# FIND THE INDICATED TERMS OF THE ARITHMETIC SEQUENCE

To find the sum of an arithmetic series, we use the formulas give below.

Sn  =  (n/2) [a + l] (or)

Sn  =  (n/2) [2a + (n - 1)d]

nth term of arithmetic progression :

an  =  a + (n - 1)d

Total number of terms :

n  =  [(l - a)/d] + 1

a = first term, d = common difference, n = number of terms and l = last term.

Example 1 :

Given an = 4, d = 2, Sn = -14 find n and a

Solution :

an = 4

a + (n - 1) d  =  4

a + (n - 1)2  =  4

a + 2n - 2  =  4

a + 2n  =  6

a  =  6 - 2n  ---(1)

Sn  =  -14

Sn  =  (n/2) [2a + (n - 1)d]

-14  =  (n/2) [2a  + (n - 1)2]

-14  =  (n/2) [2a  + 2n - 2] ----(2)

By applying the value of a in (2), we get

-14  =  (n/2)[2(6 - 2n) + 2n - 2]

-28  =  n[12 - 4n + 2n - 2]

-28  =  n[10 - 2n]

-28  =  10n - 2n2

2n2 - 10n - 28  =  0

Dividing it by 2, we get

n2 - 5n - 14  =  0

(n - 7) (n + 2)  =  0

n  =  7  or n = -2 (not admissible)

By applying the value of n in (1), we get

a  =  6 - 2(7)

a  =  6 - 14

a  =  -8

Hence the values of a and n are -8 and 7 respectively.

Example 2 :

Given a = 3, n = 8, Sn = 192 find d

Solution :

Sn  =  192

Sn  =  (n/2) [2a + (n - 1)d]

192  =  (8/2) [2(3) + (8 - 1)d]

192  =  4 [6 + 7d]

192/4  =  6 + 7d

48  =  6 + 7d

48 - 6  =  7d

7d  =  42

d  =  42/7  =  6

Hence the value of d is 6.

Example 3 :

Given l = 28 , S = 144 , and there are total 9 terms. Find a

Solution :

Let "l" be the last term or nth term of the series.

l  =  tn  =  28

tn  =  a + (n - 1)d

a + (9 - 1)d  =  28

a + 8d  =  28 ----(1)

S9  =  144

Sn  =  (n/2) [2a + (n - 1)d]

(n/2) [2a + (n - 1)d]  =  144

(9/2) [2a + 8d]  =  144

[2a + 8d]  =  144  (2/9)

2a + 8d  =  32

a + 4d  =  16  ------(2)

(1) - (2)

a + 8d  =  28

a + 4d  =  16

(-)     (-)    (-)

---------------

4d  =  12

d  =  3

By applying the value of d in (1), we get

a + 8(3)  =  28

a + 24  =  28

a  =  28 - 24  =  4

Hence the first term is 4.

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