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JEE Main · 2026 · 24 Jan 2026 · Shift 2 · Sequences and Series

(\frac{1}{3} + \frac{4}{7}) + (\frac{1}{3 ^{2}} + \frac{1}{3} \times \frac{4}{7} + \frac{4 ^{2}}{7 ^{2}}) + (\frac{1}{3 ^{3}} + \frac{1}{3 ^{2}} \times \frac{4}{7} + \frac{1}{3} \times \frac{4 ^{2}}{7 ^{2}} + \frac{4 ^{3}}{7 ^{3}}) + \dots

. upto infinite terms is equal to -

(A) $\frac{5}{2}$ ✓ Correct
(B) $\frac{7}{4}$
(C) $\frac{4}{3}$
(D) $\frac{6}{5}$

Answer: (A)

Solution

With

a = \frac{4}{7}

and

b = \frac{1}{3}

, the

n

th bracket is

\sum_{k = 0}^{n} a^{k} b^{n - k} = \frac{a ^{n + 1} - b ^{n + 1}}{a - b}

. Summing over

n \geq 1

,

S = \frac{1}{a - b} [\frac{a ^{2}}{1 - a} - \frac{b ^{2}}{1 - b}] = \frac{21}{5} [\frac{16}{21} - \frac{1}{6}] = \frac{21}{5} \cdot \frac{25}{42} = \frac{5}{2}

.

JEE Main · 2026 · 24 Jan 2026 · Shift 2 · Single correct option

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