If (1)/((20 - a)(40 - a)) + (1)/((40 - a)(60 - a)) + + (1)/((180 - a)(200 - a)) = (1)/(256) , then the maximum value of a is

Correct answer: C. 212 Given, (1)/((20 - a)(40 - a)) + (1)/((40 - a)(60 - a)) + + . . . + (1)/((180 - a)(200 - a)) = (1)/(256) So, on simplifying we get, ⇒ (1)/(20)[((1)/(20 - a) - (1)/(40 - a)) + ((1)/(40 - a) - (1)/(60 - a)). . . . . + + ((1)/(180 - a) - (1)/(200 - a))] = (1)/(256) ⇒ (1)/(20)((1)/(20 - a) - (1)/(200 - a)) = (1)/(256) ⇒ (20 - a)(200 - a) = 256 × 9 ⇒ a ^ 2 - 220a + 1696 = 0 ⇒ a = 8, 212 So, maximum value of a is 212.

JEE Main 202229 JulyMorning Shift

Exam: JEE Main
Year: 2022
Date: 29 July
Shift: Morning Shift
Subject: Mathematics
Chapter: Sequences and Series

JEE Main 2022 · 29 July · Morning Shift · Mathematics · Sequences and Series

If (1)/((20 - a)(40 - a)) + (1)/((40 - a)(60 - a)) + + (1)/((180 -...

\dfrac{1}{\left(20 - a\right)\left(40 - a\right)} + \dfrac{1}{\left(40 - a\right)\left(60 - a\right)} + \ldots \ldots +

\dfrac{1}{\left(180 - a\right)\left(200 - a\right)} = \dfrac{1}{256}

, then the maximum value of

a

Options

A $198$
B $202$
C $212$
D $218$

Correct answer

C. $212$

Step-by-step solution

Given, $\dfrac{1}{\left(20 - a\right)\left(40 - a\right)} + \dfrac{1}{\left(40 - a\right)\left(60 - a\right)} + \ldots \ldots +$ $. . . + \dfrac{1}{\left(180 - a\right)\left(200 - a\right)} = \dfrac{1}{256}$
So, on simplifying we get,
$⇒ \dfrac{1}{20}\left[\left(\dfrac{1}{20 - a} - \dfrac{1}{40 - a}\right) + \left(\dfrac{1}{40 - a} - \dfrac{1}{60 - a}\right)\right . . . . .$
\left+ \ldots + \left(\dfrac{1}{180 - a} - \dfrac{1}{200 - a}\right)\right] = \dfrac{1}{256} $$ ⇒ \dfrac{1}{20}\left(\dfrac{1}{20 - a} - \dfrac{1}{200 - a}\right) = \dfrac{1}{256} $$ ⇒ \left(20 - a\right)\left(200 - a\right) = 256 \times 9 $$ ⇒ {a}^{2} - 220a + 1696 = 0 $$ ⇒ a = 8, 212 $So, maximum value of$ a $is$ 212.

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