Question

If $6,6^2,6^3,6^4\cdots 6^n$ be n observation, then the quantity ${\left(\sqrt{6}\right)}^{n+1}$ is called

• A. G.M.
• B. H.M
• C. A.M
• D. None of these

Answer 1

The correct option is A G.M.

Clearly G.M of $6,6^2,6^3,6^4,....{.6}^n$ is $={\left(6^{1+2+3+4+.....+n}\right)}^{\frac{1}{n}}={\left(6^{\frac{n(n+1)}{2}}\right)}^{\frac{1}{n}}=(\sqrt{6}{)}^{n+1}$