JEE 2023 — Mathematics PYQ

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Question

JEE 2023 — Mathematics PYQ

JEE | Mathematics | 2023

Let the positive numbers $a_1, a_2, a_3, a_4$ and $a_5$ be in a G.P. Let their mean and variance be $\frac{31}{10}$ and $\frac{m}{n}$

Choose the correct answer:

Explanation

Let the numbers are $\frac{a}{r^2}, \frac{a}{r}, a, ar, ar^2$ then $\text{mean} = \frac{31}{10}$

\Rightarrow \frac{1}{5} \left[ \frac{a}{r^2} + \frac{a}{r} + a + ar + ar^2 \right] = \frac{31}{10}

\Rightarrow \frac{a}{r^2} [1 + r + r^2 + r^3 + r^4] = \frac{31}{10} \times 5 \quad \dots(i)

and mean of reciprocals $= \frac{31}{40}$

\Rightarrow \frac{1}{5} \left[ \frac{r^2}{a} + \frac{r}{a} + \frac{1}{a} + \frac{1}{ar} + \frac{1}{ar^2} \right] = \frac{31}{40}

\Rightarrow \frac{1}{ar^2} [r^4 + r^3 + r^2 + r + 1] = \frac{31}{40} \times 5 \quad \dots(ii)

from $(i)$ & $(ii)$

\frac{a \times ar^2}{r^2} = \frac{31 \times 5}{10} \times \frac{40}{31 \times 5} = 4

\Rightarrow a^2 = 4 \Rightarrow a = 2

also $a_3 + a_4 + a_5 = 14$

\Rightarrow a + ar + ar^2 = 14 \Rightarrow r = 2

\Rightarrow a + ar + ar^2 = 14 \Rightarrow r = 2

Choose the correct answer:

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