Question

if the sum of a certain number of term of the Ap 25,22,19... is 116 find the last term​

Answer 1

Given that,

AP 25,22,19...

First term = 25

common difference = 22-25 = -3

we know that,

$\sf Sum \: of \: n \: terms = \boxed{ \sf S_n = \dfrac{n}{2}[2a + (n-1)d]}$

By substituting the given values,

$\sf S_n = \dfrac{n}{2}[2a + (n-1)d]$

$\sf 116 = \dfrac{n}{2}[2(25) + (n-1)( - 3)]$

$\sf 116 \times 2= n[50 - 3n + 3]$

$\sf 232= - 3 {n}^{2} + 53n$

$\sf- 3 {n}^{2} + 53n - 232 = 0$

$\sf 3 {n}^{2} - 53n + 232 = 0$

$\sf 3 {n}^{2} - 29n - 24n + 232 = 0$

$\sf n(3n - 29) - 8(3n - 29) = 0$

$\sf (n - 8)(3n - 29) = 0$

$\sf n = 8 \: (or) \: \dfrac{29}{3}$

n should be positive integer so n = 8

Last term of the AP = an = a + (n-2)d

$\sf a_n = a+ (n-1)d$

$\sf a_8 = 25+ (8-1)( - 3)$

$\sf a_8 = 25+ (7)( - 3)$

$\sf a_8 = 25 - 21$

$\boxed{ \sf a_8 = 4}$