IPMAT Rohtak 2019Algebra > Easy442219None of the above✅ Correct Option: 1Related 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.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)S=(n/2)∗(a+l) where a=first term, l=last term. Choose the correct answer from the options given below.What is the sum of the infinite series: 1−12−14+18−116−132+164−1128−1256+…?1 - \frac{1}{2} - \frac{1}{4} + \frac{1}{8} - \frac{1}{16} - \frac{1}{32} + \frac{1}{64} - \frac{1}{128} - \frac{1}{256} + \dots?1−21−41+81−161−321+641−1281−2561+…?