{\left(\frac{b}{a}\right)}^{\frac{n}{n+1}}$.

14. PROPERTIES OF G.P

(1) If all the terms of a G.P. be multiplied or divided by the same non-zero constant, then it remains a G.P., with the same common ratio.

(2) The reciprocal of the terms of a given G.P. form a G.P. with common ratio as reciprocal of the common ratio of the original G.P.

(3) If each term of a G.P. with common ratio r be raised to the same power k, the resulting sequence also forms a G.P. with common ratio r^k.

(4) In a finite G.P., the product of terms equidistant from the beginning and the end is always the same and is equal to the product of the first and last term. i.e., if $a_1,a_2,a_3,\dots \dots a_n$ be in G.P.

Then $a_1{a}_n=a_2{a}_{n–1}=a_3{a}_{n–2}=a_4{a}_{n–3}=\dots \dots \dots .=a_r·a_{n–r+1}$

(5) If the terms of a given G.P. are chosen at regular intervals, then the new sequence so formed also forms a G.P.

(6) If $a_1,a_2,a_3,\dots \dots a_n........$ is a G.P. of non-zero, non-negative terms, then  is an A.P. and vice-versa.

(7) Three non-zero numbers a, b, c are in G.P., iff $b^2=ac$.

(8) If first term of a G.P. of n terms is a and last term is l, then the product of all terms of the G.P. is ${\left(al\right)}^{\raisebox{1ex}{$n$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$.

(9) If there be n quantities in G.P. whose common ratio is r and $S_m$ denotes the sum of the first m terms, then the sum of their product taken two by two is $\frac{r}{r+1}{S}_n{S}_{n–1}$.

(10) If $a^{x_1},a^{x_2},a^{x_3},\dots ,a^{x_n}$ are in G.P., then $x_1,x_2,x_3,\dots ,x_n$ will be are  in  A.P.,

15. INCREASING AND DECREASING G.P

Let a, ar, ar², ……. be G..P

a) If a > 0; r > 1 then it is an increasing G..P

b) If a > 0; 0 < r < 1 then it is decreasing G..P

c) If a < 0; r > 1 then it is decreasing G.P

d) If a < 0; 0 < r < 1 then it is an increasing G.P

Illustration -21

The third term of a G.P. is 7. Find the product of first five terms.

Solution

Let the terms be $\frac{a}{r^2},\frac{a}{r}$, a , ar, ar²

⇒ a = 7.

The product   = a⁵   = 7⁵.

Illustration -22

The sum of three numbers in GP is 21 and the sum of their squares is 189, find the numbers.

Solution

Let the three numbers be a, ar, ar²

Given a + ar + ar² = 21  ⇒ a (1 + r + r²) = 21   …(1)

And a² + a²r² + a²r⁴ = 189 ⇒ a²( 1 + r² + r⁴) = 189    …(2)

Squaring (1) and dividing (2) by it, we get

$\frac{a^2\left(1+r^2+r^4\right)}{a^2{\left(1+r+r^2\right)}^2}=\frac{189}{441}=\frac{3}{7}\Rightarrow \frac{1+2r^2+r^4–r^2}{{\left(1+r+r^2\right)}^2}=\frac{3}{7}$

$\Rightarrow \frac{{\left(1+r^2\right)}^2–r^2}{{\left(1+r+r^2\right)}^2}=\frac{3}{7}\Rightarrow \frac{1+r^2–r}{1+r+r^2}=\frac{3}{7}$ ⇒  7 ( 1+ r² – r ) = 3( 1+ r+ r²)

⇒ 4r² –10r + 4 = 0   ⇒ 2r² –5r + 2 = 0  ⇒ r = 2, $\frac{1}{2}$

Hence the three numbers are 3, 6, 12 or 12, 6, 2

Illustration -23

Find the sum $\left(1+\frac{1}{2^2}\right)+\left(\frac{1}{2}+\frac{1}{2^4}\right)+\left(\frac{1}{2^2}+\frac{1}{2^6}\right)+\dots .$ of up to infinite terms.

Solution

$\left(1+\frac{1}{2^2}\right)+\left(\frac{1}{2}+\frac{1}{2^4}\right)+\left(\frac{1}{2^2}+\frac{1}{2^6}\right)+\dots .$

$=\left(1+\frac{1}{2}+\frac{1}{2^2}+\dots \right)+\left(\frac{1}{2^2}+\frac{1}{2^4}+\frac{1}{2^6}+\dots \right)$

$=\frac{1}{1–\frac{1}{2}}+\frac{\frac{1}{2^2}}{1–\frac{1}{2^2}}=2+\frac{1}{4}\times \frac{4}{3}=2\frac{1}{3}$

Illustration -24

The first term of an infinite G..P is 1 and any term is equal to the sum of all the succeeding terms. find the series.

Solution

Given that ${\mathrm{T}}_{\mathrm{p}}=\left({\mathrm{T}}_{\mathrm{p}+1}+{\mathrm{T}}_{\mathrm{p}+2}+.....\right)$  or, ${\mathrm{ar}}^{\mathrm{p}–1}={\mathrm{ar}}^{\mathrm{p}}+{\mathrm{ar}}^{\mathrm{p}+1}+{\mathrm{ar}}^{\mathrm{p}+2}+\dots$

$\therefore {\mathrm{r}}^{\mathrm{p}–1}=\frac{{\mathrm{r}}^{\mathrm{p}}}{1–\mathrm{r}}$ [sum of an infinite G.P.]

∴  1 – r = r ⇒ r = $\frac{1}{2}$.

Hence the series is $1,\frac{1}{2},\frac{1}{4},\frac{1}{8},\dots \infty$.

Illustration -25

If a G. P. the first term is 7, the last term 448, and the sum 889; find the common ratio.

Solution

a = 7, l = ar^n-1 = 448

$S_n=\frac{a\left(r^n–1\right)}{r–1}=889$

Here  $S_n=\frac{r.\left(a{r}^{n–1}\right)–a}{r–1}=\frac{448r–a}{r–1}$

⇒ r = 2 & a = 7.

Illustration -26

Find the sum of the geometric series 2 + 6 + 18 + 54 + . . . where there are 6 terms in the series.

Solution

For this series, we have a = 2, r = 3 and n = 6. So

$s_n=\frac{a\left(1–r^n\right)}{1–r}$

$s_6=\frac{2\left(1–3^6\right)}{1–3}=\frac{2(1–729)}{–2}$

= −(−728)= 728.

Illustration -27

Find the sum of the geometric series 8 − 4 + 2 − 1 + . . . where there are 5 terms in the series.

Solution

For this series, we have a = 8, $r=–\frac{1}{2}$ and n = 5. So

$s_n=\frac{a\left(1–r^n\right)}{1–r}$

$s_5=\frac{8\left(1–{\left(–\frac{1}{2}\right)}^5\right)}{1–\left(–\frac{1}{2}\right)}=\frac{8\left(1–\left(–\frac{1}{32}\right)\right)}{\frac{3}{2}}$ = $\frac{2\times 8\times \frac{33}{32}}{3}=\frac{11}{2}=5\frac{1}{2}$

Illustration -28

How many terms are there in the geometric progression 2, 4, 8, . . . , 128 ?

Solution

In this sequence a = 2 and r = 2. We also know that the n-th term is 128. But the formula for the n-th term is arⁿ⁻¹. So

128 = 2 × 2ⁿ⁻¹

64 = 2ⁿ⁻¹

2⁶ = 2ⁿ⁻¹

6 = n – 1

n = 7. So there are 7 terms in this geometric progression.

Illustration -29

How many terms in the geometric progression 1, 1.1, 1.21, 1.331, . . . will be needed so that the sum of the first n terms is greater than 20?

Solution

The sequence is a geometric progression with a = 1 and r = 1.1. We want to find the smallest value of n such that S_n > 20. Now

$s_n=\frac{a\left(1–r^n\right)}{1–r}$

$\frac{1\times \left(1–1.1^n\right)}{1–1.1}>20$

(1.1ⁿ − 1) × 10 > 20

1.1ⁿ − 1 > 2

1.1ⁿ > 3.

If we now take logarithms of both sides, we get n ln 1.1 > ln 3 and as ln 1.1 > 0 we obtain n > ln 3/ ln 1.1 = 11.5267 . . .

and therefore the smallest whole number value of n is 12.

Illustration -30

How many terms of the G.P. $3,\frac{3}{2},\frac{3}{4},$ … are needed to give the Sum $\frac{3069}{512}$?

Solution

Let n be the number of terms needed. Given that a = 3, $r=\frac{1}{2}$and Sum $\frac{3069}{512}$

Since $s_n=\frac{a\left(1–r^n\right)}{1–r}$

Therefore $\frac{3069}{512}=\frac{3\left(1–\frac{1}{2^n}\right)}{1–\frac{1}{2}}=6\left(1–\frac{1}{2^n}\right)$

or  $\frac{3069}{3072}=1–\frac{1}{2^n}$ or $\frac{1}{2^n}=1–\frac{3069}{3072}=\frac{3}{3072}=\frac{1}{1024}$

or   2ⁿ = 1024 = 2¹⁰, which gives n = 10.

Illustration -31

The sum of first three terms of a G.P. $\frac{13}{12}$ is and their product is – 1.

Find the common ratio and the terms.

Solution

Let $\frac{a}{r}$, a, ar be the first three terms of the G.P. Then

$\frac{a}{r}+ar+a=\frac{13}{12}$ ^{_______________} (1)

and  $\left(\frac{a}{r}\right)\left(a\right)\left(ar\right)=–1$   ^{_______________} (2)

From (2), we get a³ = – 1, i.e., a = – 1 (considering only real roots)

Substituting a = –1 in (1), we have $–\frac{1}{r}–1–r=\frac{13}{12}$ or 12r² + 25r + 12 = 0.

This is a quadratic in r, solving, we get $r=–\frac{3}{4}\text{ or }–\frac{4}{3}$ .

Thus, the three terms of G.P. are : $\frac{4}{3},–1,\frac{3}{4}$ for $r=\frac{–3}{4}$ and $\frac{3}{4},–1\frac{4}{3}$ for  $r=–\frac{4}{3}$

Illustration -32

Find the sum of the sequence 7, 77, 777, 7777, … to n terms.

Solution

This is not a G.P., however, we can relate it to a G.P. by writing the terms as

S_n = 7 + 77 + 777 + 7777 + … to n terms

= $\frac{7}{9}$ [9 + 99 + 999 + 9999 +… to n terms]

= $\frac{7}{9}$ [(10 – 1) + (10² – 1) + (10³ – 1) + (10⁴ – 1) + …… n terms]

= $\frac{7}{9}$ [(10 + 10² + 10³ + … n terms) – (1+1+1+… n terms)]

= $\frac{7}{9}\left[\frac{10\left({10}^n–1\right)}{10–1}–n\right]=\frac{7}{9}\left[\frac{10\left({10}^n–1\right)}{9}–n\right]$

Illustration -33

A person has 2 parents, 4 grandparents, 8 great grandparents, and so on.

Find the number of his ancestors during the ten generations preceding his own.

Solution

Here a = 2, r = 2 and n = 10

Using the sum formula $S_n=\frac{a\left(r^n–1\right)}{r–1}$

We have S₁₀ = 2(2¹⁰ – 1) = 2046

Hence, the number of ancestors preceding the person is 2046.

Illustration -34

Insert three numbers between 1 and 256 so that the resulting sequence is a G.P

Solution

Let G₁, G₂,G₃ be three numbers between 1 and 256 such that 1, G₁,G₂, G₃ ,256 is a G.P.

Therefore  256 = r⁴ giving r = ± 4 (Taking real roots only)

For r = 4, we have G₁ = ar = 4, G₂ = ar² = 16, G₃ = ar³ = 64

Similarly, for r = – 4, numbers are – 4,16 and – 64.

Hence, we can insert, 4, 16, 64 or – 4, 16, –64, between 1 and 256 so that the resulting sequences are in G.P.

Illustration -35

If the third term of G.P is 4, then find the product of first 5terms.

Solution

Given $t_3=a{r}^2=4$

Product of first 5 terms = (a) (ar) (ar²) (ar³) (ar⁴) = a⁵ r¹⁰  $= {\left(a{r}^2\right)}^5=4^5$

= 1024

Illustration -36

If $(10{)}^9+2(11{)}^1(10{)}^8+3(11{)}^2(10{)}^7+\dots \dots +10(11{)}^9=k{\left(10\right)}^9$, then find k.

Solution

$k(10{)}^9={10}^9+2(11{)}^1(10{)}^8+3(11{)}^2(10{)}^7+\dots .+10(11{)}^9$

$=k=1+2\left(\frac{11}{10}\right)+3{\left(\frac{11}{10}\right)}^2+\dots .+10{\left(\frac{11}{10}\right)}^9\to \left(1\right)$

$\frac{11k}{10}=\frac{11}{10}+2{\left(\frac{11}{10}\right)}^2+3{\left(\frac{11}{10}\right)}^3+\dots .+10{\left(\frac{11}{10}\right)}^{10}\to \left(2\right)$

$\text{ (1) }–\left(2\right)\Rightarrow –\frac{k}{10}=1+\frac{11}{10}+{\left(\frac{11}{10}\right)}^2+\dots .+{\left(\frac{11}{10}\right)}^9–10{\left(\frac{11}{10}\right)}^{10}$

$=\frac{{\left(\frac{11}{10}\right)}^{10}–1}{\frac{11}{10}–1}–10{\left(\frac{11}{10}\right)}^{10}=10{\left(\frac{11}{10}\right)}^{10}–10–10{\left(\frac{11}{10}\right)}^{10}$

$\therefore \frac{–k}{10}=–10\Rightarrow k=100$

Illustration -37

Three Positive numbers from an increasing G.P. If the middle term in this G.P is doubled, the new number are in A.P Then find  the common ratio of the G.P.

Solution

Let a, ar, ar² be in G.P and r>1.

Given a,2ar, ar² are in A.P.

$\Rightarrow 2\left(2ar\right)=a+a{r}^2\Rightarrow r^2–4r+1=0$

$r=\frac{4\pm \sqrt{16–4}}{2}=2\pm \sqrt{3}$

$\therefore r>1\Rightarrow r=2+\sqrt{3}$

Illustration -38

Three numbers are in G.P. Whose sum is 70, if the extremes be each multiplied by 4 and the mean by 5, they will be in A.P. then find the sum of numbers.

Solution

Let the numbers be a, ar, ar² and sum=70

$\Rightarrow a\left(1+r+r^2\right)=70\to \left(1\right)$

it is given that 4a, 5ar, 4ar² are in A.P

$\Rightarrow 2\left(5ar\right)=4a+4a{r}^2\Rightarrow 5r=2+2r^2$

$\Rightarrow 2r^2–5r+2=0\Rightarrow (2r–1)(r–2)=0$

$\Rightarrow r=2,\frac{1}{2}$    put r=2 in (1), then a=10

put $r=\frac{1}{2}$ in (1), then a=40

∴ The numbers are 10,20,40 or 40,20,10.

∴ Sum of the numbers =70

Illustration -39

If the sides of a triangle are in G.P and its larger angle is twice the smallest, then find the common ratio r satisfies the inequality.

Solution

Let the sides of a triangle be a/r, a and ar, with a>0 and r>1. let α be the smallest angle. So that the largest angle is 2α. then α is opposite to the side a/r, and 2α is positive to the side ar.

Applying sine rule, we get $\frac{a/r}{\sin \alpha }=\frac{ar}{\sin 2\alpha }$

$\Rightarrow \frac{\sin 2\alpha }{\sin \alpha }=r^2\Rightarrow r^2=2\cos \alpha <2$

$\Rightarrow r^2<2\Rightarrow r<\sqrt{2}$

$\therefore 1<r<\sqrt{2}$

Illustration -40

If we insert two numbers between 3 and 81 so that the resulting sequence is G.P then find the numbers.

Solution

Let the two numbers be a and b, then 3,a,b,81 are in G.P.

$\because$ nth term $T_n=A{R}^{n–1}; 81=3R^{4–1}$ ;

$\Rightarrow R^3=\frac{81}{3}=27\Rightarrow R^3=3^3\Rightarrow R=3$

$a=AR=3\times 3=9,b=A{R}^2=3\times 3^2=27$

16. HARMONIC PROGRESSION

A progression is called a harmonic progression (H.P.) if the reciprocals of its terms are in A.P.

Standard form : $\frac{1}{a}+\frac{1}{a+d}+\frac{1}{a+2d}+$..….

Example: The sequence $1,\frac{1}{3},\frac{1}{5},\frac{1}{7},\frac{1}{9},\dots$ is a H.P., because the sequence 1, 3, 5, 7, 9, ….. is an A.P.

17. GENERAL TERMS OF H.P

If the H.P. be as $\frac{1}{a},\frac{1}{a+d},\frac{1}{a+2d}$,… then corresponding A.P. is of A.P. is $a,a+d,a+2d,\dots ..$

$T_n$ of A.P. is a+(n-1)d

∴ $T_n$ of H.P. is $\frac{1}{a+\left(n–1\right)d}$

In order to solve the question on H.P., we should form the corresponding A.P. Thus, General term :

$T_n=\frac{1}{a+(n–1)d}\text{ or }{T}_n\text{ of }\mathrm{H}.\mathrm{P}.=\frac{1}{T_n\text{ of }\mathrm{A}.\mathrm{P}.}$

18. HARMONIC MEAN

If three or more numbers are in H.P., then the numbers lying between the first and last are called harmonic means (H.M.’s) between them. For example 1, 1/3, 1/5, 1/7, 1/9 are in H.P. So 1/3, 1/5 and 1/7 are three H.M.’s between 1 and 1/9.

Also, if a, H, b are in H.P., then H is called harmonic mean between a and b.

Insertion of harmonic means

(i) Single H.M. between a and b = $\frac{2ab}{a+b}$.

(ii) H, H.M. of n non-zero numbers $a_1,a_2,a_3,.....,a_n$ is given by $\frac{1}{H}=\frac{\frac{1}{a_1}+\frac{1}{a_2}+\dots ..+\frac{1}{a_n}}{n}$.

(iii) Let a, b be two given numbers. If n numbers $H_1,H_2,......H_n$ are inserted between a and b such that the sequence $a,H_1,H_2,H_3,.....H_n,b$ is a H.P., then $H_1,H_2,......H_n$ are called n harmonic means between a and b.

Now, $a,H_1,H_2,H_3,.....H_n,b$ are in H.P.

$\Rightarrow \frac{1}{a},\frac{1}{H_1},\frac{1}{H_2},\dots ..\frac{1}{H_n},\frac{1}{b}$   are in A.P.

Let D be the common difference of this A.P. Then,

$\frac{1}{b}=(n+2{)}^{th}\text{ term }=T_{n+2}$

$\frac{1}{b}=\frac{1}{a}+(n+1)D\Rightarrow D=\frac{a–b}{(n+1)ab}$.

Thus, if n harmonic means are inserted between two given numbers a and b, then the common difference of the corresponding A.P. is given by $D=\frac{a–b}{(n+1)ab}$.

Also, $\frac{1}{H_1}=\frac{1}{a}+D, \frac{1}{H_2}=\frac{1}{a}+2D,\dots \dots ,\frac{1}{H_n}=\frac{1}{a}+nD$ ,…….,,  where $D=\frac{a–b}{(n+1)ab}$

19. PROPERTIES OF H.P

(1) No term of H.P. can be zero.

(2) If H is the H.M. between a and b, then

(i) $\frac{1}{H–a}+\frac{1}{H–b}=\frac{1}{a}+\frac{1}{b}$

(ii) $(H–2a)(H–2b)=H^2$

(iii) $\frac{H+a}{H–a}+\frac{H+b}{H–b}=2$

Illustration -41

Find the nth term of H.P. $\frac{4}{3},\frac{3}{2},\frac{12}{7},.....$

Solution

$\frac{4}{3},\frac{3}{2},\frac{12}{7},.....$are in H.P.

$\Rightarrow \frac{3}{4},\frac{2}{3},\frac{7}{12},$are in A.P.

$a=\frac{3}{4}, d=\frac{2}{3}–\frac{3}{4}=\frac{–1}{12}$

We have, ${\mathrm{t}}_{\mathrm{n}}=\frac{1}{\mathrm{a}+(\mathrm{n}–1)\mathrm{d}}$ = $\frac{1}{\frac{3}{4}+(\mathrm{n}–1)\left(\frac{–1}{12}\right)}=\frac{12}{10–\mathrm{n}}$

Illustration -42

Find the 4th and the 8th terms of the H.P. 6, 4, 3,……….

Solution

Consider $\frac{1}{6},\frac{1}{4},\frac{1}{3},\dots$

Here  T₂ – T₁ = T₃ – T₂ =  $\frac{1}{12}\Rightarrow \frac{1}{6},\frac{1}{4},\frac{1}{3},$,….. is an A.P.

4th term of this A.P. =  $\frac{1}{6}+3\times \frac{1}{12}=\frac{1}{6}+\frac{1}{4}=\frac{5}{12}$

and the 8th term = $\frac{1}{6}+7\times \frac{1}{12}=\frac{9}{12}$

Hence the 4th term of the H.P. = $\frac{12}{5}$  and the 8th term =  $\frac{12}{9}=\frac{4}{3}$

Illustration -43

Find H. P. whose 3^rd and 14^th terms are respectively $\frac{6}{7}$ and $\frac{1}{3}$.

Solution

Let a & d are first term & common difference of A.P. which is reciprocal of given H.P.

${\mathrm{t}}_3=\frac{7}{6}=\mathrm{a}+2\mathrm{d} \& {\mathrm{t}}_{14}=3=\mathrm{a}+13\mathrm{d}$   $\Rightarrow \mathrm{a}=\frac{5}{6} \& \mathrm{d}=\frac{1}{6}$

There A.P. is $\frac{5}{6},1,\frac{7}{6},\frac{8}{6},\frac{9}{6}.....$ &

H.P. is $\frac{6}{5},1,\frac{6}{7},\frac{6}{8},\frac{6}{9}$……

Illustration -44

x + y + z = 15 if a, x, y, z, b are in A.P. and $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{5}{3}$ if a, x, y, z b are in H.P. Find a and b.

Solution

Given x + y + z = 15  …..(1)

when a, x, y, z, b are in AP.

∴ sum of A.M.’s, x + y + z = $\left(\frac{a+b}{2}\right).3$

∴ 15 = $\left(\frac{a+b}{2}\right)$.3  ⇒ a + b = 10   …(2)

when a, x, y, z, b are in HP., $\frac{1}{a},\frac{1}{x},\frac{1}{y},\frac{1}{z},\frac{1}{b}$ are in AP.

$\therefore \frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{\left(\frac{1}{a}+\frac{1}{b}\right)}{2}·3$  $\Rightarrow \frac{5}{3}=\left(\frac{1}{a}+\frac{1}{b}\right)·3=\left(\frac{10}{2ab}\right)·3$

∴   ab = 9  …..(3)

Now (a –b)² + (a + b)² –4ab = 100 –36 = 64

∴  a – b = $\pm$8             …(4)

from (1) and (4) , we get a = 9, b = 1  or a = 1, b = 9

Illustration -45

Find the first term of an H.P whose second and third terms are $\frac{3}{5}$, $\frac{1}{2}$.

Solution

Let a be the first term. Then a, $\frac{3}{5}, \frac{1}{2}$are in H.P.

Using the relation, we get $\frac{a–\frac{3}{5}}{\frac{3}{5}–\frac{1}{2}}=\frac{a}{\frac{1}{2}}$, or a $–\frac{3}{5}=\frac{a}{5}$   or    5a – 3 = a,  i.e., 4a = 3, or  a = $\frac{3}{4}$

To find the nth term of an H.P.

Evidently, the nth term of an H.P is the reciprocal of nth term of the A.P. formed by the reciprocals of the H.P. Thus the nth term of the H.P.

$\frac{1}{a},\frac{1}{a+d},\frac{1}{a+2d}\frac{1}{a+3d}$,… is $\frac{1}{a+\left(n–1\right)d}$.

Illustration -46

(i) Find the harmonic mean between 2 and 3.

(ii) Insert three harmonic means between 5 and 6.

Solution

(i) Harmonic mean between a and b is $\frac{2ab}{a+b}$

Here a = 2, b = 3, ∴ H.M. = $\frac{2\times 2\times 3}{2+3}=\frac{12}{5}$.

(ii) Let us insert 3 arithmetic means between $\frac{1}{5}$ and $\frac{1}{6}$ and  then take their reciprocals.

Let A₁, A₂, A₃, be the arithmetic means in between $\frac{1}{5} \mathrm{and} \frac{1}{6}$

Then $\frac{1}{5}$, A₁, , A₂, A₃, $\frac{1}{6}$ are in A.P.

Let d be the common difference.

${\mathrm{T}}_5=\frac{1}{6},\text{ or }\frac{1}{5}+(5–1)\mathrm{d}=\frac{1}{6} \therefore \mathrm{d}=\frac{–1}{120}$

$\therefore {\mathrm{A}}_1=\frac{1}{5}+\mathrm{d}=\frac{1}{5}–\frac{1}{120}=\frac{23}{120}$;

$A_2=\frac{1}{5}+2d=\frac{1}{5}–\frac{1}{60}=\frac{11}{60};$

${\mathrm{A}}_3=\frac{1}{5}+3\mathrm{d}=\frac{1}{5}–\frac{1}{40}=\frac{7}{40};$

∴ The required harmonic means are $\frac{120}{23},\frac{60}{11},\frac{40}{7}$.

Illustration -47

Insert 6 H.M.’s between $\frac{2}{3}$ and $\frac{2}{31}$.

Solution

Let the harmonic progression be

$\frac{1}{a},\frac{1}{a+d},\frac{1}{a+2d},\dots$

Equating the first and last terms,

$\frac{1}{a}=\frac{2}{3}\Rightarrow a=\frac{3}{2}$

${\mathrm{t}}_8=\frac{1}{\mathrm{a}+7 \mathrm{d}}=\frac{2}{31}$

a + 7d = $\frac{31}{2}$

$7\mathrm{d}=\frac{31}{2}–\frac{3}{2}=14$

d = 2

$a+d=\frac{3}{2}+2=\frac{7}{2}$

$a+2d=\frac{3}{2}+2\left(2\right)=\frac{11}{2}$

$a+3d=\frac{3}{2}+3\left(2\right)=\frac{15}{2}$

$\mathrm{a}+4\mathrm{d}=\frac{3}{2}+4\left(2\right)=\frac{19}{2}$

$a+5d=\frac{3}{2}+5\left(2\right)=\frac{23}{2}$

$a+6d=\frac{3}{2}+6\left(2\right)=\frac{27}{2}$

The harmonic means are, $\frac{2}{7},\frac{2}{11},\frac{2}{15},\frac{2}{19},\frac{2}{23},\frac{2}{27}$

Illustration -48

If (b + c), (c + a) and (a + b) are in H.P. then

S.T.  $\frac{1}{a^2},\frac{1}{b^2},\frac{1}{c^2}$ will also be in H.P.

Solution

Given that,

b + c, c + a, a + b are in H.P.

$\Rightarrow \frac{1}{b+c},\frac{1}{c+a},\frac{1}{a+b}$ are in A.P.

$\Rightarrow \frac{1}{c+a}–\frac{1}{b+c}=\frac{1}{a+b}–\frac{1}{c+a}$

$\Rightarrow \frac{b+c–c–a}{(c+a)(b+c)}=\frac{c+a–a–b}{(a+b)(c+a)}$

$\Rightarrow \frac{(b–a)}{b+c}=\frac{c–b}{a+b}$

⇒ ab – a² + b² – ab = bc + c² – b² – bc

⇒ b² – a² = c² – b²

⇒ 2b² = a² + c²

⇒ a², b², c² are in A.P.

⇒ $\frac{1}{a^2},\frac{1}{b^2},\frac{1}{c^2}$ are in H.P.

20. RELATION BETWEEN A.P., G.P. AND H.P.

(1) If A, G, H be A.M., G.M., H.M. between a and b, then

$\frac{a^{n+1}+b^{n+1}}{a^n+b^n}=\left\{\begin{array}{l}A\text{ when }n=0\\ G\text{ when }n=–1/2\\ H\text{ when }n=–1\end{array}\right.$

(2) If  be two A.M.’s;  be two G.M.’s and  be two H.M.’s between two numbers a and b, then

$\frac{G_1{G}_2}{H_1{H}_2}=\frac{A_1+A_2}{H_1+H_2}$

(3) Reorganization of A.P., G.P., H.P

If a, b, c are three successive terms of a sequence.

If  $\frac{a–b}{b–c}=\frac{a}{a}$, then a, b, c are in A.P.

If, $\frac{a–b}{b–c}=\frac{a}{b}$, then a, b, c are in G.P.

If, $\frac{a–b}{b–c}=\frac{a}{c}$, then a, b, c are in H.P.

(4) If number of terms of any A.P./G.P./H.P. is odd, then A.M./G.M./H.M. of first and last terms is middle term of series.

(5) If number of terms of any A.P./G.P./H.P. is even, then A.M./G.M./H.M. of middle two terms is A.M./G.M./H.M. of first and last terms respectively.

(6) If p^th, q^th and r^th terms of a G.P. are in G.P. Then p, q, r are in A.P.

(7) If a, b, c are in A.P. as well as in G.P. then a = b = c.

(8) If a, b, c are in A.P $x^a,x^b,x^c$, then  will be in G.P.$(x\ne \pm 1)$.

21. PROPERTIES OF ARITHMETIC, GEOMETRIC, HARMONIC MEANS BETWEEN TWO GIVEN numbers

Let A, G and H be arithmetic, geometric and harmonic means of two numbers a and b.

Then, $A=\frac{a+b}{2}, G=\sqrt{ab}\text{ and }H=\frac{2ab}{a+b}$.

These three means possess the following properties :

(1) $A\ge G\ge H$

$A=\frac{a+b}{2},G=\sqrt{ab}\text{ and }H=\frac{2ab}{a+b}$

$\therefore A–G=\frac{a+b}{2}–\sqrt{ab}$ = $\frac{(\sqrt{a}–\sqrt{b}{)}^2}{2}\ge 0\Rightarrow A\ge G$   …..(i)

$G–H=\sqrt{ab}–\frac{2ab}{a+b}$ =  $\sqrt{ab}\left(\frac{a+b–2\sqrt{ab}}{a+b}\right)$  = $\frac{\sqrt{ab}}{a+b}(\sqrt{a}–\sqrt{b}{)}^2\ge 0$

$\Rightarrow G\ge H$ …..(ii)

From (i) and (ii), we get $A\ge G\ge H$.

Note that the equality holds only when a = b.

(2) A, G, H from a G.P., i.e., $G^2=AH$

$AH=\frac{a+b}{2}\times \frac{2ab}{a+b}=ab=(\sqrt{ab}{)}^2=G^2$. Hence, $G^2=AH$

(3) The equation having a and b as its roots is

$x^2–2Ax+G^2=0$

The equation having a and b its roots is

$x^2–(a+b)x+ab=0$

$\Rightarrow x^2–2Ax+G^2=0$,   $\left[\because A=\frac{a+b}{2}\text{ and }G=\sqrt{ab}\right]$.

The roots a, b are given by $A\pm \sqrt{A^2–G^2}$.

(4) If A, G, H are arithmetic, geometric and harmonic means between three given numbers a, b and c, then the equation having a, b, c as its roots is $x^3–3A{x}^2+\frac{3G^3}{H}x–G^3=0$

where $A=\frac{a+b+c}{3}, G=(abc{)}^{1/3}$ and $\frac{1}{H}=\frac{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}{3}$

$\Rightarrow a+b+c=3A, abc=G^3$  and $\frac{3G^3}{H}=ab+bc+ca$

The equation having a, b, c as its roots is  

$x^3–(a+b+c)x^2+(ab+bc+ca)x–abc=0$

$\Rightarrow x^3–3A{x}^2+\frac{3G^3}{H}x–G^3=0$

Illustration -49

Find two numbers whose arithmetic mean is 34 and geometric mean is 16.

Solution

Let the two numbers be a and b then $\frac{a+b}{2}$=34 and $\sqrt{ab}$=16

a + b=68 and ab=256

$\therefore (a–b{)}^2=(a+b{)}^2–4ab$

$=(68{)}^2–4(256)=3600\Rightarrow a–b=60$

on solving a + b=68 and a-b=60, we get a=64, and b=4. thus, the required numbers are 64 and 4.

Illustration -50

The H.M. between two numbers is 16/5, their A.M. is A and G.M. is G. If 2A+G2=26 then find the numbers.

Solution

Given H.M of a and b is $\frac{2ab}{a+b}=\frac{16}{5}$

$\Rightarrow a+b=\frac{5ab}{8}\to \left(1\right)$

Given $2A+G^2=26\Rightarrow 2\left(\frac{a+b}{2}\right)+ab=26$

$\Rightarrow (a+b)+ab=26\Rightarrow \frac{5ab}{8}+ab=26\Rightarrow ab=16$

From (1),   $a+b=\frac{5}{8}\left(16\right)\Rightarrow a+b=10\to \left(2\right)$

$\therefore (a–b{)}^2=(a+b{)}^2–4ab=100–64=36$

∴ $(a–b)=6\to \left(3\right)$ Solving (2) and (3)

∴ a=8, b=2

Illustration -51 Between 2 & 100, 13 means are inserted then find the 9th mean if means are

i) arithmetic     ii) geometric      iii) harmonic

  Solution

i. a = 2, b = 100, n = 13

$A_9=a+9d=a+9\frac{b–a}{n+1}=65$

ii. a = 2, b = 100, n = 13

${\mathrm{G}}_9={\mathrm{ar}}^9=\mathrm{a}{\left(\frac{\mathrm{b}}{\mathrm{a}}\right)}^{\frac{9}{14}}=2(50{)}^{\frac{9}{14}}$

iii. a = 2, b = 100, n = 13 reciprocal of harmonic is A.P. where   ${\mathrm{t}}_1=\frac{1}{2} \& {\mathrm{t}}_{15}=\frac{1}{100}$

$\frac{1}{{\mathrm{H}}_9}={\mathrm{t}}_{10}={\mathrm{t}}_1+9 \mathrm{d}=\frac{1}{2}+9\times \frac{–7}{200}$

Hence   ${\mathrm{H}}_9=\frac{200}{37}$

Illustration -52

If H.M. & A.M. of two numbers are 3 & 4 respectively, find the numbers.

Solution

Let numbers are a & b  

$\frac{2ab}{a+b}=3 \& \frac{a+b}{2}=4$   ⇒ ab = 12 and a+b = 8

solving there we get a = 6, b = 2 or a = 2, b = 6

Illustration -53

Let two numbers have arithmetic mean 9 and geometric mean 4. then find the numbers are the roots of the quadratic equation.

Solution

The A.M. of the two numbers is A=9 and the G.M of two numbers is G=4

The quadratic equation whose roots are the numbers having A.M and G.M. are A,G respectively is $x^2–2Ax+G^2=0$.

So, the required quadratic equation is $x^2–18x+16=0$

Illustration -54

If the arithmetic, geometric and harmonic means between two positive real numbers be A, G and H, then

(A) $A^2=GH$          (B) $H^2=AG$         (C) G = AH            (D) $G^2=AH$

Solution

(D)

Let $A=\frac{a+b}{2},G=\sqrt{ab}\text{ and }H=\frac{2ab}{a+b}$.

Then $G^2=ab$  …..(i)

and $AH=\left(\frac{a+b}{2}\right)·\frac{2ab}{a+b}=ab$ …..(ii)

From  (i) and (ii), we have $G^2=AH$.

Illustration -55

If  $a^{1/x}=b^{1/y}=c^{1/z}$ and a,b,c are in G.P., then x,y,z  will be in

(A) A.P.           (B) G.P.          (C) H.P.          (D) None of these

Solution

(A) Let $a^{1/x}=b^{1/y}=c^{1/z}$ = $k\Rightarrow a=k^x, b=k^y, c=k^z$

Now, are in G.P.

$\Rightarrow b^2=ac\Rightarrow k^{2y}=k^x.k^z=k^{x+z}\Rightarrow 2y=x+z$

⇒ x,y,z are in A.P.

Illustration -56

If $\frac{1}{b–a}+\frac{1}{b–c}=\frac{1}{a}+\frac{1}{c}$, then a,b,c  are in

(A) A.P.            (B) G.P.             (C) H.P.         (D) In G.P. and H.P. both  

Solution

(C)

Since the reciprocals of a and c occur on RHS, let us first assume that  are in H.P.

So that $\frac{1}{a},\frac{1}{b},\frac{1}{c}$ are in A.P.

$\Rightarrow \frac{1}{b}–\frac{1}{a}=\frac{1}{c}–\frac{1}{b}=d$ , say

$\Rightarrow \frac{a–b}{ab}=d=\frac{b–c}{bc}\Rightarrow a–b=abd$ andb – c = bcd

Now LHS = $–\frac{1}{a–b}+\frac{1}{b–c}=–\frac{1}{abd}+\frac{1}{bcd}$

$=\frac{1}{bd}\left(\frac{1}{c}–\frac{1}{a}\right)=\frac{1}{bd}\left(2d\right)\Rightarrow \frac{2}{b}=\frac{1}{a}+\frac{1}{c}$ = RHS

∴ a,b,c are in H.P. is verified.

Illustration -57

If $a^2,b^2,c^2$ are in A.P., then $(b+c{)}^{–1},(c+a{)}^{–1}\text{ and }(a+b{)}^{–1}$ will be in

(A) H.P.             (B) G.P.           (C) A.P.          (D) None of these

Solution

(C)

$a^2,b^2,c^2$ are in A.P. Therefore  $a^2+(ab+bc+ca), b^2+(ab+bc+ca), c^2+(ab+bc+ca)$ will be in A.P. will be in A.P.

$\Rightarrow \left\{\alpha \right(a+b)+c(a+b\left)\right\}, \left\{b\right(b+a)+c(b+a\left)\right\}, c(c+b)+a(b+c)$ will be in A.P.

$\Rightarrow \frac{1}{b+c},\frac{1}{c+a},\frac{1}{a+b}$ will be in A.P.  {Dividing each term by $(a+b)(b+c)(c+a)$}

22. ARITHMETICO-GEOMETRIC PROGRESSION(A.G.P.)

The combination of arithmetic and geometric progression is called arithmetico-geometric progression.

1 n^th TERM OF A.G.P

If $a_1,a_2,a_3,\dots \dots ,a_n,\dots \dots$ is an A.P. and $b_1,b_2,\dots \dots ,b_n,\dots \dots$ is a G.P., then the sequence $a_1{b}_1,a_2{b}_2,a_3{b}_3,\dots \dots ,a_n{b}_n,\dots .$ is said to be an arithmetico-geometric sequence.

Thus, the general form of an arithmetico geometric sequence is $a,(a+d)r,(a+2d)r^2,(a+3d)r^3$

From the symmetry we obtain that the n^th term of this sequence is $[a+(n–1\left)d\right]r^{n–1}$.

Also, let $a, (a+d)r, (a+2d)r^2, (a+3d)r^3,\dots .$ be an arithmetico-geometric sequence.

Then, $a+(a+d)r+(a+2d)r^2+(a+3d)r^3+\dots$ is an arithmetico-geometric series.

2 SUM OF N TERMS OF A.G.P

Sum of n terms : The sum of n terms of an arithmeti co-geometric sequence $a,(a+d)r,(a+2d)r^2$, $(a+3d)r^3,\dots$ is given by

${\mathrm{S}}_{\mathrm{n}}=\left\{\begin{array}{l}\frac{\mathrm{a}}{1–\mathrm{r}}+\mathrm{dr}\frac{\left(1–{\mathrm{r}}^{\mathrm{n}–1}\right)}{(1–\mathrm{r}{)}^2}–\frac{\{\mathrm{a}+(\mathrm{n}–1\left)\mathrm{d}\right\}{\mathrm{r}}^{\mathrm{n}}}{1–\mathrm{r}},\text{ when }\mathrm{r}\ne 1\\ \frac{\mathrm{n}}{2}[2\mathrm{a}+(\mathrm{n}–1\left)\mathrm{d}\right],\text{ when }\mathrm{r}=1\end{array}\right.$

3 SUM OF INFINITE TERMS OF A.G.P

Let |r|< 1. Then  $r^n, r^{n–1}\to 0\text{ as }n\to \infty$ and it can also be shown that $n.r^n\to 0\text{ as }n\to \infty$. So, we obtain that $S_n\to \frac{a}{1–r}+\frac{dr}{(1–r{)}^2},\text{ as }n\to \infty$.

In other words, when |r|< 1 the sum to infinity of an arithmetico-geometric series is $S_{\infty }=\frac{a}{1–r}+\frac{dr}{(1–r{)}^2}$.

Illustration -58

The sum of the first n terms of the series $\frac{1}{2}+\frac{3}{4}+\frac{7}{8}+\frac{15}{16}+$……. is

(A) $2^n–n–1$         (B) $1–2^{–n}$        (C)  $n+2^{–n}–1$           (D) $2^n–1$ 

Solution

(C)

The sum of the first n terms is

$S_n=\left(1–\frac{1}{2}\right)+\left(1–\frac{1}{2^2}\right)+\left(1–\frac{1}{2^3}\right)+\left(1–\frac{1}{2^4}\right)+\dots \dots .+\left(1–\frac{1}{2^n}\right)$

$=n–\left\{\frac{1}{2}+\frac{1}{2^2}+\dots .+\frac{1}{2^n}\right\}$

$=n–\frac{1}{2}\left(\frac{1–\frac{1}{2^n}}{1–\frac{1}{2}}\right)=n–\left(1–\frac{1}{2^n}\right)=n–1+2^{–n}$.

Illustration -59

The sum of $1+\frac{2}{5}+\frac{3}{5^2}+\frac{4}{5^3}+$….. up to n terms is

(A) $\frac{25}{16}–\frac{4n+5}{16\times 5^{n–1}}$        (B) $\frac{3}{4}–\frac{2n+5}{16\times 5^{n+1}}$       (C) $\frac{3}{7}–\frac{3n+5}{16\times 5^{n–1}}$       (D) $\frac{1}{2}–\frac{5n+1}{3\times 5^{n+2}}$ 

Solution

(A)

Given series, let $S_n=1+\frac{2}{5}+\frac{3}{5^2}+\frac{4}{5^3}+\dots \dots \dots +\frac{n}{5^{n–1}}$

$\frac{1}{5}{S}_n=\frac{1}{5}+\frac{2}{5^2}+\frac{3}{5^3}+\dots \dots .+\frac{n}{5^n}$

Subtracting,

$\left(1–\frac{1}{5}\right)S_n=1+\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+\dots \dots +$ upto n times $–\frac{n}{5^n}$

$\Rightarrow \frac{4}{5}{S}_n=\frac{1–\frac{1}{5^n}}{\frac{4}{5}}–\frac{n}{5^n}\Rightarrow S_n=\frac{25}{16}–\frac{4n+5}{16\times 5^{n–1}}$

Illustration -60

$2^{1/4}·4^{1/8}·8^{1/16}·{16}^{1/32}$……. is equal to

(A) 1                   (B) 2            (C) $\frac{3}{2}$               (D) $\frac{5}{2}$

Solution

(B)

$2^{1/4}·4^{1/8}·8^{1/16}·{16}^{1/32}.....\infty$

= $2^{1/4+2/8+3/16+\dots ..}=2^S$, where S is given by

$S=\frac{1}{4}+2\frac{1}{8}+3\frac{1}{16}+4\frac{1}{32}+\dots \dots \dots .\infty$……(i)

$\Rightarrow \frac{1}{2}S=\frac{1}{8}+\frac{2}{16}+\frac{3}{32}+\frac{4}{64}+\dots \dots \dots .\infty$……(ii)

Subtracting (ii) from (i), we get S=1.

Hence required product = $2^1=2$.

Illustration -61

Find the sum of the series 1.2 + 2.2² + 3.2³ + ….. + 100.2¹⁰⁰.

Solution

S = 1.2 + 2.2² + 3.2³ + …. + 100.2¹⁰⁰

2S =1.2² + 2.2³ +  …. + 99.2¹⁰⁰ + 100.2¹⁰¹

⇒ –S = 1.2 + 1.2² + 1.2³ + …. + 1.2¹⁰⁰ – 100.2¹⁰¹

⇒ -S = 1.2 – 100.2¹⁰¹ 

⇒ S = -2¹⁰¹ + 2 + 100.2¹⁰¹  = 99.2¹⁰¹ + 2.

Illustration -62

Find 1 + 2.2 + 3.2² + 4. 2³ + …….+ 100.2⁹⁹.

Solution

Let S_n = 1 + 2.2 + 3.2² + ……+ t_n    ……(1)

then 2S_n = 1.2 + 2.2² + 3.2³ + ……+ 2.t_{n –1}  + 2. t_n    ……(2)

Subtracting (2) from (1), we get

-S_n = 1 + [(2 –1) 2 + (3 –2) 2² + …to (n –1) terms] –2. t_n

= 1 + 2 + 2² + ….to n terms – 2. t_n =     …..(3)

Now t_n = (nth term of AP 1, 2, 3, 4, ….) . (nth term of GP 1, 2, 2², …)

= {1 + (n –1) .1} {1. 2^{n –1} } = n2^{n –1}

from (3), S_n = -(2ⁿ –1) + 2n2^{n –1}  = 1 –2ⁿ –n2ⁿ = 1- (1 + n)2ⁿ  

Illustration -63

Find the sum of infinity of the series $1+\frac{2.1}{3}+\frac{3.1}{3^2}+\frac{4.1}{3^3}+\dots \dots \dots$

Solution

$\mathrm{S}=1+2·\frac{1}{3}+3·\frac{1}{3^2}+4·\frac{1}{3^3}+\dots \dots \dots$

$\frac{1}{3} \mathrm{S}=\frac{1}{3}+2·\frac{1}{3^2}+3·\frac{1}{3^3}+\dots \dots \dots \dots$

$\frac{2}{3} \mathrm{S}=1+\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\dots \dots \dots \text{ upto in infinite }\right)=1+\frac{1/3}{1–\frac{1}{3}}=\frac{3}{2}$

Illustration -64

Find sum to n terms of the series, 1 + 4x + 7x² + 10x³ + …. when | x | < 1.

Solution

T_n = (3n – 2) x^n-1

S_n = 1 + 4x + 7x² + 10x³ + ……….+ (3n–2) x^n-1      ………………(1)

x S_n = x + 4x² + 7x³ + ……..+ (3n–5) x^n-1 + (3n–2)xⁿ    ………….(2)

On subtracting (2) from (1)

(1 – x) S_n = 1 + (3x + 3x² + 3x³ + ……….. up to (n – 1) term) – (3n–2) xⁿ

(1 – x)S_n = $1+3\mathrm{x}\frac{\left(1–{\mathrm{x}}^{\mathrm{n}–1}\right)}{1–\mathrm{x}}–(3\mathrm{n}–2){\mathrm{x}}^{\mathrm{n}}$

${\mathrm{S}}_{\mathrm{n}}=\frac{1}{1–\mathrm{x}}\left(1+\frac{3\mathrm{x}}{1–\mathrm{x}}\left(1–{\mathrm{x}}^{\mathrm{n}–1}\right)–(3\mathrm{n}–2){\mathrm{x}}^{\mathrm{n}}\right)$

23. SUMMATION OF SOME SERIES OF NATURAL NUMBERS

$\Sigma 1=\sum _{k=1}^{n}1=1+1+\dots \dots \dots \dots +1\left(n\text{ terms }\right)=n$

$\sum _{k=1}^{n}k=1+2+3+\dots +n=\frac{1}{2}n(n+1)$

$\sum _{\mathrm{k}=1}^{\mathrm{n}}{\mathrm{k}}^2=1^2+2^2+\dots +{\mathrm{n}}^2=\frac{1}{6}\mathrm{n}(\mathrm{n}+1)(2\mathrm{n}+1)$

$\sum _{\mathrm{k}=1}^{\mathrm{n}}{\mathrm{k}}^3=1^3+2^3+\dots +{\mathrm{n}}^3={\left(\sum _{\mathrm{k}=1}^{\mathrm{n}}\mathrm{k}\right)}^2$ = $(1+2+3+\dots +\mathrm{n}{)}^2={\left[\frac{1}{2}\mathrm{n}(\mathrm{n}+1)\right]}^2=\frac{1}{4}{\mathrm{n}}^2(\mathrm{n}+1{)}^2$

1+3+5+….n terms = $\sum _{k=1}^n(2k–1)=n^2$

2+4+6+….n terms = $\sum _{k=1}^n\left(2k\right)=n(n+1)$

$1^2+3^2+5^2+\dots .+n\text{ terms }$= $\sum _{k=1}^n(2k–1{)}^2=\frac{n}{3}\left(4n^2–1\right)$

$2^2+4^2+6^2+\dots .+n\text{ terms }$ = $\sum _{k=1}^n(2k{)}^2=\frac{2}{3}n(n+1\left)\right(2n+1)$

$1^3+3^3+5^3+\dots .+n\text{ terms }$= $\sum _{k=1}^n(2k–1{)}^3=n^2\left(2n^2–1\right)$

Sum of n terms of series

$1^2–2^2+3^2–4^2+5^2–\dots ..=\left\{\begin{array}{l}\frac{n(n+1)}{2},\text{ if n is odd }\\ \frac{–n(n+1)}{2},\text{ if n is even }\end{array}\right.$

Method For Finding Sum

This method is applicable for both sum of n terms and sum of infinite number of terms. First suppose that sum of the series is S, then multiply it by common ratio of the G.P. and subtract. In this way, we shall get a G.P., whose sum can be easily obtained.

Method of Difference

If the differences of the successive terms of a series are in A.P. or G.P., we can find n^th term of the series by the following steps :

Step I: Denote the n^th term by $T_n$ and the sum of the series upto n terms by S_n.

Step II: Rewrite the given series with each term shifted by one place to the right.

Step III: By subtracting the later series from the former, find $T_n$.

Step IV: From $T_n$, S_n can be found by appropriate summation.

Example : Consider the series 1+ 3 + 6 + 10 + 15 +…..to n terms. Here differences between the successive terms are  6 – 3, 10 – 6, 15 – 10, …….i.e.,  2, 3, 4, 5,…… which are in A.P. This difference could be in G.P. also.

Now let us find its sum

S = 1+3+6+10+15+…..+T_n-1 + T_n

S = 1+3+6+10+……+T_n-1 + T_n

Subtracting, we get

0=1+2+3+4+5+…..+(T_n – T_n-1) – T_n

⇒ T_n = 1+2+3+4+……to n terms.

$\Rightarrow {\mathrm{T}}_{\mathrm{n}}=\frac{1}{2}\mathrm{n}(\mathrm{n}+1) \therefore {\mathrm{S}}_{\mathrm{n}}={\mathrm{\Sigma T}}_{\mathrm{n}}=\frac{1}{2}\left[{\mathrm{\Sigma n}}^2+\mathrm{\Sigma n}\right]$

$=\frac{1}{2}\left[\frac{n(n+1)(2n+1)}{6}+\frac{n(n+1)}{2}\right]=\frac{n(n+1)(n+2)}{6}$.

Illustration -65

Find the sum of n terms of the series 1+ 5 + 11 + 19 + 29+ ……….

Solution

Here the difference of consecutive terms of the series are 4, 6, 8, 10… which are in AP.

Let S_n = 1 + 5 + 11 + 19 + ……+ t_{n –1} + t_n          …(1)

And S_n =      1 + 5 + 11 + 19 + ……+ t_{n –1} + t_n    …(2)

Subtracting (2) from (1), we get

0 = 1 + {4 + 6 + 8 + …..to (n –1) terms} –t_n

or t_n  = 1 + {4 + 6 + 8 + …..to (n –1) terms}

= 1 + $\frac{n–1}{2}${2.4 + (n –1 –1)2} = 1 + (n –1) (n + 2) = 1 + n² +n –2 = n² + n – 1

∴ S_n = $\Sigma$ t_n = $\Sigma$(n² + n –1) = $\Sigma$n² + $\Sigma$n – $\Sigma$1

= $\left[\frac{\mathrm{n}(\mathrm{n}+1)(2\mathrm{n}+1)}{6}+\frac{\mathrm{n}(\mathrm{n}+1)}{2}–\mathrm{n}\right]=\frac{\mathrm{n}(\mathrm{n}+1)}{2}\left[\frac{2\mathrm{n}+1}{3}+1\right]–\mathrm{n}$ = $\frac{n(n+1)(2n+4)}{6}–n$

= $\frac{\mathrm{n}(\mathrm{n}+1)(\mathrm{n}+2)}{3}–\mathrm{n}=\frac{\mathrm{n}\left({\mathrm{n}}^2+3\mathrm{n}–1\right)}{3}$

Illustration -66

Find the sum of 1st n terms of the series 5, 7, 11, 17, 25,……..

Solution

Let S = 5 + 7 + 11 + 17 + 25+……….+ t_r

S = 5 + 7 + 11 + 17 + ….…… + t_{r – 1} + t_r

Subtracting, we get   0 = 5 + 2 + 4 + 6 + 8 + ……..+ r^th term – t_r

⇒ t_r = 5 + 2 $\left\{\frac{r\left(r–1\right)}{2}\right\}$ ⇒  t_r   = r² – r + 5

S_n = $\sum _{\mathrm{r}=1}^{\mathrm{n}}{\mathrm{r}}^2–\sum _{\mathrm{r}=1}^{\mathrm{n}}\mathrm{r}+5\sum _{\mathrm{r}=1}^{\mathrm{n}}1$

= $\frac{n}{6}${(n + 1)(2n + 1) – 3(n + 1) + 30}

= $\frac{n}{3}$(n² + 14).

Illustration -67

Find the n^th term and the sum of n terms of the series 1.2.4 + 2.3.5 + 3.4.6 + ………

Solution

r^th term of the series = r.(r + 1).(r + 3) =  r³ + 4r² + 3r

Hence the sum of n terms = + 4 + 3 =  = {3n² + 19n + 26} 

= $\sum _{\mathrm{r}=1}^{\mathrm{n}}{\mathrm{r}}^3+4\sum _{\mathrm{r}=1}^{\mathrm{n}}{\mathrm{r}}^2+3\sum _{\mathrm{r}=1}^{\mathrm{n}}\mathrm{r}$

= ${\left(\frac{n(n+1)}{2}\right)}^2+4\frac{n(n+1)(2n+1)}{6}+\frac{3n(n+1)}{2}$ = $\frac{n(n+1)}{12}\left\{3n^2+19n+26\right\}$

Illustration -68

Find the sum of “n” terms in the series 2 . 3 . 4 + 3 . 4. 5 + 4 . 5 . 6 + . . . . . .

Solution

Consider

2, 3, 4, . . . . . . .

a = 2, d = 3 – 2 = 1

t_n = a + (n–1) d

= 2 + (n–1) (1)

= 2 + n–1

= n + 1

Again consider,

3, 4, 5, 6, . . . . . . . . .

a = 3, d = 4 – 3 = 1

t_n = a + (n–1) d

= 3 + (n–1) (1)

= n + 2

Consider

4, 5, 6, . . . . . . . .

a = 4, d = 5 – 4 = 1

t_n = a + (n – 1) d

= 4 + (n – 1) (1)

= n + 3

The n^th term in the given series is,

t_n = (n + 1) (n + 2) (n + 3)

= (n + 1) (n² + 5n + 6)

= n³ + 5n² + 6n + n² + 5n + 6

= n³ + 6n² + 11n + 6

S_n = ${\mathrm{\Sigma t}}_{\mathrm{n}}$

= $\sum \left(n^3+6n^2+11n+6\right)$

=  ${\mathrm{\Sigma n}}^3+\mathrm{\Sigma }6{\mathrm{n}}^2+\mathrm{\Sigma }11\mathrm{n}+\mathrm{\Sigma }6$

= $\frac{n^2(n+1{)}^2}{4}+6\times \frac{n(n+1)(2n+1)}{6}+11\times \frac{n(n+1)}{2}+6n$

= $\frac{n}{4}\left[n(n+1{)}^2+4(n+1\left)\right(2n+1)+22(n+1)+24\right]$

= $\frac{n}{4}\left[n\left(n^2+2n+1\right)+4\left(2n^2+3n+1\right)+22(n+1)+24\right]$

= $\frac{\mathrm{n}}{4}\left[{\mathrm{n}}^3+2n^2+\mathrm{n}+8{\mathrm{n}}^2+12\mathrm{n}+4+22\mathrm{n}+22+24\right]$

= $\frac{\mathrm{n}}{4}\left[{\mathrm{n}}^3+10{\mathrm{n}}^2+35\mathrm{n}+50\right]$

Illustration -69

Find sum to “n” terms of 1² + (1² + 2²) + (1² + 2² + 3²) + . . . . . . . . .

Solution

Given series is

1² + (1² + 2²) + (1² + 2² + 3²) + (1² + 2² + 3² + 4²) + . . . . . . . . .

The n^th term in the given series is,

t_n = 1² + 2² + 3² + . . . . . .+n²

$t_n=\frac{n(n+1)(2n+1)}{6}$

We have,

${\mathrm{S}}_n=\Sigma {\mathrm{t}}_n$

= $\sum \frac{n(n+1)(2n+1)}{6}$

= $\frac{1}{6}\sum n(n+1)(2n+1)$

= $\frac{1}{6}\left[\frac{2n^2(n+1{)}^2}{4}+\frac{3n(n+1)(2n+1)}{6}+\frac{n(n+1)}{2}\right]=\frac{1}{6}\Sigma \left(2n^2+3n^2+n\right)$

= $\frac{1}{6}\times \frac{\mathrm{n}(\mathrm{n}+1)}{2}\left[\mathrm{n}\right(\mathrm{n}+1)+(2\mathrm{n}+1)+1]$

= $\frac{n(n+1)}{12}\left[n^2+n+2n+1+1\right]=\frac{n(n+1)\left(n^2+3n+2\right)}{12}$

Illustration -70

What is the sum of the first 2nterms of the series [2 + 1 + 4 + 2 + 6 + 4 + 8 + …]?

(A) n²+ n + 2ⁿ + 1        (B) n²+ 2n + 2ⁿ – 1    (C) n²+ n + 2ⁿ – 1       (D) n²+ 2n+2ⁿ+ 1  

Solution.

C

S_2n = 2 + 1 + 4 + 2 + 6 + 4 + … +T_2n

= (2 + 4 + 6 + …T_n ) + (1 + 2 + 4 + …T_n)

= $\frac{\mathrm{n}}{2}[4+(\mathrm{n}–1)\times 2]+\left(2^{\mathrm{n}}–1\right)$

= n (2 + n − 1) + 2ⁿ – 1

=2n + n² − n + 2ⁿ −1

= n² + n + 2ⁿ – 1

Thus, the sum of first 2n terms of the given series is n² + n + 2ⁿ − 1.

Illustration -71

The sum of the series 6 + 12 + 20 + 30 + 42 + … to 221 terms is given by

(A) $\frac{222\times 443}{3}+\frac{3\times 224\times 442}{2}$                        (B) $\frac{4\times 222\times 443}{6}+\frac{3\times 221\times 222}{2}$

(C) $\frac{221\times 222\times 443}{6}+\frac{3\times 221\times 222}{2}+442$   (D) $\frac{221\times 222\times 223}{6}+\frac{221\times 443}{2}+442$

Solution

C

The given series is 6 + 12 + 20 + 30 + 42 + …

S_n = 6 + 12 + 20 + 30 + 42 + ……… + a_n–1 + a_n ^{___________________} (1)

S_n = 6 + 12 + 20 + 30 + …….. + a_n–2 + a_n–1 + a_n ^{___________________} (2)

Subtracting (2) from (1), we obtain

0 = 6 + (6 + 8 + 10 + 12 + ……. (n – 1) terms) – a_n

a_n = 6 + (6 + 8 + 10 + 12 + ……. (n – 1) terms)

$a_n=6+\frac{(n–1)[2\times 6+(n–2\left)2\right]}{2}$ [6 + 8 + 10 + 12 + …….. (n – 1) terms is an A.P.]

= $6+\frac{(n–1)[12+2n–4]}{2}$ = 6 + $\frac{(n–1)[2n+8]}{2}$ = 6+(n-1)(n+4)

= 6 + n² + 3n – 4 = n² + 3n + 2

$\therefore {\mathrm{S}}_{221}=\sum _{k=1}^{221}{\mathrm{a}}_k=\sum _{k=1}^{221}{k}^2+3\sum _{k=1}^{221}k+2\sum _{k=1}^{221}1=\sum _{k=1}^{221}{k}^2+3\sum _{k=1}^{221}k+2\times 221$

= $\frac{221(221+1)(2\times 221+1)}{6}+3\left(\frac{221(221+1)}{2}\right)+442$

$\left[\sum _{k=1}^n{k}^2=\frac{n(n+1)(2n+1)}{6},\sum _{k=1}^{n}k=\frac{n(n+1)}{2}\right]$

24. MISCELLANEOUS CONCEPTS AND PROBLEMS

1. INEQUALITIES

(i) Let a₁, a₂, ……….., a_n be n positive real numbers, then we define their arithmetic mean (A), geometric mean (G) and harmonic mean (H) as ,

G = (a₁, a₂, ……….., a_n)^1/2 and H = $\frac{n}{\left(\frac{1}{a_1}+\frac{1}{a_2}+\frac{1}{a_3}+\dots \dots \frac{1}{a_n}\right)}$

It can be shown that  Moreover equality holds at either place if and only if a₁ = a₂ = …………… = a_n

(ii) Weighted Means

Let a₁, a₂, ……….., a_n be n positive real numbers and W₁, W₂, ……….., W_n be n positive  rational numbers. Then we define weighted Arithmetic mean (A*), weighted Geometric mean (G*) and weighted harmonic mean (H*) as

$A^{*}=\frac{a_1{w}_1+a_2{w}_2+\dots +a_n{w}_n}{w_1+w_2+\dots +w_n}$

and ${\mathrm{H}}^{*}=\frac{{\mathrm{w}}_1+{\mathrm{w}}_2+\dots +{\mathrm{w}}_{\mathrm{n}}}{\frac{{\mathrm{W}}_1}{{\mathrm{a}}_1}+\frac{{\mathrm{w}}_2}{{\mathrm{a}}_2}+\dots +\frac{{\mathrm{w}}_{\mathrm{n}}}{{\mathrm{a}}_{\mathrm{n}}}}$

A*  $\ge$G* $\ge$H* More over equality holds at either place if & only if a₁ = a₂i …………=.a_n

(iii) Cauchy’s Schwartz Inequality:

If a₁, a₂, ……….., a_n and b₁, b₂, ……….., b_n are 2n real numbers, then ${\left(a_1{b}_1+a_2{b}_2+\dots \dots \dots \dots +a_n{b}_n\right)}^2$ $\left({\mathrm{a}}_1{}^2+{\mathrm{a}}_2{}^2+\dots \dots \dots \dots +{\mathrm{a}}_{\mathrm{n}}^2\right)\left({\mathrm{b}}_1^2+{\mathrm{b}}_2^2+\dots \dots \dots +{\mathrm{b}}_{\mathrm{n}}^2\right)$ with the    equality holding if and only if $\frac{a_1}{b_1}=\frac{a_2}{b_2}=\dots \dots \dots .=\frac{a_n}{b_n}$.

Illustration -72

Prove that ${\left(\frac{\mathrm{a}+\mathrm{b}}{2}\right)}^{\mathrm{a}+\mathrm{b}}>{\mathrm{a}}^{\mathrm{b}}·{\mathrm{b}}^{\mathrm{a}}, \mathrm{a}, \mathrm{b}\in \mathrm{N}; \mathrm{a}\ne \mathrm{b}$

Solution

Let us consider b quantities each equal to a and a quantities each  equal to b. Then since A.M. > G.M.

$\frac{(a+a+a+\dots b\text{ times })+(b+b+b+\dots a\text{ times })}{a+b}$ > [(a.a.a…..b times) (b.b.b….a times)^1/(a+b)]

$\Rightarrow \frac{ab+ab}{a+b}>{\left(a^b{b}^2\right)}^{1/(a+b)}\Rightarrow \frac{2ab}{a+b}>{\left(a^b{b}^2\right)}^{1/(a+b)}$

Now $\frac{\mathrm{a}+\mathrm{b}}{2}>\frac{2\mathrm{ab}}{\mathrm{a}+\mathrm{b}} (\text{ A.M. }>\text{ H.M. })\Rightarrow {\left(\frac{\mathrm{a}+\mathrm{b}}{2}\right)}^{2+\mathrm{b}}>{\mathrm{a}}^{\mathrm{b}}.{\mathrm{b}}^2$

2. ARITHMETIC MEAN OF m^th POWER

Let a₁, a₂, ……….., a_n be n positive real numbers  and let m be a real number, then

$\frac{{\mathrm{a}}_1^{\mathrm{m}}+{\mathrm{a}}_2^{\mathrm{m}}+\dots +{\mathrm{a}}_{\mathrm{n}}^{\mathrm{m}}}{\mathrm{n}}\ge {\left(\frac{{\mathrm{a}}_1+{\mathrm{a}}_2+\dots +{\mathrm{a}}_{\mathrm{n}}}{\mathrm{n}}\right)}^{\mathrm{m}},\text{ if }\mathrm{m}\in \mathrm{R}–[0,1]$

However if  (0, 1), then $\frac{{\mathrm{a}}_1^{\mathrm{m}}+{\mathrm{a}}_2^{\mathrm{m}}+\dots +{\mathrm{a}}_{\mathrm{n}}^{\mathrm{m}}}{\mathrm{n}}\le {\left(\frac{{\mathrm{a}}_1+{\mathrm{a}}_2+\dots +{\mathrm{a}}_{\mathrm{n}}}{\mathrm{n}}\right)}^{\mathrm{m}}$

Obviously if $\mathrm{m}\in \{0,1\},\text{ then }\frac{{\mathrm{a}}_1^{\mathrm{m}}+{\mathrm{a}}_2^{\mathrm{m}}+\dots +{\mathrm{a}}_{\mathrm{n}}^{\mathrm{m}}}{\mathrm{n}}={\left(\frac{{\mathrm{a}}_1+{\mathrm{a}}_2+\dots +{\mathrm{a}}_{\mathrm{n}}}{\mathrm{n}}\right)}^{\mathrm{m}}$

Illustration -73

Prove that a⁴+ b⁴ + c⁴   abc (a + b + c),   [a, b, c > 0]

Solution

Using m^th power inequality, we get $\frac{a^4+b^4+c^4}{3}\ge {\left(\frac{a+b+c}{3}\right)}^4$

$\left(\frac{\mathrm{a}+\mathrm{b}+\mathrm{c}}{3}\right){\left(\frac{\mathrm{a}+\mathrm{b}+\mathrm{c}}{3}\right)}^3\ge \left(\frac{\mathrm{a}+\mathrm{b}+\mathrm{c}}{3}\right){\left[(\mathrm{abc}{)}^{1/3}\right]}^3$   (  A.M  G.M)

or $\frac{a^4+b^4+c^4}{3}\ge \left(\frac{a+b+c}{3}\right)abc$

∴   a⁴+ b⁴ + c⁴  $\ge$abc (a + b + c).

Illustration -74

Prove that $\frac{\mathrm{s}}{\mathrm{s}–\mathrm{a}}+\frac{\mathrm{s}}{\mathrm{s}–\mathrm{b}}+\frac{\mathrm{s}}{\mathrm{s}–\mathrm{c}}\ge \frac{9}{2}$ , if s = a + b + c,   [a, b, c > 0]

Solution

We have to prove that $\frac{1}{b+c}+\frac{1}{c+a}+\frac{1}{a+b}\ge \frac{9}{2(a+b+c)}$

for the proof , using m^th power theorem of inequality, we get

$\frac{(a+b{)}^{–1}+(b+c{)}^{–1}+(c+a{)}^{–1}}{3}\ge {\left[\frac{a+b+b+c+c+a}{3}\right]}^{–1}$  or, $\frac{1}{b+c}+\frac{1}{c+a}+\frac{1}{a+b}\ge \frac{9}{2(a+b+c)}$

Aliter : A.M. $\ge$  H.M.

$\Rightarrow \frac{(a+b)+(b+c)+(c+a)}{3}\ge \frac{3}{\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}}$ $\Rightarrow \frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\ge \frac{9}{2(a+b+c)}$

3. GENERAL RULE FOR FINDING THE VALUES OF RECURRING DECIMAL

Let X denote the figure which do not recur and assume they are l in number. Let Y denote recurring period of consisting of m figures. Let R denote the value of recurring decimal then R = XYYY…. (or)

$R=X\stackrel{\bullet }{Y}$

$\because {10}^{l}R=X.YYY\text{ and }{10}^{l+m}R=XY.YYY$

Subtracting  we get $R=\frac{XY–X}{{10}^{l+m}–{10}^l}$

Example:  $0.6\stackrel{\bullet }{2}\stackrel{\bullet }{3}$ = $\frac{623–6}{990}=\frac{617}{990}$

Example: $1.2\stackrel{\bullet }{4}\stackrel{\bullet }{3}=1+\frac{243–2}{990}=1+\frac{241}{990}=\frac{1231}{990}$

4. SUM OF THE PRODUCTS OF TWO TERMS OF A SEQUENCE

To obtain the sum $\sum _{\mathrm{i}<\mathrm{j}} {\mathrm{a}}_{\mathrm{i}}{\mathrm{a}}_{\mathrm{j}}$, we use the identity

$2\sum _{\mathrm{i}<\mathrm{j}}{\mathrm{a}}_{\mathrm{i}} {\mathrm{a}}_{\mathrm{j}}={\left({\mathrm{a}}_1+{\mathrm{a}}_2+\dots +{\mathrm{a}}_{\mathrm{n}}\right)}^2–\left({\mathrm{a}}_1^2+{\mathrm{a}}_2^2+\dots +{\mathrm{a}}_{\mathrm{n}}^2\right)$