Question

How many terms of the sequence -12,-9,-6,-3,…,must be taken to make the sum 54?

Answer 1

Given Ap is -12,-9,-6,-3.

Let a be the first term and d be the common difference. 

First term a = -12.

Common difference d = -9 + 12

                                       = 3.


Sn = 54

We know that Sum of n terms sn = n/2(2a + (n - 1) * d)

= > 54 = n/2(2(-12) + (n - 1) * 3)

= > 54 = n/2(-24 + 3n - 3)

= > 54 = n/2(3n - 27)

= > 108 = n(3n - 27)

= > 108 = 3n^2 - 27n

= > 3n^2 - 27n - 108 = 0

= > n^2 - 9n - 36 = 0

= > n^2 - 12n + 3n - 36 = 0

= > n(n - 12) + 3(n - 12) = 0

= > (n + 3)(n - 12) = 0

= > n = -3, 12.



Therefore the number of terms = 12.


Hope this helps!

Answer 2

Hey !!!

a ' first term = - 12

a₂ = -9

d = a₂ - a₁ = -9 - (-12 )

d = 3

Sn = 54 (given )

∴ sn = n /2 { 2a + (n - 1) d

hence ,

54 = n /2 ( 2* (-12) + ( n - 1 ) 3

54 = n/2 ( -27 + 3n )

108 = 3n² - 27n

0 = 3n² -27n -108

3n² - 36n + 9n -108 =0

3 n ( n - 12 ) + 9 ( n - 12 ) =0

(3n + 9 ) ( n - 12 ) =0

3n + 9 = 0

n = -3 ( neglected )

Again....... ( n - 12 ) = 0

n = 12

hence no. of terms in the sequence is 12

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Hope it helps you !!

@Rajukumar111