((1)/(3)+(4)/(7))+ ( 1 3^ 2 +(1)/(3) × (4)/(7)+ 4^ 2 7^ 2 )+ ( 1 3^ 3 + 1 3^ 2 × (4)/(7)+(1)/(3) × 4^ 2 7^ 2 + 4^ 3 7^ 3 )+ upto infinite terms, is equal to

Correct answer: C. (5)/(2) The n -th bracket ( n ≥ 1 ) is ∑_ k=0 ^ n ((1)/(3))^ n-k ((4)/(7))^k = (4/7)^ n+1 - (1/3)^ n+1 4/7 - 1/3 = (21)/(5)[((4)/(7))^ n+1 - ((1)/(3))^ n+1 ] . Total sum = (21)/(5)[∑_ n=2 ^ ∞ ((4)/(7))^n - ∑_ n=2 ^ ∞ ((1)/(3))^n] = (21)/(5)[(16/49)/(3/7) - (1/9)/(2/3)] = (21)/(5)[(16)/(21) - (1)/(6)] = (21)/(5) · (25)/(42) = (5)/(2) .

JEE Main 202624 JanuaryEvening Shift

Exam: JEE Main
Year: 2026
Date: 24 January
Shift: Evening Shift
Subject: Mathematics
Chapter: Sequences and Series

JEE Main 2026 · 24 January · Evening Shift · Mathematics · Sequences and Series

((1)/(3)+(4)/(7))+ ( 1 3^ 2 +(1)/(3) × (4)/(7)+ 4^ 2 7^ 2 )+ ( 1 3^...

\left(\frac{1}{3}+\frac{4}{7}\right)+

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

\left(\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}}\right)+\ldots

upto infinite terms, is equal to

Options

A $\frac{4}{3}$
B $\frac{6}{5}$
C $\frac{5}{2}$
D $\frac{7}{4}$

Correct answer

C. $\frac{5}{2}$

Step-by-step solution

The

n

-th bracket (

n \geq 1

) is

\sum_{k=0}^{n} \left(\frac{1}{3}\right)^{n-k}\left(\frac{4}{7}\right)^k = \frac{(4/7)^{n+1} - (1/3)^{n+1}}{4/7 - 1/3} = \frac{21}{5}\left[\left(\frac{4}{7}\right)^{n+1} - \left(\frac{1}{3}\right)^{n+1}\right]

.
Total sum

= \frac{21}{5}\left[\sum_{n=2}^{\infty}\left(\frac{4}{7}\right)^n - \sum_{n=2}^{\infty}\left(\frac{1}{3}\right)^n\right] = \frac{21}{5}\left[\frac{16/49}{3/7} - \frac{1/9}{2/3}\right]

= \frac{21}{5}\left[\frac{16}{21} - \frac{1}{6}\right] = \frac{21}{5} \cdot \frac{25}{42} = \frac{5}{2}

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