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A geometric sequence has all positive terms. The sum of the first two terms is 15 and the sum to infinity is 27. Find the value of a)the common ratio |
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Answer» \(\because\) a + ar = 15 ⇒ a(1 + r) = 15 ⇒ a \(=\frac{15}{1+r}\) And sum of infinite term = 27 (Given) \(\therefore \frac{a}{1-r}=27\) \(\Rightarrow \frac{15}{(1+r)(1-r)}=27\) \(\Rightarrow 1-r^2 = \frac{15}{27}=\frac{5}{9}\) \(\Rightarrow r^2=1-\frac{5}{9}=\frac{4}{9}\) \(\Rightarrow r=\pm\frac{2}{3}\) \(\Rightarrow \) But r \(\ne \frac{-2}{3}\) \((\because \) all terms of g.p are positive) \(\therefore\) common ratio \(=r=\frac{2}{3}\) |
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