Question

Find 9th term of G.P √2, 1/√2 ,1/2√2​

Answer 1

Step-by-step explanation:

nth term of GP is given by ar$^{n-1}$, where a is the first term and r is the common ratio.

Here,

a = √2 ; r = (1/√2)/√2 = 1/(√2√2)

a = √2 ; r = 1/2

Therefore,

= > 9th term is:

= > √2.$(\small{\frac{1}{2}})^{9-1}$

= > √2.$\small{\frac{1}{2^8}}$

= > √2.$\small{\frac{1}{\sqrt2^{16}}}$

= > √2$^{1-16}$

= > √2$^{-15}$