2 + 4 + 6 + 8 + ...= 10100. Find the number of terms' Share with your friends Share 3 Varun.Rawat answered this The given sequence is,2, 4, 6, 8, ........The above sequence is in AP with first term, a = 2 and common difference, d = 2Now, let n be the number of terms of the above AP whose sum is 10100.Now, sum of first n terms of an AP is given bySn = n22a + n-1d⇒10100 = n22×2+n-1×2⇒20200 = n4+2n-2⇒20200 = n2n+2⇒nn+1 = 10100⇒n2 + n - 10100 = 0⇒n2 + 101n - 100n - 10100 = 0⇒nn + 101 - 100n + 101 = 0⇒n - 100n + 101 = 0⇒n = 100 or n = -101 rejectedSo, n = 100So, number of terms whose sum is 10100 is 100 -2 View Full Answer Jayanth answered this a = 2, d = 2 sum of n terms = n/2 [2a +(n-1)d] 10100 = n/2 [4 +(n-1)2] 10100 = 2n + n(n-1) n2 - n + 2n -10100 = 0 n2 +n -10100 = 0 n2 + 101n -100n -10100 = 0 n(n+101) -100(n+101)=0 (n-100)(n+101)=0 n = 100 or n = -101 (negative number of terms is not possible ) n = 100 hope it helped 1
