If the sum of the series $$\frac{1}{1 \cdot(1+\mathrm{d})}+\frac{1}{(1+\mathrm{d})(1+2 \mathrm{~d})}+\ldots+\frac{1}{(1+9 \mathrm{~d})(1+10 \mathrm{~d})}$$ is equal to 5, then $50 \mathrm{~d}$ is equal to :
Solution
<p>$\frac{1}{1 \cdot(1+d)}+\frac{1}{(1+d)(1+2 d)}+\ldots+\frac{1}{(1+9 d)(1+10 d)}=5$</p>
<p>Multiply and divide by $d$</p>
<p>$$\begin{aligned}
& \frac{1}{d}\left[\frac{d}{1 \times(1+d)}+\frac{d}{(1+d)(1+2 d)}+\ldots+\frac{1}{(1+9 d)(1+10 d)}\right]=5 \\
& \frac{1}{d}\left[\left(1-\frac{1}{1+d}\right)+\left(\frac{1}{1+d}-\frac{1}{1+2 d}\right)+\ldots+\left(\frac{1}{1+9 d}-\frac{1}{1+10 d}\right)\right]=5 \\
& \frac{1}{d}\left[1-\frac{1}{1+10 d}\right]=5 \\
& \frac{1}{d}\left[\frac{1+10 d-1}{1+10 d}\right]=5
\end{aligned}$$</p>
<p>$$\begin{aligned}
& \frac{10}{1+10 d}=5 \\
& 1+10 d=2 \\
& d=\frac{1}{10} \\
& 50 d=50 \times \frac{1}{10}=5
\end{aligned}$$</p>
About this question
Subject: Mathematics · Chapter: Sequences and Series · Topic: Arithmetic Progression
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