JEE Main 20209 Jan, Evening shiftMathematicsMediumSequence and Series179× since 2002Let 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 :AOption A. 150BOption B. 175COption C. 225DOption D. 300CheckSkip · show solution
