Question

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

Answer 1

Answer:

4

Step-by-step explanation:

Sn = n/2 [ 2a + (n-1)d ]

atq, a = 25, d= 22-25 = -3 and Sn = 116

∴ 116 = n/2 [ 2*25 + (n-1)* -3 ]

or, 116 x 2 = n [ 50 - 3n + 3]

or, 232 = -3n² + 50n + 3n

or, 232 = -3n² + 53n

or, 3n²-53n+232=0

or, 3n²-24n-29n+232=0

or, 3n(n-8)-29(n-8)=0

∴ n= 8, 29/3

As "n" is an integer n=8

∴ 8th term = a + (n-1)d = 25 + (8-1)*-3 = 25 - 21 = 4