Let a1, a2, a3,.... be in an A.P. such that sum from k=1 to 12 of a(2k-1) = -72/5 a1, a1 != 0. If sum from k=1 to n of ak = 0, then n is:

The correct answer is Option 2. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Maths Q19 – JEE-MAIN 2025

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Home Maths Sequences and SeriesQuestion #405

Let

a_{1}, a_{2}, a_{3}, . . . .

be in an A.P. such that

\sum_{k = 1}^{12} a_{2 k - 1} = - \frac{72}{5} a_{1}, a_{1} \neq 0

. If

\sum_{k = 1}^{n} a_{k} = 0

, then n is:

Clue 1 of 3: Properties of Arithmetic Progression

Recall the formula for the general term and the sum of terms in an A.P.

🧠 Full Solution Path

● Video Walkthrough

Step 1: Determine the relationship between

a_{1}

and

d

✦ Active

The terms $a_{2 k - 1}$ for $k = 1, \dots, 12$ are $a_{1}, a_{3}, \dots, a_{23}$ . This sequence forms an A.P. with first term $A = a_{1}$ and common difference $D = a_{3} - a_{1} = (a_{1} + 2 d) - a_{1} = 2 d$ . The sum of these 12 terms is given by the A.P. sum formula:

S_{12}^{'} = \frac{12}{2} [2 A + (12 - 1) D] = 6 [2 a_{1} + 11 (2 d)] = 12 (a_{1} + 11 d)

Given that $S_{12}^{'} = - \frac{72}{5} a_{1}$ , we set up the equation:

12 (a_{1} + 11 d) = - \frac{72}{5} a_{1}

Dividing by 12 and simplifying:

a_{1} + 11 d = - \frac{6}{5} a_{1} ⟹ 11 d = - \frac{6}{5} a_{1} - a_{1} = - \frac{11}{5} a_{1}

Since $a_{1} \neq 0$ , we can divide by $11 a_{1}$ (if $11 d = 0$ , then $a_{1} = 0$ , which is not allowed), yielding the relationship:

d = - \frac{1}{5} a_{1}

Step 2: Set up the equation for the sum of the first

n

terms○ Expand

The sum of the first $n$ terms of the original A.P. ( $a_{1}, a_{2}, \dots, a_{n}$ ) is given by:

S_{n} = \frac{n}{2} [2 a_{1} + (n - 1) d]

We are given that $S_{n} = 0$ . Therefore:

\frac{n}{2} [2 a_{1} + (n - 1) d] = 0

Since $n$ must be a positive integer (number of terms), $n \neq 0$ . Thus, the term in the square brackets must be zero:

2 a_{1} + (n - 1) d = 0

💡 Teacher's Secret Hint

Remember that $n$ cannot be zero for a sum of terms.

Step 3: Solve for

n

○ Expand

Substitute the relationship $d = - \frac{1}{5} a_{1}$ (found in Step 1) into the equation from Step 2:

2 a_{1} + (n - 1) (- \frac{1}{5} a_{1}) = 0

Since $a_{1} \neq 0$ , we can divide the entire equation by $a_{1}$ :

2 - \frac{n - 1}{5} = 0

Multiply by 5 to clear the denominator:

10 - (n - 1) = 0

Simplify and solve for $n$ :

10 - n + 1 = 0 ⟹ 11 - n = 0 ⟹ n = 11

💡 Teacher's Secret Hint

Ensure you correctly handle the negative sign when distributing $(n - 1)$ .

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Maths Question 19 – JEE-MAIN 2025

Clue 1 of 3: Properties of Arithmetic Progression

🧠 Full Solution Path