JEE Mains · Maths · STD 11 - 8. sequence and series
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.
- A \(198\)
- B \(202\)
- C \(212\)
- D \(218\)
Answer & Solution
Correct Answer
(C) \(212\)
Step-by-step Solution
Detailed explanation
By splitting\(\frac{1}{20}\left[\left(\frac{1}{20-a}-\frac{1}{40-a}\right)+\left(\frac{1}{40-a}-\frac{1}{60-a}\right)\right.\) \(\left.+\ldots+\left(\frac{1}{180-a}-\frac{1}{200-a}\right)\right]\) \(\frac{1}{20}\left(\frac{1}{20-a}-\frac{1}{200-a}\right)=\frac{1}{256}\)…
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