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