# Formulae for Sum of A.P.

### Formulae

- Sum of first n terms of an A.P. is

S_{n}= (n/2)[2a +(n -1)d] - If S
_{n}Is sum of n terms of an A.P. whose first term is a and last term is*l*,

then S_{n}= (n/2)(a +*l*) - If common difference is d, number of terms n and the last term
*l*,

then S_{n}= (n/2)[2*l*-(n -1)d]

### Remark

Sometimes the question involves 3, 4 or 5 terms of an A.P.

If the sum of the numbers is given, then in an A.P.,

- three numbers are taken as a -d, a, a +d
- four numbers are taken as a -3d, a -d, a +d, a +3d.
- five numbers are taken as a -2d, a -d, a, a +d, a +2d.

This considerably simplifies the calculations of a and d.

## Illustrative Examples

### Example

- Sum the series 2 +4 +6 +... upto 40 terms.
- Find the sum of first 19 terms of the A.P. whose nth term is 2n +1.
- Find the sum of the series 1 +3 +5 +... +99.

### Solution

- We see that given series is an A.P. with first term a = 2,

common difference d = 2 and number of terms n = 40

Hence S_{40}= n[2a +(n -1)d]/2 = 40[2.2 +(40 -1)2]/2

= 20(4 +78) = 1640 - Here first term a = T
_{1}= 2n +1 = 2(1) +1 = 3,

and last term = T_{19}= 2(19) +1 = 39

S_{19}= n(a +l)/2 = 19(3 +39)/2 = 19.21 = 399 - Given series is an A.P. with a = 1,
*l*= 99, d = 2

To find the number of terms, we use*l*= a +(n -1)d

=> 99 = 1 +(n -1)2 => n -1 = 49 => n = 50

S_{n}= n(a +*l*)/2 = 50(1 +99)/2 = 50.50 = 2500

### Example

- Sum up 0·7 +0·71 +0·72 +... upto 100 terms.
- How many terms of the A.P. 17 +15 +13 +... must be taken so that sum is 72? Explain the double answer.

### Solution

- We see that given series is an A.P. with first term

a = 0·7 and common difference d = 0·01

S_{100}= n[2a +(n -1)d]/2 = 100[2(0·7) +(100 -1)(0·01)]/2

= 50 (1· 4 +0·99) = 119·5 - We are given a = 17, d = -2, S
_{n}= 72 and we have to find n.

Using S_{n}= n[2a +(n -1)d]/2, we get

72 = n[2.17 +(n -1)(-2)]/2 = n(36 -2n)/2 = n(18 -n)

=> n² -18n +72 = 0 => (n -6)(n -12) = 0

=> n = 6, 12

Both values of n, being positive integers are valid. We get double answer because sum of 7th to 12th terms is zero, as some terms are positive and some are negative.

### Example

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

### Solution

The sum of all exterior angles of a polygon = 360°. If the interior angles
are in A.P., then exterior angles are also in A.P.; largest exterior angle is
180°-120° = 60° and common difference is -5°.

Sum = 360° = n [2 a +(n -1) d]/2 = [2. 60° +(n -1)(-5°)]/2

=> 720 = 120 n -5n² +5n => 5n² -125n +720 = 0

=> n² -25 n +144 = 0 => (n -9)(n -16) = 0

=> n = 9 or n = 16.

However if n = 16, then internal angles would vary from 120° to 120° +(16
-1)(5°) = 195°; so one of the internal angles will be 180° which is not
possible in a polygon.

Hence the only correct solution is n = 9.

### Example

The sum of three numbers in A.P. is 18 and their product is 192. Find the numbers.

### Solution

Let the three numbers in A.P. be a -d, a, a +d. By given conditions,

(a -d) +a +(a +d) = 18, (a -d)a(a +d) = 192

=> 3a = 18, a(a² -d²) = 192 => a = 6, 6(36 -d²) = 192

=> d² = 36 -192/6 = 36 -32 = 4 => d = ±2.

With a = 6, d = 2, we get three numbers as 4, 6, 8.

With a = 6, d = -2, we get three numbers as 8, 6, 4.

Hence the required numbers are 4, 6, 8.

## Exercise

- Find the sum of

(i) 3 +7 +11 +15 ... to 30 terms

(ii) 1 +2/3 +1/3 +0 +.. to 19 terms

(iii) 5/2, 10/3, 25/6, 5 ... upto 24 terms

(iv) 51 +50 +49 +... +21

(v) 72 +70 +68 +... +40. - Find the sum of an A.P. of

(i) 25 terms whose nth term is 2n+5

(ii) 19 terms whose nth term is (2n+1)/3 - If the sum of n terms of a series is an²+bn, where a, b are constants,
show that it is an A.P. Find the first term and the common difference. Also
find an A.P. whose sum of any number of terms is equal to the square of the
number of terms.

[**Hint.**an²+bn=n² for all n => a = 1, b = 0.] - Of a, l, d, n, S
_{n}determine the ones which are missing for the following arithmetic progressions:

(i) a = -2, d = 5, S_{n}= 568

(ii) l = 8, n = 8, S_{8}= -20

(iii) d = 2/3, l = 10, n = 20. - (i) How many terms of series 13 +11 +9 +... make the sum 45? Explain the double answer.

(ii) How many terms of the sequence 18, 16, 14,... should be taken so that their sum is zero? - (i) Determine the sum of first 35 terms of an A.P. if T
_{2}= 2, T_{7}= 22.

(ii) It the third term of an A.P. is 1 and 6th term is -11, find the sum of first 32 terms.

(iii) If the first term of an A.P. is 2 and the sum of first 4 terms is equal to one fourth of the sum of the next five terms, find the sum of first 30 terms. - If the sum of an n terms of two arithmetic series are in ratio

(i) (14 -4n):(3n +5), find the ratio of their 8th terms

(ii) (2 +3n):(3 +2n), find the ratio of their 7th terms. - (i) If the ratio of sum of m terms of an A.P. to the sum of n terms is
m² : n², show that ratio of the pth and qth term is (2p -1) : (2q -1).

(ii) The sums of m and n terms of an A.P. are in ratio (2m +1) : (3n +1). Find the ratio of its 7th and 10th terms. - (i) If in an A.P., S
_{1}= 6 and S_{7}= 105, prove that

S_{n}: S_{n -3}: : (n +3) : (n -3).

(ii) If in an A.P., S_{3}= 6 and S_{6}= 3, prove that

2(2n +1)S_{n +4}= (n +4)S_{2n +1}. - (i) Find the sum of all two digit numbers.

(ii) Find the sum of all natural numbers between 100 and 1000 which are divisible by 2 as well as by 5.

(iii) Find the sum of all two digit numbers which leave 1 as remainder when divided by 3.

(iv) Find the sum of all odd integers between 2 and 100 which are divisible by 3.

[**Hint.**(iv) The odd integers between 2 and 100 which are divisible by 3 are 3, 9, 15, 21, ..., 99.] - (i) The sum of three numbers in A.P. is -3 and their product is 8. Find the numbers.

(ii) The sum of three consecutive numbers in an A.P. is 24 and the sum of their squares is 194. Find the numbers.

(iii) The sum of three numbers in an A.P. is 30, and the ratio of first to third is 3 : 7. Find the numbers.

(iv) Find five numbers in an A.P. whose sum is 25 and the ratio of the first to the last is 2 : 3.

(v) Divide 32 into four parts which are in A.P. such that the product of extremes is to the product of means is 7 : 15.

## Answers

**1.**(i) 1830 (ii) -38 (iii) 290 (iv) 1116 (v) 952

**2.**(i) 775 (ii) 133

**3.**a +b; 2a; 1, 3, 5, 7, ....

**4.**(i) n = 16,

*l*= 73 (ii) a = -13, d = 3

(iii) a = -8/3, s

_{1}= 220/3

**5.**(i) 5 or 9 (ii) 19

**6.**(i) 2310 (ii) -1696 (iii) 1655

**7.**(i) -23/25 (ii) 41/29

**8.**(ii) 513/754

**10.**(i) 4905 (ii) 98450 (iii) 1605 (iv) 867

**11.**(i) -4, -1, 2 (ii) 7, 8, 9 (iii) 6, 10, 14 (iv) 4, 9/2, 5, 11/2, 6

(v) 2, 6, 10, 14