Question

If $\frac{1}{(20-a)(40-a)} + \frac{1}{(40-a)(60-a)} + \ldots + \frac{1}{(180-a)(200-a)} = \frac{1}{256}$, then the maximum value of a is :

• A. 198
• B. 202
• C. 212
• D. 218

Answer 1

Answer: (C)

212

Answer 2

$\frac{1}{20}\left(\frac{1}{20-a} - \frac{1}{40-a} + \frac{1}{40-a} - \frac{1}{60-a} + \ldots + \frac{1}{180-a} - \frac{1}{200-a}\right) = \frac{1}{256}$

$⇒$ $\frac{1}{20}\left(\frac{1}{20-a} - \frac{1}{200-a}\right) = \frac{1}{256}$

$⇒$ $\frac{1}{20} \cdot \frac{180}{(20-a)(200-a)} = \frac{1}{256}$

$⇒$ $(20 - a)(200 - a) = 9.256$

$⇒$_ _ $a^2 - 220a + 1696 = 0$

$⇒$ $a = 212, 8$
Then the maxixum value of $a = 212$
So, the correct option is (C): 212