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 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 2021The 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) isJIPMAT 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.