FIND THE INDICATED TERMS OF THE ARITHMETIC SEQUENCE

Example 1 :

Given a_n = 4, d = 2, S_n = -14 find n and a 

Solution :

a_n = 4 

a + (n - 1) d  =  4

a + (n - 1)2  =  4

a + 2n - 2  =  4

a + 2n  =  6

a  =  6 - 2n  ---(1)

S_n  =  -14

S_n  =  (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, S_n = 192 find d

Solution :

S_n  =  192

S_n  =  (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 n^th term of the series.

l  =  t_n  =  28

t_n  =  a + (n - 1)d

a + (9 - 1)d  =  28

a + 8d  =  28 ----(1)

S₉  =  144

S_n  =  (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.

Example 4 :

Here are the nth terms of 4 sequences.

For each sequence state whether the numbers in the sequence are

A    Always multiples of 7

S    Sometimes multiples of 7

N    Never multiples of 7

Sequence 1 ...............

Sequence 2 ...............

Sequence 3 ...............

Sequence 4 ...............

Solution :

Sequence 1 :

nth term = 4n + 3

Some times it must be the multiple of 7.

Sequence 2 :

nth term = 7n + 1

This will never become a multiple of 7.

Sequence 3 :

nth term = 14n

Some times it may a multiple of 7.

Sequence 4 :

nth term = 8n - 1

It will never become multiple of 7.

Example 5 :

Can you spot any mistakes?

8, 15, 22, 29, ................

A sequence of numbers is shown below.

a)  Find the expression for the n^th term of the sequence.

b)  Explain why 96 will not be the term of the sequence.

Solution :

a)  8, 15, 22, 29, ................

t_n  =  a + (n - 1)d

a = 8, d = 15 - 8 ==> 7

t_n = 8 + (n - 1)7

= 8 + 7n - 7

t_n = 1 + 7n

b) Let 96 be one of the terms of the sequence.

96 = 8 + (n - 1)7

96 - 8 = (n - 1) 7

88/7 = n - 1

n - 1 = 12.57

n = 12.57 + 1

n = 13.57

Since the value of n is a decimal, 96 cannot be a one of the terms of the sequence.

Example 6 :

Is 205 a term in the sequence 1, 5, 9, 13, ... ... ?

Solution :

t_n  =  a + (n - 1)d

a = 1, d = 5 - 1 ==> 4

205 = 1 + (n - 1)4

205 - 1 = (n - 1) 4

4(n - 1) = 204

n - 1 = 204/4

n - 1 = 51

n = 52

Yes 205 is one of the terms of the sequence.

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