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 :

Correct answer: B. 5 aligned & 1 1 ·(1+d) + 1 (1+d)(1+2 ~d) + . . & 1 (1+9 ~d)(1+10 ~d) =5 aligned aligned & (1)/(d)[((1+d)-1)/(1 ·(1+d))+((1+2 d)-(1-d))/((1+d)(1+2 d))]+ . . . & ((1+10 d)-(1+9 d))/((1+9 d)(1+10 d))=5 & (1)/(d)[(1-(1)/(1+d))+((1)/(1+d)-(1)/(1+2 d))+ . .. & .((1)/(1+9 d)-(1)/(1+10 d))]=5 & (1)/(d)[1-(1)/((1+10 d))]=5 & (10 d)/(1+10 d)=5 d & 50 d=5 aligned

JEE Main 20249 AprilMorning Shift

Exam: JEE Main
Year: 2024
Date: 9 April
Shift: Morning Shift
Subject: Mathematics
Chapter: Sequences and Series

JEE Main 2024 · 9 April · Morning Shift · Mathematics · Sequences and Series

If the sum of the series 1 1 ·(1+d) + 1 (1+d)(1+2 ~d) + + 1 (1+9...

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 :

Options

Correct answer

B. 5

Step-by-step solution

\begin{aligned} & \frac{1}{1 \cdot(1+\mathrm{d})}+\frac{1}{(1+\mathrm{d})(1+2 \mathrm{~d})}+\ldots \ldots \ldots \ldots . . \\ & \frac{1}{(1+9 \mathrm{~d})(1+10 \mathrm{~d})}=5\end{aligned}

\begin{aligned} & \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 \ldots \ldots \ldots . . . \\ & \frac{(1+10 d)-(1+9 d)}{(1+9 d)(1+10 d)}=5 \\ & \frac{1}{d}\left[\left(1-\frac{1}{1+d}\right)+\left(\frac{1}{1+d}-\frac{1}{1+2 d}\right)+\ldots \ldots \ldots \ldots . .\right. \\ & \left.\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{10 d}{1+10 d}=5 d \\ & 50 d=5\end{aligned}

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