If the sum of the series \frac11 \cdot(1+d)+\frac1(1+d)(1+2 ~d)+\ldots+\frac1(1+9 ~d)(1+10 ~d) is equal to 5, then 50 ~d is equal to :

Sequences and Series PYQ — JEE Main 2024 · 9 Apr · Shift 1 · Q37 with Solution

View written solutionFree

Correct answer: A

Write the given sum

The series is

S=\frac{1}{1(1+d)}+\frac{1}{(1+d)(1+2d)}+\cdots+\frac{1}{(1+9d)(1+10d)}

So the general term is

\frac{1}{(1+kd)(1+(k+1)d)}, \qquad k=0,1,2,\dots,9

Hence,

S=\sum_{k=0}^{9}\frac{1}{(1+kd)(1+(k+1)d)}

Use partial fraction / telescoping form

Observe that

\frac{1}{(1+kd)(1+(k+1)d)} =\frac{1}{d}\left(\frac{1}{1+kd}-\frac{1}{1+(k+1)d}\right)

because

\frac{1}{d}\left(\frac{1}{1+kd}-\frac{1}{1+(k+1)d}\right) =\frac{1}{d}\cdot \frac{(1+(k+1)d)-(1+kd)}{(1+kd)(1+(k+1)d)} =\frac{1}{(1+kd)(1+(k+1)d)}

Substitute into the sum

Therefore,

S=\frac{1}{d}\sum_{k=0}^{9}\left(\frac{1}{1+kd}-\frac{1}{1+(k+1)d}\right)

This is a telescoping series:

S=\frac{1}{d}\left(1-\frac{1}{1+10d}\right)

Simplify

S=\frac{1}{d}\cdot \frac{(1+10d)-1}{1+10d} =\frac{1}{d}\cdot \frac{10d}{1+10d} =\frac{10}{1+10d}

Use the given condition $S=5$

\frac{10}{1+10d}=5

So,

10=5(1+10d)

10=5+50d

50d=5

Match with the options

Thus,

50d=5

So the correct option is A.

Comparison with stored answer

Stored correct answer: A

Our derived answer: A

They agree.

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View written solutionFree

Correct answer: A

Write the given sum

The series is

S=\frac{1}{1(1+d)}+\frac{1}{(1+d)(1+2d)}+\cdots+\frac{1}{(1+9d)(1+10d)}

So the general term is

\frac{1}{(1+kd)(1+(k+1)d)}, \qquad k=0,1,2,\dots,9

Hence,

S=\sum_{k=0}^{9}\frac{1}{(1+kd)(1+(k+1)d)}

Use partial fraction / telescoping form

Observe that

\frac{1}{(1+kd)(1+(k+1)d)} =\frac{1}{d}\left(\frac{1}{1+kd}-\frac{1}{1+(k+1)d}\right)

because

\frac{1}{d}\left(\frac{1}{1+kd}-\frac{1}{1+(k+1)d}\right) =\frac{1}{d}\cdot \frac{(1+(k+1)d)-(1+kd)}{(1+kd)(1+(k+1)d)} =\frac{1}{(1+kd)(1+(k+1)d)}

Substitute into the sum

Therefore,

S=\frac{1}{d}\sum_{k=0}^{9}\left(\frac{1}{1+kd}-\frac{1}{1+(k+1)d}\right)

This is a telescoping series:

S=\frac{1}{d}\left(1-\frac{1}{1+10d}\right)

Simplify

S=\frac{1}{d}\cdot \frac{(1+10d)-1}{1+10d} =\frac{1}{d}\cdot \frac{10d}{1+10d} =\frac{10}{1+10d}

Use the given condition $S=5$

\frac{10}{1+10d}=5

So,

10=5(1+10d)

10=5+50d

50d=5

Match with the options

Thus,

50d=5

So the correct option is A.

Comparison with stored answer

Stored correct answer: A

Our derived answer: A

They agree.

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