JIPMAT 2022Algebra > 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 2022If 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 2022Given 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.JIPMAT 2022The 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}})(9−81−8−71+7−61−6−51+5−41) is