Let a n be the n t h term of a G.P. of positive terms. If ∑ n = 1 100 a 2 n + 1 = 200 and ∑ n = 1 100 a 2 n = 100 , then ∑ n = 1 200 a n is equal to:

Step-by-step solution

Let GP is a , a r , a r 2 … ∑ n = 1 100 a 2 n + 1 = a 3 + a 5 + … a 201 = 200 ⇒ a r 2 ( r 200 - 1 ) r 2 - 1 = 200       … 1 ∑ n = 1 100 a 2 n = a 2 + a 4 + … a 200 = 100 = a r r 200 - 1 r 2 - 1 = 100           … 2 Form 1 and 2 , r = 2 adding both, ⇒ a 2 + a 3 + … a 200 + a 201 = 300 ⇒ r a 1 + … a 200 = 300 ∑ n = 1 200 300 r = 150