Q32 / 75 · JEE Main 2026 (04 April Shift 1) · Mathematics
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:
- Correct answerA$\dfrac{5}{87}$
- B$\dfrac{25}{83}$
- C$\dfrac{15}{29}$
- D$\dfrac{5}{29}$
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: $\dfrac{5}{87}$
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: $\dfrac{5}{87}$
Hence, the correct answer is option: 'A'.