JEE Math Practice 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 $$\mathrm{a}$$ is :
- A.198
- B.202
- C.212
- D.218
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