JEE Main · 2021 · 26 Feb 2021 · Shift 1 · Sequences and Series

In an increasing geometric series, the sum of the second and the sixth term is

\frac{25}{2}

and the product of the third and fifth term is 25. Then, the sum of

4^{th}, 6^{th}

and

8^{th}

terms is equal to :

Answer: (B)

Solution

Let the first term be

a

and the common ratio

r

, with

r > 1

because the series increases.
From

a r^{2} \cdot a r^{4} = 25

(a r^{3})^{2} = 25

, and the terms of an increasing positive geometric series are positive, so

a r^{3} = 5

.
Then

a r + a r^{5} = a r^{3} (\frac{1}{r ^{2}} + r^{2}) = \frac{25}{2} ⟹ \frac{1}{r ^{2}} + r^{2} = \frac{5}{2}

2 r^{4} - 5 r^{2} + 2 = 0

, giving

r^{2} = 2

r^{2} = \frac{1}{2}

; the increasing condition selects

r^{2} = 2

T_{4} + T_{6} + T_{8} = a r^{3} (1 + r^{2} + r^{4}) = 5 (1 + 2 + 4) = 35

The sum of the fourth, sixth and eighth terms is

35

JEE Main · 2021 · 26 Feb 2021 · Shift 1 · Single correct option