JIPMATAlgebra > Conceptual1mn\frac{1}{mn}mn1mn\frac{m}{n}nmnm\frac{n}{m}mn1✅ Correct Option: 4Related questions:Given 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.The sum of n−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 \ldots1×21+2×31+3×41……. IsAssertion [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.