JEE Mains 29-July-2022 Shift 1 Solved Paper

Section: Mathematics

© examsiri.com

Question : 4 of 90

Marks: +1, -0

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 :

[29-Jul-2022-Shift-1]

198
202
212
218

Go to Question:

Prev Question

Next Question