In an AP 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.

$\Rightarrow 10a+45d=210$

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

And its $35^{th}$  term, $a_{35} =a+( 35-1) d$

$a_{35} =a+34d$

Similarly ,its $50^{th}$ term $a_{50} =a+49d$

Here for the last 15 term, $a_{35}$ is the first term and $a_{50}$ is the last term.

common difference will be the same.

sum of the last 15 terms$=\frac{n}{2}( a+l) \ \ \ \ \ \ \ \ \ \ \ ( l\ stands\ for\ last\ term\ of\ the\ A.P.)$

$\Rightarrow \frac{15}{2}( a+35d+a+49d) =2565$

$\Rightarrow 2a+84d=171\times 2$

$\Rightarrow 2a+84d=342$

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

On solving $( 1)$  and $( 2)$ ,

We have $a=3$ and $d=4$

Therefore the A.P. $3,\ 7,\ 11,\ 15,............199