Sum of the first 14 terms of an A.P. is 1505 and its first term is 10. Find its 25th term.

Given:

Sum of the first 14 terms of an A.P. is 1505 and its first term is 10.

To do:

We have to find the 25th term of the A.P.

Solution:

Let the first term of the A.P. be $a$ and the common difference be $d$.

$a=10$

We know that,

nth term of an A.P. $a_n=a+(n-1)d$

Sum of $n$ terms of an A.P. $S_n=\frac{n}{2}(2a+(n-1)d)$

Therefore,

$S_{14}=\frac{14}{2}(2a+(14-1)d$

$1505=7(2(10)+13d)$

$215=20+13d$

$13d=215-20$

$13d=195$

$d=15$

$\Rightarrow a_{25}=a+(25-1)d$

$=10+24(15)$

$=10+360$

$=370$

Hence, the 25th term of the given A.P. is $370$.