Question -
Answer -
Given A.P.,
25, 22, 19, …
Here,
First term, a = 25 and
Common difference, d =22 – 25 = -3
Also given, sum ofcertain number of terms of the A.P. is 116
The number of terms ben
So, we have
Sn =n/2 [2a + (n-1)d] = 116
116 = n/2 [2(25) +(n-1)(-3)]
116 x 2 = n [50 – 3n +3]
232 = n [53 – 3n]
232 = 53n – 3n2
3n2 –53n + 232 = 0
3n2 –24n – 29n+ 232 = 0
3n(n – 8) – 29(n – 8)= 0
(3n – 29) (n – 8) = 0
Hence,
n = 29/3 or n = 8
As n can only be anintegral value, n = 8
Thus, 8th termis the last term of the A.P.
a8 =25 + (8 – 1)(-3)
= 25 – 21
= 4