JIPMATAlgebra > Easy1n+1\frac{1}{n+1}n+111n\frac{1}{n}n1n+1n\frac{n+1}{n}nn+1nn+1\frac{n}{n+1}n+1n✅ Correct Option: 4Related questions:JIPMAT 2021If the mthm^{\text{th}}mth term of an arithmetic progression is 1n\frac{1}{n}n1 and the nthn^{\text{th}}nth term is 1m\frac{1}{m}m1, then the mnthmn^{\text{th}}mnth term of this progression will beJIPMAT 2023Assertion [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)S=(n/2)∗(a+l) where a=first term, l=last term. Choose the correct answer from the options given below.JIPMAT 2024Given below are two statements, one is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion (A) : The sum of nnn terms of the Progression 1+12+122+123+1+\frac{1}{2}+\frac{1}{2^{2}}+\frac{1}{2^{3}}+1+21+221+231+ is 2n−1−12n−1\frac{2^{n-1}-1}{2^{n-1}}2n−12n−1−1. Reason (R) : Sum of a geometric series having nnn terms is given by Sn=a(1−rn)1−rS_{n}=\frac{a\left(1-r^{n}\right)}{1-r}Sn=1−ra(1−rn), where aaa is the 1st 1^{\text {st }}1st term and rrr is the common ratio.