JEE Main 2025 — Sequences And Series Question with Solution

First, let us denote the first term of the G.P. by $a_1=A$ and the common ratio (which is $>1$, since the G.P. is increasing) by $r$. Then the terms are:

$a_1=A,a_2=Ar,a_3=A{r}^2,a_4=A{r}^3,a_5=A{r}^4,a_6=A{r}^5,\dots$

We are given:

$a_1\cdot a_5=28$, i.e.

$A\cdot (A{r}^4)=A^2{r}^4=28.(1)$

$a_2+a_4=29$, i.e.

$Ar+A{r}^3=A(r+r^3)=29.(2)$

We want to find $a_6=A{r}^5$.

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1. Solve for $r$

From $\text{(1)}$:

$A^2{r}^4=28⟹A^2=\frac{28}{r^4}.$

From $\text{(2)}$:

$A(r+r^3)=29⟹A=\frac{29}{r+r^3}.$

Plug $A$ from $\text{(2)}$ into $\text{(1)}$. After some algebra (or by a systematic approach), one finds that

$r^2=28⟹r=\sqrt{28}=2\sqrt{7}$

(since $r>1$, we take the positive root).

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2. Solve for $A$

From (2), using $r=2\sqrt{7}$:

$r+r^3=2\sqrt{7}+(2\sqrt{7}{)}^3=2\sqrt{7}+56\sqrt{7}=58\sqrt{7}.$

Hence,

$A=\frac{29}{58\sqrt{7}}=\frac{1}{2\sqrt{7}}.$

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3. Find $a_6$

$a_6=A{r}^5=\frac{1}{2\sqrt{7}}\times (2\sqrt{7}{)}^5.$

Compute $(2\sqrt{7}{)}^5$. First, $(2\sqrt{7}{)}^2=28$; thus

$(2\sqrt{7}{)}^5=(2\sqrt{7}{)}^4\cdot (2\sqrt{7})=784\times (2\sqrt{7})=1568\sqrt{7}.$

Therefore,

$a_6=\frac{1}{2\sqrt{7}}\cdot 1568\sqrt{7}=\frac{1568}{2}(\text{since }\sqrt{7}\text{ cancels})=784.$

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Answer: $a_6=784$.

Hence, the correct option is Option B.