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Find:The 10th term of the G.P. √2, \(\frac{1}{\sqrt{2}}, \frac{1}{2\sqrt{2}}\),........ |
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Answer» Tn= arn-1 a = √2, r = \(\frac{\frac{1}{\sqrt{2}}}{\sqrt{2}}\) = \(\frac{1}{2}\) \(\therefore\) T10 = √2\((\frac{1}{2})^{10-1}\) = \(\frac{\sqrt{2}}{512}\) = \(\frac{1}{256\sqrt{2}}\) \(\therefore\) The 10 terms is \(\frac{1}{256\sqrt{2}}\) |
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