Let $a_{1}, a_{2}, a_{3}, \ldots$ be a G.P. of increasing positive numbers. Let the sum of its $6^{\text{th}}$ and $8^{\text{th}}$ terms be 2 and the product of its $3^{\text{rd}}$ and $5^{\text{th}}$ terms be $\frac{1}{9}$. Then $6(a_{2}+a_{4})(a_{4}+a_{6})$ is equal to

JEE Main · 2023 · 13 Apr 2023 · Shift 2 · Sequences and Series

Let

a_{1}, a_{2}, a_{3}, \dots

be a G.P. of increasing positive numbers. Let the sum of its

6^{th}

and

8^{th}

terms be 2 and the product of its

3^{rd}

and

5^{th}

terms be

\frac{1}{9}

. Then

6 (a_{2} + a_{4}) (a_{4} + a_{6})

is equal to

Answer: (D)

Given:

a r^{5} + a r^{7} = 2

and

(a r^{2}) (a r^{4}) = \frac{1}{9}

From the product:

a^{2} r^{6} = \frac{1}{9} \Rightarrow a r^{3} = \pm \frac{1}{3}

Since the G.P. consists of increasing positive numbers,

a r^{3} = \frac{1}{3}

From the sum:

a r^{5} (1 + r^{2}) = 2

Substituting

a r^{3} = \frac{1}{3}

\frac{r ^{2}}{3} (1 + r^{2}) = 2 \Rightarrow r^{2} (1 + r^{2}) = 6

Solving:

r^{4} + r^{2} - 6 = 0 \Rightarrow (r^{2} + 3) (r^{2} - 2) = 0

Since

r > 0

r^{2} = 2 \Rightarrow r = 2

Then

a = \frac{1}{3 r ^{3}} = \frac{1}{6 2}

Now:

6 (a_{2} + a_{4}) (a_{4} + a_{6}) = 6 (a r + a r^{3}) (a r^{3} + a r^{5}) = 6 a^{2} r^{4} (1 + r^{2})^{2} = 6 \cdot \frac{1}{72} \cdot 4 \cdot 9 = 3

JEE Main · 2023 · 13 Apr 2023 · Shift 2 · Single correct option