JEE Main 2026 — Apr 4, Shift 1 — Mathematics solution

JEE Main2026Apr 4, Shift 1Mathematics

The first term of an A.P. of $30$ non-negative terms is $\frac{10}{3}$ . If the sum of this A.P. is the cube of its last term, then its common difference is :

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

Solution

With $a = \frac{10}{3}$ , the last term is $T_{30} = a + 29 d = \frac{10}{3} + 29 d$ and the sum is

$S_{30} = \frac{30}{2} (2 a + 29 d) = 15 (\frac{20}{3} + 29 d) = 15 (\frac{10}{3} + \frac{10}{3} + 29 d) .$

The condition $S_{30} = T_{30}^{3}$ reads

$15 (\frac{10}{3} + T_{30}) = T_{30}^{3} .$

By inspection $T_{30} = 5$ satisfies $15 \cdot \frac{25}{3} = 125$ . Then

$\frac{10}{3} + 29 d = 5 \Rightarrow 29 d = \frac{5}{3} \Rightarrow d = \frac{5}{87} .$

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