Question

Sum of the first $14$ terms of and AP is $1505$ and its first term is $10$. Find its ${25}^{th}$ term.

Answer 1

Let $d$ be the common difference of the AP.

Here, $a=10$ and $n=14$

Now, $S_{14}=1505$

$\Rightarrow \frac{14}{2}[2\times 10+(14-1)\times d]=1505$

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

$\Rightarrow 20+13d=215$

$\Rightarrow d=\frac{215-20}{13}=\frac{195}{13}=15$

$\therefore {25}^{th}$ term of the A.P using $t_n=a+(n-1)d$

$=10+(25-1)\times 15$

$=10+360=370$

Hence, the required term is $370$.