Question

In an A.p the sum of n terms .common difference and last term are 136, 4 and 31 respectively .find the value of n.

Answer 1

Sn = 136, d= 4, an = 31

we know that

a+ (n-1)d = an

a + (n-1)4= 31

a = 31-(n-1)4.......(1)

and

n/2[ a + l]= Sn

n/2[a + 31] = 136

put value of a from (1)

n/2[31-4n+4 + 31] = 136

-4n^2 + 66n = 136×2

4n^2 -66n = -272

4n^2-66n +272= 0

2n^2 - 33n + 136 = 0

2n^2 - 16n - 17n + 136 = 0

2n(n-8)-17(n- 8)= 0

(n-8)(2n-17)= 0

n= 8 and n = 17/2

but number of terms can not be in fraction and negativv

hence number of terms are 8

n =8 is answer