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AP Sum equals Cube of Last Term: d = 5/87 | JEE Main 2026

JEE Maths question with a full step-by-step solution.

Question
The first term of an A.P. of 3030 non-negative terms is 103\dfrac{10}{3}. If the sum of this A.P. is the cube of its last term, then its common difference is
A587\dfrac{5}{87}correct
B2583\dfrac{25}{83}
C1529\dfrac{15}{29}
D529\dfrac{5}{29}
Solution
Step 1: With a=103a=\dfrac{10}{3} and 3030 terms, the sum is
S30=302[2⋅103+29d]=15(203+29d),S_{30}=\frac{30}{2}\left[2\cdot\frac{10}{3}+29d\right]=15\left(\frac{20}{3}+29d\right),
 and the last term is T30=103+29dT_{30}=\dfrac{10}{3}+29d. Given S30=(T30)3S_{30}=(T_{30})^3:
15(203+29d)=(103+29d)3.15\left(\frac{20}{3}+29d\right)=\left(\frac{10}{3}+29d\right)^3.
 Step 2: Try 103+29d=5\dfrac{10}{3}+29d=5 (so RHS =125=125). Then 203+29d=203+53=253\dfrac{20}{3}+29d=\dfrac{20}{3}+\dfrac53=\dfrac{25}{3}, and LHS =15⋅253=125=15\cdot\dfrac{25}{3}=125. ✓ So the equation is satisfied by
103+29d=5.\frac{10}{3}+29d=5.
 Step 3: Solve:
29d=5−103=53 ⇒ d=587.29d=5-\frac{10}{3}=\frac53\ \Rightarrow\ d=\frac{5}{87}.
 Correct answer: (1)
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