If Sn=1/2+1/6+1/12+… to n Terms, A.P. Sum is √(2026·S2025)

JEE Main 2025 — 24 January, Shift 1. Previous Year Question.

Problem

Let $S_n = \dfrac12+\dfrac16+\dfrac{1}{12}+\cdots$ up to $n$ terms. The sum of the first six terms of an A.P. with first term $-p$ and common difference $p$ is $\sqrt{2026\cdot S_{2025}}$. The absolute value of the difference between the 20th and 15th terms of this A.P. is:

Key insight. The two halves of this problem are independent: $S_n$ is a self-contained telescoping series that simplifies to a clean formula in $n$, completely unrelated to the A.P. until its final numeric value is plugged in as the A.P.’s given sum. Solving them in that order — series first, then A.P. — keeps the algebra manageable.

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Approach

The denominators $2, 6, 12, \ldots$ are each of the form $k(k+1)$ (i.e. $1\times2, 2\times3, 3\times4,\ldots$), which is the standard signal for a telescoping series: each term $\dfrac{1}{k(k+1)}$ splits into $\dfrac1k-\dfrac{1}{k+1}$, and almost everything cancels when the terms are summed. Once $S_{2025}$ is known as a plain number, it feeds directly into the A.P. condition, which is then just standard A.P. term-formula work.

Solution

Step 1 — Rewrite each term of the series

The $k$-th term is $\dfrac{1}{k(k+1)}$, which splits via partial fractions:

$$\frac{1}{k(k+1)} = \frac1k – \frac{1}{k+1}$$

Step 2 — Sum the telescoping series

$$S_n = \left(1-\frac12\right)+\left(\frac12-\frac13\right)+\left(\frac13-\frac14\right)+\cdots+\left(\frac1n-\frac{1}{n+1}\right)$$

Every middle term cancels with its neighbour, leaving only the first and last pieces:

$$S_n = 1 – \frac{1}{n+1} = \frac{n}{n+1}$$

Step 3 — Evaluate $S_{2025}$

$$S_{2025} = \frac{2025}{2026}$$

Step 4 — Compute the A.P.’s given sum

$$\sqrt{2026\cdot S_{2025}} = \sqrt{2026\times\frac{2025}{2026}} = \sqrt{2025} = 45$$

Step 5 — Use the sum of the first six A.P. terms

For an A.P. with first term $a=-p$ and common difference $d=p$, the sum of the first six terms is:

$$\frac{6}{2}\big[2a+(6-1)d\big] = 3\big[2(-p)+5p\big] = 3(3p) = 9p$$

Setting this equal to $45$:

$$9p = 45 \implies p = 5$$

Step 6 — Find the difference between the 20th and 15th terms

Using $a_n = a+(n-1)d$ with $a=-p$, $d=p$:

$$a_{20} = -p+19p = 18p, \qquad a_{15} = -p+14p = 13p$$

$$|a_{20}-a_{15}| = |18p-13p| = |5p| = 5\times5=25$$

Answer

Common mistakes

• Not recognising the $k(k+1)$ pattern in the denominators, and instead trying to sum the series term-by-term numerically — this works for small $n$ but is completely impractical for $n=2025$.
• Forgetting the square root applies to the whole product $2026\cdot S_{2025}$, not just $S_{2025}$ — the $2026$ is specifically there to cancel the denominator of $S_{2025}=\tfrac{2025}{2026}$, and skipping that cancellation leaves a much messier number.

Practise next

• If $S_n = \dfrac13+\dfrac{1}{15}+\dfrac{1}{35}+\cdots$ up to $n$ terms (denominators of the form $(2k-1)(2k+1)$), find $S_{50}$ using the same telescoping technique.