Sequence Series – Vn-Vn-1 Method – JEE Main 24 January 2025 Shift 1

Solution:

First, let’s analyze the denominators of the series $S_n$:
$2 = 1 \times 2$
$6 = 2 \times 3$
$12 = 3 \times 4$
$20 = 4 \times 5$

Observing this pattern, the $k^{th}$ term $T_k$ can be written as:
$$ T_k = \frac{1}{k(k+1)} $$

To simplify the sum, we split this term using partial fractions:
$$ T_k = \frac{(k+1)-k}{k(k+1)} = \frac{k+1}{k(k+1)} – \frac{k}{k(k+1)} = \frac{1}{k} – \frac{1}{k+1} $$

Now, find the sum $S_{2025}$:
$$ S_{2025} = \sum_{k=1}^{2025} \left( \frac{1}{k} – \frac{1}{k+1} \right) $$
$$ S_{2025} = \left(1 – \frac{1}{2}\right) + \left(\frac{1}{2} – \frac{1}{3}\right) + \dots + \left(\frac{1}{2025} – \frac{1}{2026}\right) $$
Using the telescoping property, intermediate terms cancel out:
$$ S_{2025} = 1 – \frac{1}{2026} = \frac{2025}{2026} $$

Calculate the value of the expression given for the A.P.:
$$ \text{Sum Value} = \sqrt{2026 S_{2025}} = \sqrt{2026 \times \frac{2025}{2026}} = \sqrt{2025} = 45 $$

The sum of the first 6 terms of the A.P. ($S’_6$) is 45.
Let the A.P. have first term $a = -p$ and common difference $d = p$.
$$ S’_6 = \frac{6}{2} [2a + (6-1)d] = 3 [2(-p) + 5p] $$
$$ 45 = 3 [-2p + 5p] $$
$$ 45 = 3 [3p] = 9p $$
$$ p = 5 $$

We need to find the absolute difference between the $20^{th}$ and $15^{th}$ terms:
$$ |T_{20} – T_{15}| = |(a + 19d) – (a + 14d)| $$
$$ = |5d| = |5p| $$

Substitute $p=5$:
$$ |T_{20} – T_{15}| = 5 \times 5 = 25 $$

Ans. (1)