JEE Mathematics 2025 Question with Solution

Given: The first term is $a=3$. The sum of the first four terms is one-fifth of the sum of the next four terms.

Find: The sum of the first $20$ terms, namely $S_{20}$.

For an A.P.,

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

The sum of the first four terms is

$$S_4=\frac{4}{2}\left(2a+3d\right)$$

Substituting $a=3$,

$$S_4=2(6+3d)=12+6d$$

The sum of the next four terms, that is the $5^{\text{th}}$ to $8^{\text{th}}$ terms, is

$$S_{5-8}=\frac{4}{2}\left(2(a+4d)+3d\right)$$

So,

$$S_{5-8}=2(6+11d)=12+22d$$

Using the given condition,

$$12+6d=\frac{1}{5}(12+22d)$$

Multiply by $5$:

$$5(12+6d)=12+22d$$ $$60+30d=12+22d$$ $$8d=-48$$ $$d=-6$$

Now find $S_{20}$:

$$S_{20}=\frac{20}{2}(2\times 3+19\times(-6))$$ $$S_{20}=10(6-114)$$ $$S_{20}=10(-108)=-1080$$

Therefore, the sum of the first $20$ terms is $-1080$. The correct option is B.

Given: The first term is $a=3$ and the sum of the first four terms is one-fifth of the sum of the next four terms.

Find: $S_{20}$.

The sum of the next four terms is $S_8-S_4$, so the condition becomes

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

Multiply by $5$:

$$5S_4=S_8-S_4$$ $$6S_4=S_8$$

Now,

$$S_4=\frac{4}{2}(2a+3d)=2(2a+3d)$$

and

$$S_8=\frac{8}{2}(2a+7d)=4(2a+7d)$$

Using $6S_4=S_8$,

$$6\cdot 2(2a+3d)=4(2a+7d)$$

Substitute $a=3$:

$$12(6+3d)=4(6+7d)$$ $$72+36d=24+28d$$ $$8d=-48$$ $$d=-6$$

Then,

$$S_{20}=\frac{20}{2}(2a+19d)=10(6-114)=-1080$$

Therefore, the required sum is $-1080$, so the correct option is B.