JEE 2023 — Mathematics PYQ

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Question

JEE 2023 — Mathematics PYQ

JEE | Mathematics | 2023

Let $a_1, a_2, ..., a_n$ be in A.P. If $a_5 = 2a_7$ and $a_{11} = 18$ , then find the value of: $12 (\frac{1}{a _{10} + a _{11}} + \frac{1}{a _{11} + a _{12}} + ... + \frac{1}{a _{17} + a _{18}})$

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Explanation

Solution:

$a$ aur $d$ nikalna:

$a_5 = 2a_7 \implies a + 4d = 2(a + 6d) \implies a + 8d = 0$ .

$a_{11} = 18 \implies a + 10d = 18$ .

Subtract karne par: $2d = 18 \implies d = 9$ aur $a = -72$ .
Series ko simplify karna:

Har term ko rationalize karne par: $\frac{\sqrt{a_{n+1}} - \sqrt{a_n}}{d}$ .

$S = \frac{12}{d} \left( (\sqrt{a_{11}} - \sqrt{a_{10}}) + (\sqrt{a_{12}} - \sqrt{a_{11}}) + ... + (\sqrt{a_{18}} - \sqrt{a_{17}}) \right)$

$S = \frac{12}{d} (\sqrt{a_{18}} - \sqrt{a_{10}})$
Values rakhein:

$a_{10} = a + 9d = -72 + 81 = 9$

$a_{18} = a + 17d = -72 + 153 = 81$

$S = \frac{12}{9} (\sqrt{81} - \sqrt{9}) = \frac{4}{3} (9 - 3) = \frac{4}{3} \times 6 = 8$

Answer: Iska sahi maan 8 hai.

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