Which term of the G.P. :√2, \(\frac{1}{\sqrt{2}},\)\(\frac{1}{2\sqrt

1.	Which term of the G.P. :√2, \(\frac{1}{\sqrt{2}},\)\(\frac{1}{2\sqrt{2}}\),\(\frac{1}{4\sqrt{2}}\) .... is \(\frac{1}{512\sqrt{2}}\)?
Answer» T_n = ar^n-1 a = √2, r = \(\frac{\frac{1}{\sqrt{2}}}{\sqrt{2}}\) = \(\frac{1}{2}\), T_n = \(\frac{1}{512\sqrt2}n\) = ? \(\therefore\)\(\frac{1}{512\sqrt2}\) = \(\sqrt{2} \times (\frac{1}{2})^{n-1}\) \(\Rightarrow\) \(\frac{1}{1024}\) = \((\frac{1}{2})^n\) \(\times 2\) \(\Rightarrow\) \(\frac{1}{2048}\) = \((\frac{1}{2})^n\) \(\Rightarrow\) 2ⁿ = 2048 \(\Rightarrow\) n = 11

Which term of the G.P. :√2, \(\frac{1}{\sqrt{2}},\)\(\frac{1}{2\sqrt{2}}\),\(\frac{1}{4\sqrt{2}}\) .... is \(\frac{1}{512\sqrt{2}}\)?

Answer»

T_n = ar^n-1

a = √2, r = \(\frac{\frac{1}{\sqrt{2}}}{\sqrt{2}}\) = \(\frac{1}{2}\), T_n = \(\frac{1}{512\sqrt2}n\) = ?

\(\therefore\)\(\frac{1}{512\sqrt2}\) = \(\sqrt{2} \times (\frac{1}{2})^{n-1}\)

\(\Rightarrow\) \(\frac{1}{1024}\) = \((\frac{1}{2})^n\) \(\times 2\)

\(\Rightarrow\) \(\frac{1}{2048}\) = \((\frac{1}{2})^n\)

\(\Rightarrow\) 2ⁿ = 2048

\(\Rightarrow\) n = 11