In the geometric series 2+4+8+...., starting from the first term how many consecutive terms are needed to yield the sum 1022?
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Solution
Given the geometric series is 2+4+8+... Let n be the number of terms required to get the sum. Here a=2,r=2,Sn=1022. To find n, let us consider Sn=a[rn−1]r−1 if r≠1 =(2)[2n−11]=2(2n−1). But Sn=1022 and hence 2(2n−1)=1022 ⇒2n−1=511 ⇒2n=512=29