MathematicsMedium178×since 2002Q5257Let a_n be the n^th term of a G.P. of positive terms. ∑n=1100a2n+1=200\sum\limits_{n = 1}^{100} {{a_{2n + 1}} = 200}n=1∑100a2n+1=200 and ∑n=1100a2n=100\sum\limits_{n = 1}^{100} {{a_{2n}} = 100}n=1∑100a2n=100, then ∑n=1200an\sum\limits_{n = 1}^{200} {{a_n}}n=1∑200an is equal to :A150B175C225D300Check answerSkip