Let a_ 1 , a_ 2 , a_ 3 , be G.P. of increasing positive terms such that a_ 2 · a_ 3 · a_ 4 =64 and a_ 1 +a_ 3 +a_ 5 =(813)/(7) . Then a_ 3 +a_ 5 +a_ 7 is equal to :

Correct answer: C. 3252 Let first term be a and common ratio r > 1 . From a_2 · a_3 · a_4 = 64 : (ar)(ar^2)(ar^3) = a^3r^6 = 64 ar^2 = 4 . From a_1 + a_3 + a_5 = (813)/(7) : a(1 + r^2 + r^4) = (813)/(7) . Substituting a = (4)/(r^2) : 4((1)/(r^2) + 1 + r^2) = (813)/(7) . Let u = r^2 : 28u^2 - 785u + 28 = 0 gives u = 28 (since r > 1 ). a_3 + a_5 + a_7 = ar^2(1 + r^2 + r^4) = 4(1 + 28 + 784) = 3252 .

JEE Main 202621 JanuaryMorning Shift

Exam: JEE Main
Year: 2026
Date: 21 January
Shift: Morning Shift
Subject: Mathematics
Chapter: Sequences and Series

JEE Main 2026 · 21 January · Morning Shift · Mathematics · Sequences and Series

Let a_ 1 , a_ 2 , a_ 3 , be G.P. of increasing positive terms such...

Let

a_{1}, a_{2}, a_{3}, \ldots

be G.P. of increasing positive terms such that

a_{2} \cdot a_{3} \cdot a_{4}=64

and

a_{1}+a_{3}+a_{5}=\frac{813}{7}

. Then

a_{3}+a_{5}+a_{7}

is equal to :

Options

A $3244$
B $3248$
C $3252$
D $3256$

Correct answer

C. $3252$

Step-by-step solution

Let first term be

a

and common ratio

r > 1

.
From

a_2 \cdot a_3 \cdot a_4 = 64

(ar)(ar^2)(ar^3) = a^3r^6 = 64 \Rightarrow ar^2 = 4

.
From

a_1 + a_3 + a_5 = \frac{813}{7}

a(1 + r^2 + r^4) = \frac{813}{7}

.
Substituting

a = \frac{4}{r^2}

4\left(\frac{1}{r^2} + 1 + r^2\right) = \frac{813}{7}

.
Let

u = r^2

28u^2 - 785u + 28 = 0

gives

u = 28

(since

r > 1

a_3 + a_5 + a_7 = ar^2(1 + r^2 + r^4) = 4(1 + 28 + 784) = 3252

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