If the sum of the series [11 (1+d)+1(1+d)(1+2 ~d)++1(1+9 ~d)(1+10 ~d)] is equal to 5, then…

Question

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 :

Accepted Answer

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

Multiply and divide by $d$

$$\begin{aligned}
& \frac{1}{d}\left[\frac{d}{1 \times(1+d)}+\frac{d}{(1+d)(1+2 d)}+\ldots+\frac{1}{(1+9 d)(1+10 d)}\right]=5 \
& \frac{1}{d}\left[\left(1-\frac{1}{1+d}\right)+\left(\frac{1}{1+d}-\frac{1}{1+2 d}\right)+\ldots+\left(\frac{1}{1+9 d}-\frac{1}{1+10 d}\right)\right]=5 \
& \frac{1}{d}\left[1-\frac{1}{1+10 d}\right]=5 \
& \frac{1}{d}\left[\frac{1+10 d-1}{1+10 d}\right]=5
\end{aligned}$$

$$\begin{aligned}
& \frac{10}{1+10 d}=5 \
& 1+10 d=2 \
& d=\frac{1}{10} \
& 50 d=50 \times \frac{1}{10}=5
\end{aligned}$$

Solutions · 1