jee-main 2022 Q62

jee-main · India · session2_29jul_shift1 Arithmetic Sequences and Series Telescoping or Non-Standard Summation Involving an AP

If $\frac { 1 } { ( 20 - a ) ( 40 - a ) } + \frac { 1 } { ( 40 - a ) ( 60 - a ) } + \ldots\ldots + \frac { 1 } { ( 180 - a ) ( 200 - a ) } = \frac { 1 } { 256 }$, then the maximum value of $a$ is
(1) 198
(2) 202
(3) 212
(4) 218

If $\frac { 1 } { ( 20 - a ) ( 40 - a ) } + \frac { 1 } { ( 40 - a ) ( 60 - a ) } + \ldots\ldots + \frac { 1 } { ( 180 - a ) ( 200 - a ) } = \frac { 1 } { 256 }$, then the maximum value of $a$ is\\
(1) 198\\
(2) 202\\
(3) 212\\
(4) 218

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