Q1:

JIPMAT 2024

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 nn terms of the Progression

1+12+122+123+1+\frac{1}{2}+\frac{1}{2^{2}}+\frac{1}{2^{3}}+ is 2n−1−12n−1\frac{2^{n-1}-1}{2^{n-1}}.

Reason (R) :

Sum of a geometric series having nn terms is given by Sn=a(1−rn)1−rS_{n}=\frac{a\left(1-r^{n}\right)}{1-r}, where aa is the 1st 1^{\text {st }} term and rr is the common ratio.

Answer options
Option 4
Correct Answer
Explanation for JIPMAT 2024 QA question 1

Q2:

JIPMAT 2023

Progression & Series

Easy

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

Answer options
Option 4
Correct Answer
Explanation for JIPMAT 2023 QA question 2

Q3:

JIPMAT 2022

Progression & Series

Medium

The sum of n−n- terms of sequence 11×2+12×3+13×4……\frac{1}{1 \times 2}+\frac{1}{2 \times 3}+\frac{1}{3 \times 4} \ldots \ldots. Is

Answer options
Option 4
Correct Answer
Explanation for JIPMAT 2022 QA question 3

Q4:

JIPMAT 2021

Progression & Series

Medium

The value of (19−8−18−7+17−6−16−5+15−4)(\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

Answer options
Option 4
Correct Answer
Explanation for JIPMAT 2021 QA question 4

Q5:

JIPMAT 2021

Progression & Series

Medium

If the mthm^{\text{th}} term of an arithmetic progression is 1n\frac{1}{n} and the nthn^{\text{th}} term is 1m\frac{1}{m}, then the mnthmn^{\text{th}} term of this progression will be

Answer options
Option 4
Correct Answer
Explanation for JIPMAT 2021 QA question 5