JEE Main 2026 · 4 Apr · Shift 1 · Mathematics Q7

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MathematicsSingle correct+4 / -1Question 7 of 75

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:

\dfrac5{87}

\dfrac{25}{83}

\dfrac{15}{29}

\dfrac5{29}

Answer and worked solution

Official answer: A

Condition. $S_{30}=15(a+l)=l^3$ with $a=\frac{10}3$ and $l=a+29d$ .

Check (A), $d=\frac5{87}$ : $l=\frac{10}3+\frac53=5$ , and $15\left(\frac{10}3+5\right)=125=5^3$ ✓.

(A)

Topics:Arithmetic progression

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JEE Main 2026 · 4 Apr · Shift 1, Mathematics question 7