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QJEEM24Ma67f4: JEE Main 2024 Mathematics - Class 11 JEEM24Ma67f4 Solution: If the sum of the series is equal to 5, then is equal to :

If the sum of the series 11⋅(1+d)+1(1+d)(1+2 d)+…+1(1+9 d)(1+10 d)\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 d50 \mathrm{~d} is equal to :

Multiple Choice Options

  • A 5
  • B 10
  • C 15
  • D 20

Step-by-Step Verified Solution

11⋅(1+d)+1(1+d)(1+2d)+…+1(1+9d)(1+10d)=5\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 dd

amp;1d[d1×(1+d)+d(1+d)(1+2d)+…+1(1+9d)(1+10d)]=5amp;1d[(1−11+d)+(11+d−11+2d)+…+(11+9d−11+10d)]=5amp;1d[1−11+10d]=5amp;1d[1+10d−11+10d]=5\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}

amp;101+10d=5amp;1+10d=2amp;d=110amp;50d=50×110=5\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}

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