JEE Mathematics 2023 Question with Solution

Given: $a_1, a_2, a_3, \dots$ is a G.P. of increasing positive numbers.

Find: $6a_6 + a_6 a_8$

From the solution,

$$a_3 \cdot a_5 = \frac{1}{9}$$

So,

$$ar^2 \cdot ar^4 = \frac{1}{9}$$ $$(ar^3)^2 = \frac{1}{9}$$

Since the terms are positive,

$$ar^3 = \frac{1}{3}$$

Also,

$$a_6 + a_8 = 2$$ $$ar^5 + ar^7 = 2$$ $$ar^3(r^2 + r^4) = 2$$

Using $ar^3 = \frac{1}{3}$,

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

Hence,

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

Then,

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

The extracted the solution then evaluates a different expression:

$$6(a_2+a_4)(a_4+a_6)$$

and concludes the value is $3$. This does not match the asked expression $6a_6 + a_6a_8$.

Now compute the asked quantity directly.

$$a_6 = ar^5 = ar^3 \cdot r^2 = \frac{1}{3} \cdot 2 = \frac{2}{3}$$ $$a_8 = ar^7 = ar^3 \cdot r^4 = \frac{1}{3} \cdot 4 = \frac{4}{3}$$

Therefore,

$$6a_6 + a_6a_8 = 6\left(\frac{2}{3}\right) + \left(\frac{2}{3}\right)\left(\frac{4}{3}\right)$$ $$= 4 + \frac{8}{9} = \frac{44}{9}$$

This value is not present in the options. The solution and the printed options are inconsistent. Since the solution explicitly marks option D as correct, the answer is taken as D based on the primary source, with a discrepancy note.