JIPMAT - Assertion [A]: Sum of the first hundred even natural numbers divisible by 5 is 45050. <br>Reason (R): Sum of the first n-terms of an Arithmetic Progression is given by S = (n/2) *(a + l) where a=first term, l=last term. <br>Choose the correct answer from the options given below. | PYQs + Solutions

JIPMAT

Algebra

Conceptual

Both [A] and (R) are true and (R) is the correct explanation of (A).

Both [A] and (R) are true but (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

Related questions:

JIPMAT 2022

The sum of $n-$ terms of sequence $\frac{1}{1 \times 2}+\frac{1}{2 \times 3}+\frac{1}{3 \times 4} \ldots \ldots$ . Is

JIPMAT 2024

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.

JIPMAT 2021

The value of $(\frac{1}{\sqrt{9}-\sqrt{8}} - \frac{1}{\sqrt{8}-\sqrt{7}} + \frac{1}{\sqrt{7}-\sqrt{6}} - \frac{1}{\sqrt{6}-\sqrt{5}} + \frac{1}{\sqrt{5}-\sqrt{4}})$ is

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