JEE Main · 2024 · 08 Apr 2024 · Shift 2 · Sequences and Series

In an increasing geometric progression of positive terms, the sum of the second and sixth terms is

\frac{70}{3}

and the product of the third and fifth terms is 49. Then the sum of the

4^{th}

6^{th}

and

8^{th}

terms is:-

Answer: (C)

Solution

Let the GP have first term

a

, ratio

r

a r + a r^{5} = \frac{70}{3}

a r^{2} \cdot a r^{4} = 49 ⟹ (a r^{3})^{2} = 49 ⟹ a r^{3} = 7

(positive terms).

a r (1 + r^{4}) = \frac{70}{3}

; using

a r^{3} = 7 \Rightarrow a r = 7/ r^{2}

\frac{7}{r ^{2}} (1 + r^{4}) = \frac{70}{3} ⟹ \frac{1}{r ^{2}} + r^{2} = 10

Let

t = r^{2}

t^{2} - 10 t + 1 = 0

, giving

t = 3

t = 1/3

— since the GP is increasing,

r^{2} = 3

T_{4} + T_{6} + T_{8} = a r^{3} + a r^{5} + a r^{7} = a r^{3} (1 + r^{2} + r^{4}) = 7 (1 + 3 + 9) = 91

JEE Main · 2024 · 08 Apr 2024 · Shift 2 · Single correct option