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

Let d be the common difference of the AP.

Here, a = 10 and n = 14
Now,

$S_{14}=1505$                                       (Given)

$\Rightarrow \frac{14}{2}[2 \times 10+(14-1) \times d]=1505 \quad\left\{S_{n}=\frac{n}{2}[2 a+(n-1) d]\right\}$

$\Rightarrow 7(20+13 d)=1505$

$\Rightarrow 20+13 d=215$

$\Rightarrow 13 d=215-20=195$

$\Rightarrow d=15$

∴ 25th term of the AP, a25

$=10+(25-1) \times 15 \quad\left[a_{n}=a+(n-1) d\right]$

$=10+360$

$=370$

Hence, the required term is 370.