Question

The $2^{nd},{31}^{st}$ and the last term of an A.P. are : $\frac{31}{4},\frac{1}{2},\frac{-13}{2}$ respectively. Find the number of terms and the sum of the terms of the given A.P.

Answer 1

Formula,

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

Given,

$a_2=a+d=\frac{31}{4}$........(1)

$a_{31}=a+30d=\frac{1}{2}$.........(2)

solving (1) and (2), we get,

$d=-\frac{1}{4},a=8$

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

$-\frac{13}{2}=8+(n-1)(-\frac{1}{4})$

$\frac{32-n+1}{4}=-\frac{13}{2}$

$\therefore n=59$

$S_n=\frac{n}{2}[2a+(n-1\left)d\right]$

$=\frac{59}{2}\left[2\right(8)+(59-1\left)(-\frac{1}{4})\right]$

$=\frac{59}{2}\times \frac{6}{4}$

$=\frac{177}{4}$