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:

The correct answer is Option 1. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Maths Q7 – JEE-MAIN 2026

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Home Maths Sequences and SeriesQuestion #393

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:

Clue 1 of 3: Arithmetic Progression Properties

Recall the formulas for the $n$ -th term ( $a_{n}$ ) and the sum of $n$ terms ( $S_{n}$ ) of an A.P.

🧠 Full Solution Path

Step 1: Formulate expressions for the last term and sum✦ Active

Given the first term $a = \frac{10}{3}$ and the number of terms $n = 30$ . Let $d$ be the common difference.

The last term $a_{30}$ is given by $a_{30} = a + (n - 1) d = \frac{10}{3} + (30 - 1) d = \frac{10}{3} + 29 d$ .

The sum of the A.P., $S_{30}$ , is given by $S_{30} = \frac{n}{2} (a + a_{30}) = \frac{30}{2} (\frac{10}{3} + a_{30}) = 15 (\frac{10}{3} + a_{30})$ .

Step 2: Set up and solve the cubic equation for the last term○ Expand

We are given that the sum of the A.P. is the cube of its last term, i.e., $S_{30} = (a_{30})^{3}$ .

Substituting the expressions from Step 1:

15 (\frac{10}{3} + a_{30}) = (a_{30})^{3}

Let $x = a_{30}$ . The equation becomes $15 (\frac{10}{3} + x) = x^{3}$ , which simplifies to $50 + 15 x = x^{3}$ , or $x^{3} - 15 x - 50 = 0$ .

By testing integer factors of 50, we find that $x = 5$ is a root: $5^{3} - 15 (5) - 50 = 125 - 75 - 50 = 0$ .

Factoring the cubic equation using $(x - 5)$ as a factor gives $(x - 5) (x^{2} + 5 x + 10) = 0$ . The quadratic factor $x^{2} + 5 x + 10$ has a discriminant $Δ = 5^{2} - 4 (1) (10) = 25 - 40 = - 15 < 0$ , so it has no other real roots.

Thus, the only real value for the last term is $a_{30} = 5$ .

💡 Teacher's Secret Hint

Remember to check for all real roots of the cubic equation and ensure they satisfy any given conditions (e.g., non-negative terms).

Step 3: Calculate the common difference○ Expand

Using the formula for the last term $a_{30} = a + 29 d$ and the value $a_{30} = 5$ found in Step 2:

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

Subtract $\frac{10}{3}$ from both sides:

29 d = 5 - \frac{10}{3} = \frac{15 - 10}{3} = \frac{5}{3}

Divide by 29 to find $d$ :

d = \frac{5}{3 \times 29} = \frac{5}{87}

Since $a_{1} = \frac{10}{3} > 0$ and $d = \frac{5}{87} > 0$ , all terms in the A.P. will be positive, satisfying the condition of non-negative terms.

💡 Teacher's Secret Hint

Always verify that the calculated common difference leads to terms that satisfy all conditions mentioned in the problem statement.

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Maths Question 7 – JEE-MAIN 2026

Clue 1 of 3: Arithmetic Progression Properties

🧠 Full Solution Path