Question

Find :

the 10^th term of the G.P.

\[\sqrt{2}, \frac{1}{\sqrt{2}}, \frac{1}{2\sqrt{2}}, . . .\]

Answer 1

Here,

\[\text { First term }, a = \sqrt{2}\]

\[\text { Common ratio, } r = \frac{a_2}{a_1} = \frac{\frac{1}{\sqrt{2}}}{\sqrt{2}} = \frac{1}{2}\]

\[ \therefore 10th\text { term  }= a_{10} = a r^{(10 - 1)} = \sqrt{2} \left( \frac{1}{2} \right)^9 = \frac{1}{\sqrt{2}} \times \frac{1}{2^8}\]

\[\text { Thus, the 10th term of the given GP is } \frac{1}{\sqrt{2}} \times \frac{1}{2^8} .\]