1.Let ana_nan be the nthn^{\text{th}}nth term of a G.P. of positive terms. If ∑n=1100a2n−1=200\sum\limits_{n=1}^{100} a_{2n-1} = 200n=1∑100a2n−1=200 and ∑n=1100a2n=100\sum\limits_{n=1}^{100} a_{2n} = 100n=1∑100a2n=100, then ∑n=1200an\sum\limits_{n=1}^{200} a_nn=1∑200an is equal toa.300300300b.225225225c.175175175d.150150150Login to continueOnly logged in users canattempt or see the solution.
