Mathematics · Sequence and Series

JEE Main 2025 — 7 April, Evening Shift — Question 31

Let $a_{n}$ be the $n^{\text {th }}$ term of an A.P. If $S_{n}=a_{1}+a_{2}+a_{3}$ $+\ldots+a_{n}=700, a_{6}=7$ and $S_{7}=7$ ,

then $a_{n}$ is equal to :

Answer: B

$a_{6}=7=a+5 d=7 ..(i)$

S_{7}=\frac{7}{2}(2 a+6 d)=7(a+3 d)=7

\Rightarrow a+3 d=1 …(ii)

Using (i) & (ii)

$\Rightarrow 2 d=6 \Rightarrow d=3, a_{1}=-8$

$S_{n}=\frac{n}{2}(2 a+(n-1) d)=700$

$=\frac{n}{2}(-16+(n-1) 3)=700$

$=n(3 n-19)=1400, n>0 \Rightarrow n=25$

$\Rightarrow a_{25}=(-8)+24 \cdot(3)=72-8=64$

Answer key and solution verified before publishing.

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