Question

If the sum of the AP; 48, 44, 40, 36, ......... upto n terms is –168, then the value of n is

Answer 1

Hii dear this

Is simple question

Answer 2

GIVEN :–

• An A.P. Series : 48 , 44 , 40 , 36 , ...........up to n terms.

• Sum of given A.P. = - 168

TO FIND :–

• Value of 'n' = ?

SOLUTION :–

• If first term of A.P. is 'a' and common difference is 'd' , Then sum of n terms –

$\\ \implies \large{ \boxed{ \bold{ S_{n} = \dfrac{n}{2}[2a + (n - 1)d] }}}$

• Put the values –

$\\ \implies { \bold{ - 168 =\dfrac{n}{2}[2(48) + (n - 1)( - 4)]}}$

$\\ \implies { \bold{ - 168 =\dfrac{n}{2}[96 - 4n + 4]}}$

$\\ \implies { \bold{ - 168 =n( 50 - 2n )}}$

$\\ \implies { \bold{ 0= 168 + 50n - 2n^{2} }}$

$\\ \implies { \bold{n^{2} - 25 n - 84=0}}$

$\\ \implies { \bold{n^{2} - 28 n + 3n - 84=0}}$

$\\ \implies { \bold{n(n-28)+3(n-28)}}=0$

$\\ \implies { \bold{(n+3)(n-28)=0}}$

$\\ \implies { \bold{n=28,n=-3(rejected)}}$

$\\ \implies \large{ \boxed { \bold{n= 28}}}$