If the sum of the series 1 1 · ( 1 + d ) + 1 ( 1 + d ) ( 1 + 2 d ) + ? + 1 ( 1 + 9 d ) ( 1 + 10 d ) is equal to 5, then 50 d is equal to : [2024]

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Solution

Q.
If the sum of the series $\frac{1}{1 \cdot (1 + d)} + \frac{1}{(1 + d) (1 + 2 d)} + \dots + \frac{1}{(1 + 9 d) (1 + 10 d)}$ is equal to 5, then $50 d$ is equal to : [2024]

1 5

2 15

3 10

4 20

Ans.
(1)

We have,

$\frac{1}{1 \cdot (1 + d)} + \frac{1}{(1 + d) (1 + 2 d)} + \dots + \frac{1}{(1 + 9 d) (1 + 10 d)} = 5$

$\Rightarrow \sum_{r = 1}^{10} [\frac{1}{{1 + (r - 1) d} {1 + r d}}] = 5$

$\Rightarrow \sum_{r = 1}^{10} \frac{1}{d} [\frac{1}{1 + (r - 1) d} - \frac{1}{1 + r d}] = 5$

$\Rightarrow \frac{1}{d} \sum_{r = 1}^{10} [\frac{1}{1 + (r - 1) d} - \frac{1}{1 + r d}] = 5$

$\Rightarrow [\frac{1}{1} - \frac{1}{1 + d}] + [\frac{1}{1 + d} - \frac{1}{1 + 2 d}] + \dots + [\frac{1}{1 + 9 d} - \frac{1}{1 + 10 d}] = 5 d$

$\Rightarrow 1 - \frac{1}{1 + 10 d} = 5 d \Rightarrow \frac{10 d}{1 + 10 d} = 5 d$

$\Rightarrow 5 + 50 d = 10 \Rightarrow 50 d = 5$

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