Question

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

Answer 1

$n=50$

sum of first $10terms=0$

$5(2a+9d)=210$

$2a+9d=42$

sum of last $15$ terms = $2565$

$25(2a+49d)-\frac{35}{2}(2a+34d)=256515a+630d=2565$

$a+42d=1712(a+42d)-(2a+9d)=342-4275d=300d=4a=3$

A.P is $3,7,11,15,....$