Let an be the nth term of the G.P. of positive numbers. Let ∑$$n=1100$$ $$a2n=$$α and ∑$$n=1100$$ $$a2n$$−$$1=$$β, such that α≠β, then the common ratio is

Question

Let an be the nth term of the G.P. of positive numbers.  Let ∑$$n=1100$$ $$a2n=$$α and ∑$$n=1100$$ $$a2n$$−$$1=$$β, such that α≠β, then the common ratio is

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Let a be the first term and r, the common ratio of the  given G.P. Thenα−∑$$n=1100$$ $$a2n$$⇒α−$$a2+a4+$$…$$+a200$$⇒ α$$=ar+ar3+$$…$$+ar199$$⇒ α$$=ar1+r2+r4+$$…$$+r198$$        (1) and  β$$=$$∑$$n=1100$$ $$a2n$$−$$1$$⇒β$$=a1+a3+$$…$$+a199$$⇒ β$$=a+ar2+$$…$$+ar198$$              (2)⇒ β$$=a1+r2+$$…$$+r198From$$ Eqs (1) and (2), we get   αβ$$=r$$.

Let a_n be the nth term of the G.P. of positive numbers. $Let \sum_{n = 1}^{100} a_{2 n} = α and \sum_{n = 1}^{100} a_{2 n - 1} = β, such that α \neq β, then$ the common ratio is

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Let an be the nth term of the G.P. of positive numbers. Let ∑n=1100 a2n=α and ∑n=1100 a2n−1=β, such that α≠β, then the common ratio is

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Let a_n be the nth term of the G.P. of positive numbers. $Let \sum_{n = 1}^{100} a_{2 n} = α and \sum_{n = 1}^{100} a_{2 n - 1} = β, such that α \neq β, then$ the common ratio is