Question

An an arithmetic progression of $50$ terms, the sum of first ten terms is $210$ and the sum of last fifteen terms is $2565$. Find the arithmetic progression.

Answer 1

Sum of first 10 terms $S_{10}$

Sum of last 15 terms $S_{50-15+1}=S_{36}$

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

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

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

Sum of last 15 terms $a_{36}=a+35d$

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

$=15[a+35d+7d]=2565$

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

From (1) and (2) we get,

$d=4,a=3$

Therefore the series is:

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