JEE Main 2025 — Sequences And Series Question with Solution

The first term, $a=3$

Common difference, $d$

The formula for the sum of the first $n$ terms of an A.P. is:

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

Given:

$S_4=\frac{1}{5}(S_8-S_4)$

This implies:

$5S_4=S_8-S_4\Rightarrow 6S_4=S_8$

Substituting the sum formulas:

$6\cdot \frac{4}{2}[2\times 3+(4-1)d]=\frac{8}{2}[2\times 3+(8-1)d]$

Simplifying:

$6\times 2[6+3d]=4[6+7d]$

$12(6+3d)=4(6+7d)$

$72+36d=24+28d$

$36d-28d=24-72$

$8d=-48$

$d=-6$

Now, to find $S_{20}$:

$S_{20}=\frac{20}{2}[2\times 3+(20-1)(-6)]$

$S_{20}=10[6+19\times (-6)]$

$S_{20}=10[6-114]$

$S_{20}=10\times (-108)$

$S_{20}=-1080$

Thus, the sum of the first 20 terms is $-1080$.