Question

In an $A.P$ of $50$ terms, the sum of first $10$ terms is $210$ and the sum of its last $15$ terms is $2565.$ find the $A.P$

Answer 1

Given,

$n=50$

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

$S_{10}=\frac{10}{2}[2a+(10-1\left)d\right]=210$

$2a+9d=42$.....(1)

15th term from last $=50-15+1=36$th term

$a_{36}=a+35d$

sum of last 15 terms,

$\frac{15}{2}[2a_{36}+(15-1\left)d\right]=2565$

$\frac{15}{2}\left[2\right(a+35d)+14d]=2565$

$a+42d=171$.......(2)

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

$d=4,a=3$

Therefore the series is,

$3,7,11,15,............,199$

Hence, this is the answer.