1.

100100Let an be the nth term of a G.P. of positive numbers. Let azn = a & a2n-1 = ßsuch that a # ß. Then the common ratio of the G.P.is -n=1​

Answer»

Step-by-step EXPLANATION:

\sum_{n = 1}^{100} a _{2n} = \alpha \: \: and \: \: \sum_{n = 1}^{100} a _{2n - 1} = \beta \\

Now,

a _{2} + a _{4} + a _{6} + .... +a _{200} = \alpha \: \: and \: \: a _{1} + a _{3} + a _{5} + ... +a _{199} = \beta \\

\frac{a_{2} +a_{4} + a_{6} + ... +a_{200}}{a_{1} +a_{3} +a_{5} + ... +a_{199}} = \frac{ \alpha }{ \beta } \\

\implies \frac{ a_{2}( \frac{(2r) ^{100} - 1 }{r - 1} )}{ a_{1} ( \frac{(2r)^{100} - 1 }{r - 1}) } = \frac{ \alpha }{ \beta }

\implies \frac{a_{2}}{a_{1}} = \frac{ \alpha }{ \beta } \\


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