JEE Main 2022 · 29 Jul · Shift 1 · Mathematics Q4

MathematicsSingle correct+4 / -1Question 4 of 90

If $\dfrac1{(20-a)(40-a)}+\dfrac1{(40-a)(60-a)}+\dots+\dfrac1{(180-a)(200-a)}=\dfrac1{256}$ , then the maximum value of a is :

198

202

212

218

Answer and worked solution

Official answer: C

Telescope. Each term is $\frac1{20}\left(\frac1{20k-a}-\frac1{20k+20-a}\right)$ , so the sum is $\dfrac{9}{(20-a)(200-a)}=\dfrac1{256}$ .

Solve $a^2-220a+1696=0$ : $a=8$ or 212. Maximum 212.

(C)

Topics:Special series sums Roots and coefficients

JEE Main 2022 · 29 Jul · Shift 1, Mathematics question 4