JEE Mains · Maths · STD 11 - 8. sequence and series
If the sum of series \(\frac{1}{1 \cdot(1+d)}+\frac{1}{(1+d)(1+2 d)}+\ldots \ldots+\frac{1}{(1+9 d)(1+10 d)}\) is equal to \(5\) , then \(50 \mathrm{~d}\) is equal to :
- A \(20\)
- B \(5\)
- C \(15\)
- D \(10\)
Answer & Solution
Correct Answer
(B) \(5\)
Step-by-step Solution
Detailed explanation
\( \frac{1}{1 \cdot(1+d)}+\frac{1}{(1+d)(1+2 d)}+\ldots \ldots \) \( \frac{1}{(1+9 d)(1+10 d)}=5\) \( \frac{1}{d}\left[\frac{(1+d)-1}{1 \cdot(1+d)}+\frac{(1+2 d)-(1-d)}{(1+d)(1+2 d)}\right]+\ldots . . . . \) \( \frac{(1+10 d)-(1+9 d)}{(1+9 d)(1+10 d)}=5 \)…
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