Sum of first $14$ terms of an $A P$ is $1505$ and it first term is $10$ . Find its $25^{t h}$ term.

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Solution

Sum, $S_{n} = \frac{n}{2} [2 a + (n - 1) d]$

Given $a = 10$

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

$\Rightarrow$ $1505 = 7 [20 + 13 d]$

$\Rightarrow$ $1505 = 140 + 91 d$

$\Rightarrow$ $d = 15$

$\Rightarrow$ $n^{t h}$ term $T_{n} = a + (n - 1) d$

$\Rightarrow$ $T_{25} = 10 + (25 - 1) 15$

$= 10 + 24 \times 15$

$= 370$

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