Let upto n terms. If the sum of the first six terms of an A.P. with first term –p and common difference p is , then the absolute difference between and terms of the A.P. is [2025]

Question

Let $S_{n} = \frac{1}{2} + \frac{1}{6} + \frac{1}{12} + \frac{1}{20} + . . .$ upto n terms. If the sum of the first six terms of an A.P. with first term –p and common difference p is $\sqrt{2026 S_{2025}}$ , then the absolute difference between $20^{th}$ and $15^{th}$ terms of the A.P. is [2025]

Smart Achievers · Accepted Answer

(3)

$S_{n} = \frac{1}{2} + \frac{1}{6} + \frac{1}{12} + \frac{1}{20} + . . .$ upto n terms

$= \frac{1}{1 \times 2} + \frac{1}{2 \times 3} + \frac{1}{3 \times 4} + \frac{1}{4 \times 5} + . . . + \frac{1}{n (n + 1)}$

$= (1 - \frac{1}{2}) + (\frac{1}{2} - \frac{1}{3}) + (\frac{1}{3} - \frac{1}{4}) + (\frac{1}{4} - \frac{1}{5}) + . . . (\frac{1}{n} - \frac{1}{n + 1})$

$= 1 - \frac{1}{n + 1} = \frac{n}{n + 1}$

$∴ \sqrt{2026 S_{2025}} = \sqrt{2026 \times \frac{2025}{2025 + 1}} = 45$

$S_{6} = 45$ , where $S_{6}$ denotes the sum of first six terms of an A.P.

$∴ \frac{6}{2} (- 2 p + 5 p) = 45 \Rightarrow p = 5$

$| a_{20} - a_{15} | = | (- 5 + 19 \times 5) - (- 5 + 14 \times 5) |$

$= | 90 - 65 | = 25$ .

Solution

1 20

2 45

3 25

4 90

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