The first term of an A.P. of 30 non-negative terms is (10)/(3) . If the sum of this A.P. is the cube of its last term, then its common difference is:

Correct answer: A. (5)/(87) Given a = (10)/(3) and n = 30 . Let the last term be l and the common difference be d . The sum of the A.P. is given by S_ 30 = (30)/(2)(a + l) = 15((10)/(3) + l) . According to the given condition, S_ 30 = l^3 . 15((10)/(3) + l) = l^3 50 + 15l = l^3 l^3 - 15l - 50 = 0 (l - 5)(l^2 + 5l + 10) = 0 Since l^2 + 5l + 10 = 0 has no real roots, we get l = 5 . The last term of the A.P. is l = a + 29d . (10)/(3) + 29d = 5 29d = 5 - (10)/(3) = (5)/(3) d = (5)/(87) Answer: (5)/(87)

JEE Main 20264 AprilMorning Shift

Exam: JEE Main
Year: 2026
Date: 4 April
Shift: Morning Shift
Subject: Mathematics
Chapter: Sequences and Series

JEE Main 2026 · 4 April · Morning Shift · Mathematics · Sequences and Series

The first term of an A.P. of 30 non-negative terms is (10)/(3) . If...

The first term of an A.P. of

30

non-negative terms is

\dfrac{10}{3}

. If the sum of this A.P. is the cube of its last term, then its common difference is:

Options

A $\dfrac{5}{87}$
B $\dfrac{25}{83}$
C $\dfrac{15}{29}$
D $\dfrac{5}{29}$

Correct answer

A. $\dfrac{5}{87}$

Step-by-step solution

Given

a = \dfrac{10}{3}

and

n = 30

.

Let the last term be

l

and the common difference be

d

.

The sum of the A.P. is given by

S_{30} = \dfrac{30}{2}(a + l) = 15\left(\dfrac{10}{3} + l\right)

.

According to the given condition,

S_{30} = l^3

15\left(\dfrac{10}{3} + l\right) = l^3

50 + 15l = l^3

l^3 - 15l - 50 = 0

(l - 5)(l^2 + 5l + 10) = 0

Since

l^2 + 5l + 10 = 0

has no real roots, we get

l = 5

.

The last term of the A.P. is

l = a + 29d

\dfrac{10}{3} + 29d = 5

29d = 5 - \dfrac{10}{3} = \dfrac{5}{3}

d = \dfrac{5}{87}

Answer:

\dfrac{5}{87}

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