Question 19.ShareReportConsider the arithmetic progression 3,7,11,… and let An denote the sum of the first n terms of this progression. Then the value of 125∑n=125An\frac{1}{25} \sum_{n=1}^{25} A_n251∑n=125An isA404B442C455D415Backspace7894561230.-Clear AllSubmit Previous QuestionRate this SolutionNext Question Question ExplanationText ExplanationSum of n terms in an A.P = n2(2a+(n−1)d)\frac{n}{2}(2a + (n - 1)d)2n(2a+(n−1)d)An=n2(6+(n−1)4)=n(2n+1)A_n = \frac{n}{2}(6 + (n - 1)4) = n(2n + 1)An=2n(6+(n−1)4)=n(2n+1)∑An=∑n(2n+1)=2∑n2+∑n=2n(n+1)(2n+1)6+n(n+1)2\sum A_n = \sum n(2n + 1) = 2\sum n^2 + \sum n = \frac{2n(n+1)(2n+1)}{6} + \frac{n(n+1)}{2}∑An=∑n(2n+1)=2∑n2+∑n=62n(n+1)(2n+1)+2n(n+1)Substituting n=25n = 25n=25, we get125∑n=125An=125(2(25)(25+1)(50+1)6+25(25+1)2)\frac{1}{25}\sum_{n=1}^{25} A_n = \frac{1}{25}\left(\frac{2(25)(25+1)(50+1)}{6} + \frac{25(25+1)}{2}\right)251∑n=125An=251(62(25)(25+1)(50+1)+225(25+1))125∑n=125An=26(17)+13=455\frac{1}{25}\sum_{n=1}^{25} A_n = 26(17) + 13 = 455251∑n=125An=26(17)+13=455Video Explanation
