JIPMAT 2024 (QA) - Given below are two statements, one is labelled as Assertion (A) and the other is labelled as Reason (R). (A) : sum of n terms of the Progression1+1/2+12^2+12^3+ is 2^n-1-12^n-1. (R) : of a geometric series having n terms is given by S_n=a(1-r^n)1-r, where a is the 1^st term and r is the common ratio. | PYQs + Solutions

JIPMAT 2024

Algebra

Progression & Series

Medium

Given below are two statements, one is labelled as Assertion (A) and the other is labelled as Reason (R).
Assertion (A) :
The sum of $n$ terms of the Progression
$1+\frac{1}{2}+\frac{1}{2^{2}}+\frac{1}{2^{3}}+$ is $\frac{2^{n-1}-1}{2^{n-1}}$ .
Reason (R) :
Sum of a geometric series having $n$ terms is given by $S_{n}=\frac{a\left(1-r^{n}\right)}{1-r}$ , where $a$ is the $1^{\text {st }}$ term and $r$ is the common ratio.

Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of (A).

Both Assertion (A) and Reason (R) are true but Reason (R) is NOT the correct explanation of (A).

(A) is true but (R) is false.

(A) is false but (R) is true.

Correct Option: 4

1) First, let's verify the sum in Assertion (A):

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- We're given:

1 + \frac{1}{2} + \frac{1}{2^2} + \frac{1}{2^3} + ...

up to n terms

- And it's claimed this equals

\frac{2^{n-1}-1}{2^{n-1}}

2) Reason (R) states:

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- For geometric series with n terms,

S_n = \frac{a(1-r^n)}{1-r}

\newline

- Where a is first term and r is common ratio

3) Let's apply R's formula to the series in A:

a = 1

r = \frac{1}{2}

S_n = \frac{1(1-(\frac{1}{2})^n)}{1-\frac{1}{2}}

= \frac{1-\frac{1}{2^n}}{\frac{1}{2}}

= 2(1-\frac{1}{2^n})

= 2-\frac{2}{2^n}

Which is NOT equal to

\frac{2^{n-1}-1}{2^{n-1}}

4) Therefore:

\newline

- A is false

\newline

- R is true (it's a standard formula)