**8.1. Introduction**

Federation Internationale de Football Association (FIFA) is organizing the Football World Cup from . . . . . . . . .. 1904, 1908, 1912,………,1994, 1998,2002, 2006, _________.*Can you predict the year of the next FIFA ?*

It is 2010. How is this obtained?

It can be observed that each successive year is obtained by adding 4 to the previous year i.e., successive year = previous year + 4. Such a set of numbers is called a **sequence**.**Sequence**

An arrangement of numbers in a definite order according to some rule is called a sequence . It is also called a **progression or pattern**. Every element of a sequence is called a **term**. For example, in a cricket match number of balls bowled in an innings increase by 6 balls every over. 6, 12, 18, . . . .. balls.

**8.2. Arithmetic progression**

Let us observe the following sequence.

Friday will come after every 7 days. 7, 14, 21, 28 ……….. days.*What rule does the above sequence follow?*

In the above sequence, numbers are obtained by adding a fixed number 7 to the preceding (Except first). Such kind of sequences are called Arithmetic sequences or **Arithmetic progressions**.*A sequence in which every term is obtained by adding a fixed constant (positive or negative) to its preceding term except the first term is called an Arithmetic progression(A.P).*

*A sequence in which the difference between any two successive terms is constant is called an Arithmetic Progression(A.P).***Examples :** Leap year comes after every four years (if we start from 2000). 2000, 2004, 2008, 2012 ……..

**8.3. Terms related to A.P.**

Let us consider an A.P : 4, 7, 10, 13 ………

The terms of an A.P. are generally denoted by ${t}_{1}$,${t}_{2}$,${t}_{3}$……… Here, the first term (${t}_{1}$) is generally denoted a .

First term (a) = 4 and common difference (d)

= ${t}_{2}$ – ${t}_{1}$= 7 – 4 = 3

${t}_{1}$ = 4 ; ${t}_{2}$ = 7 = 4 + 3 = ${t}_{1}$ + 3

$\Rightarrow $ ${t}_{2}$$\u2013$${t}_{1}$ = 3

$\Rightarrow $ ${t}_{2}$ – ${t}_{1}$ = 3

$\Rightarrow $ ${t}_{3}$ – ${t}_{2}$ = 3

$\Rightarrow $ ${t}_{4}$= 13 =10 + 3 = ${t}_{3}$ + 3

$\Rightarrow $ ${t}_{4}$ – ${t}_{3}$ = 3

${t}_{n\u20131}$ = …………= ${t}_{n\u20132}$ + 3 $\Rightarrow $ ${t}_{n\u20131}$ – ${t}_{n\u20132}$ = 3

${t}_{n}$ = ………… = ${t}_{n\u20131}$ + 3 $\Rightarrow $ ${t}_{n}$ – ${t}_{n\u20131}$ = 3

${t}_{2}$ = 7 = 4 + 3 = ${t}_{1}$ + 3 $\Rightarrow $ ${t}_{2}$ – ${t}_{1}$ = 3

${t}_{3}$ = 10 = 7 + 3 = ${t}_{2}$ + 3 $\Rightarrow $ ${t}_{3}$ – ${t}_{2}$= 3

${t}_{4}$ = 13 =10 + 3 = ${t}_{3}$ + 3 $\Rightarrow $ ${t}_{4}$ – ${t}_{3}$ = 3

${t}_{n}$ = …………= ${t}_{n\u20131}$+ 3 $\Rightarrow $ ${t}_{n}$ – ${t}_{n\u20131}$= 3

*What do you observe ?*

Here, the difference between successive and preceding is always constant i.e.. 3*What the above constant is called ?*

The constant is called common difference and it is denoted by **d**

$\therefore $ In the sequence 4, 7, 10, 13,…….**First term** = ${t}_{1}$ = a = 4 and**Common difference** = d = 3**General representation of AP**

${n}^{th}$term of A.P is ${t}_{n}$ = a + ( n–1)d

The first term is **‘a’** and common difference is **‘d’.****Note:** Sum of ‘n’ terms of A.P ${S}_{n}$ = $\frac{n}{2}$ [ 2a + ( n– 1) d] and ${\mathrm{S}}_{\mathrm{n}}=\frac{\mathrm{n}}{2}(\mathrm{a}+1)$

where $l$ is the last term.**8.4. General Term of an A.P****Theorem**

Let ‘a’ be the first term and ‘d’ be the common difference of an A.P. Then, its nth term or general term is given by an = a + (n – 1)d

**Proof:**

Let ${a}_{1}$ , ${a}_{2}$ , ${a}_{3}$, …,${a}_{n}$, … be the given A.P.

Then, ${a}_{1}$ = a $\Rightarrow $ ${a}_{1}$ = a + (1 – 1)d ——- (1)

Since each term of an A.P. is obtained by adding common difference to the preceding term.

Therefore, ${a}_{2}$ = a + d $\Rightarrow $ ${a}_{2}$ = a + (2 – 1)d —— (2)

Similarly, we have,

${a}_{3}$ = ${a}_{2}$ + d

$\Rightarrow $ ${a}_{3}$ = (a + d) + d

$\Rightarrow $ ${a}_{3}$ = a + 2d

$\Rightarrow $ ${a}_{3}$ = a + (3 – 1) d ——— (3)

and ${a}_{4}$ = ${a}_{3}$ + d

$\Rightarrow $ ${a}_{4}$ = (a + 2d) + d

$\Rightarrow $ ${a}_{4}$ = a + 3d

$\Rightarrow $ ${a}_{4}$ = a + (4 – 1)d ——– (4)

Observing the pattern in equations (1), (2), (3) and (4), we find that ${a}_{n}$ = a + (n – 1)d**Remark :** It is evident from the above theorem that**General term of an A.P. = First term + (Term number – 1) × (Common difference)**

From the above, we can conclude that, a sequence in which each term differ from its preceding term by a constant is called an Arithmetic progression, written as A. P. This constant term is called common difference of the A.P.

The general form of an A.P is a, a + d, a + 2d, a+3d, ….a + [n-1]d.

Here ‘a’ is called first term ‘d’ is the common difference and a + [n-1]d is nth term or general term of A.

**8.5. Properties of Arithmetic Progressions**

1. If a constant is added or subtracted from each term of an A.P., then the resulting sequence is also an A.P. with the same common difference. For example, if ‘k’ be added to each term of an A.P. a, a + d, a + 2d, .., then the resulting sequence

(a + k), (a + k + d), (a + k + 2d), …., is also an A.P. in which the first term is (a + k), which is different from the first term of the given sequence, but in both the sequences, the common difference is the same.

Similarly, if ‘k’ be subtracted from each term of an A.P. a, a + d, a + 2d, .., then the new sequence (a – k), (a – k + d), (a – k + 2d), …, is an A.P., whose first term is (a – k), which is different from the first term of the given sequence but the common difference is the same for both the sequences.

2. If each term of a given A.P. (with common difference ‘d’) is multiplied or divided by a non-zero constant ‘k’, then the resulting sequence is also an A.P. with common difference kd or d/k. For example, if each term of an A.P. a, a + d,

a + 2d, …, be multiplied by k ($k\ne 0$), then the new sequence ak, (a + d)k, (a + 2d)k, …, is also an A.P., whose first term = ak and the common difference = kd.

Again, if each term of the A.P. i.e., a, a + d, a + 2d, …, is divided by k($k\ne 0$), then the new sequence $\frac{a}{k},\frac{a+d}{k},\frac{a+2d}{k}$, … is also an A.P., whose first term = a/k and the common difference = d/k.

3. If the corresponding terms of two A.P.’s be added or subtracted, the resulting sequence is also an A.P.

For example, Let the two A.P’s be ${a}_{1}$ ,${a}_{1}$+${d}_{1}$,${a}_{1}$+ 2${d}_{1}$, … and ${a}_{2}$,${a}_{2}$+${d}_{2}$,${a}_{2}$+ 2${d}_{2}$, …

Adding the corresponding terms of the above two A.P.’s, the new sequence ${a}_{1}$ +${a}_{2}$, ${a}_{1}$+ ${a}_{2}$+${d}_{1}$ + ${d}_{2}$ ,${a}_{1}$+${a}_{2}$+ 2${d}_{1}$+ 2${d}_{2}$, … is also an A.P., whose first term = (${a}_{1}$ + ${a}_{2}$) and the common difference = (${d}_{1}$+${d}_{2}$).

Similarly, subtracting the terms of the second A.P. from the corresponding terms of the first, we get the new sequence

${a}_{1}$ – ${a}_{2}$ , ${a}_{1}$ – ${a}_{2}$ + ${d}_{1}$–${d}_{2}$,${a}_{1}$–${a}_{2}$+ 2${d}_{1}$– 2${d}_{2}$, …

which is also an A.P. having first term a1 – a2 and common difference ${d}_{1}$ – ${d}_{2}$.**Note:** The sequence, formed by multiplying or dividing the terms of one A.P. by the corresponding terms of another A.P., is not necessarily an A.P.

For example, the sequence ${a}_{1}$ ${a}_{2}$, (${a}_{1}$+${d}_{1}$) (${a}_{2}$+${d}_{2}$), (${a}_{1}$+ 2${d}_{1}$) (${a}_{2}$+ 2${d}_{2}$), …, the terms of which are formed by the product of the corresponding terms of two A.P’s, is not an A.P.

Similarly, the sequence, the terms of which are the quotients of the corresponding terms of two A.P’s, is also not necessarily an A.P. in general.

4. If ${a}_{1}$ ,${a}_{2}$, …, ${a}_{n}$are in A.P., then

a) ${a}_{1}$+${a}_{n}$= ${a}_{2}$ +${a}_{n\u20131}$ = ${a}_{3}$ +${a}_{n\u20132}$ = …

b) ${a}_{r}=\frac{{a}_{r\u2013k}+{a}_{r+k}}{2},0\le k\le n\u2013r$

5. If ${n}^{th}$ term of a sequence is a linear expression in n, then the sequence is an A.P.

6. Three numbers ${a}_{1}$, ${a}_{2}$, ${a}_{3}$ are in A.P. if and only if 2${a}_{2}$= ${a}_{1}$ +${a}_{3}$.

**8.6. Sum of n terms of an A.P**

Let a be the first term, d, the common difference and l, the nth term of the given A.P. If Sn be the sum of n terms, then

${S}_{n}$ = a + (a + d) + (a + 2d) + . + ($l$– 2d) + ($l$– d) + $l$———- (1)

Rewriting Sn in the reverse order, we have

${S}_{n}$= $l$+ ($l$– d) + ($l$– 2d) + .. + (a + 2d) + (a + d) +a——— (2)

Adding (1) and (2) column wise, we get

2${S}_{n}$=$\left(a+l\right)+(a+l)+(a+l)+...+(a+l)(ntimes)=n(a+l)$

$\Rightarrow {S}_{n}=\frac{n}{2}\left(a+l\right)\u2013\frac{n}{2}\left[a+a+\left(n\u20131\right)d\right]$

($\because $ $l$= a + (n – 1)d)

$\therefore $ ${S}_{n}$ = $\frac{n}{2}$ [2a + (n – 1)d]**Note**

1. In the above formula, four quantities ${S}_{n}$, a, n, d are involved. If any three of these are known, the fourth can be found. If two of the quantities are known, then the other two can be related by an equation.

2. If ${S}_{n}$ be the sum of n terms of an A.P., whose first term is a and the last term is l, then ${S}_{n}$ = $\frac{n}{2}$ (a + $l$)

3. In any sequences ${S}_{n}$–${S}_{n\u20131}$ = (${t}_{1}$+${t}_{2}$+${t}_{3}$+ … +${t}_{n\u20131}$+${t}_{n}$) – (${t}_{1}$+${t}_{2}$+${t}_{3}$+ … +${t}_{n\u20131}$) = ${t}_{n}$