JEE Main · 2023 · 15 Apr 2023 · Shift 1 · Sequences and Series

Let

A_{1}

and

A_{2}

be two arithmetic means and

G_{1}, G_{2}, G_{3}

be three geometric means of two distinct positive numbers. Then

G_{1}^{4} + G_{2}^{4} + G_{3}^{4} + G_{1}^{2} G_{3}^{2}

is equal to

Answer: (B)

Solution

Let the two numbers be

a, b

. Then

a, A_{1}, A_{2}, b

in AP gives

A_{1} + A_{2} = a + b

.

For

a, G_{1}, G_{2}, G_{3}, b

in GP with common ratio

r = (b / a)^{1/4}

G_{1} = (a^{3} b)^{1/4}

G_{2} = (a^{2} b^{2})^{1/4} = ab

G_{3} = (a b^{3})^{1/4}

.

Note

G_{1} G_{3} = a^{3} b \cdot a b^{3} = ab

.

Compute:

G_{1}^{4} + G_{2}^{4} + G_{3}^{4} + G_{1}^{2} G_{3}^{2} = a^{3} b + a^{2} b^{2} + a b^{3} + a^{2} b^{2} = ab (a^{2} + 2 ab + b^{2}) = ab (a + b)^{2}

.

Since

G_{1} G_{3} = ab

and

A_{1} + A_{2} = a + b

, the answer is

(A_{1} + A_{2})^{2} G_{1} G_{3}

JEE Main · 2023 · 15 Apr 2023 · Shift 1 · Single correct option