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Find:The 10th term of the G.P. √2, \(\frac{1}{\sqrt{2}}, \frac{1}{2\sqrt{2}}\),........

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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