Question
Mathematics Question on geometric progression
If an be the nth term of the GP of positive numbers. Let ∑n=100100a2n=α and ∑n=1100a2n−1=β such that α≠β then the common ratio is:
A
(A) αβ
B
(B) βα
C
(C) αβ
D
(D) βα
Answer
(A) αβ
Explanation
Solution
Explanation:
Given:A geometric progression such that nth term is an and∑n=1100a2n=α,∑n=1100a2n−1=βWe have to find the common ratio of given GP.Let a be the first term, r be the common ratio of the given GP (r>0).Then a,ar,ar2,ar3,….,ar199 are in GP.We shall use the formula of general term of GP.α=∑n=1100a2n⇒a=a2+a4+…….+a200⇒a=ar+ar3+……..+ar199=ar(1+r2+……+r198)and β=∑n=1100a2n−1⇒β=a1+a3+………+a199⇒β=a+ar2+……..+ar198⇒β=a(1+r2+…….+r198)Clearly, a/β=r= common ratioHence, the correct option is (A).