JEE Main · 2022 · 28 Jun 2022 · Shift 1 · Sequences and Series

Let

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

be an increasing geometric progression of positive real numbers. If

A_{1} A_{3} A_{5} A_{7} = \frac{1}{1296}

and

A_{2} + A_{4} = \frac{7}{36}

, then, the value of

A_{6} + A_{8} + A_{10}

is equal to

Answer: (C)

Solution

In a GP,

A_{1} A_{3} A_{5} A_{7} = (A_{4})^{4}

(using the property that in a GP, the product of terms equidistant from the middle equals the middle term raised to appropriate power).

Given

(A_{4})^{4} = \frac{1}{1296} = \frac{1}{6 ^{4}}

, so

A_{4} = \frac{1}{6}

.

From

A_{2} + A_{4} = \frac{7}{36}

, we get

A_{2} = \frac{7}{36} - \frac{1}{6} = \frac{7 - 6}{36} = \frac{1}{36}

.

The common ratio

r = \frac{A _{4}}{A _{2}} = \frac{1/6}{1/36} = 6

.

Then:

Therefore,

A_{6} + A_{8} + A_{10} = 1 + 6 + 36 = 43

JEE Main · 2022 · 28 Jun 2022 · Shift 1 · Single correct option