Write the first five terms of the sequence given by $(\sqrt{3})^n$, n ∈ N.

(a) Is the sequence an A.P. or G.P.?

(b) If the sum of its first ten terms is $p(3 + \sqrt{3})$, find the value of p.

Terms of the sequence given by $(\sqrt{3})^n$ are :

⇒ $(\sqrt{3})^1, (\sqrt{3})^2, (\sqrt{3})^3, (\sqrt{3})^4, (\sqrt{3})^5$, …….

⇒ $\sqrt{3}, 3, 3\sqrt{3}, 9, 9\sqrt{3}$

(a) Ratio between terms = $\dfrac{3}{\sqrt{3}} = \sqrt{3}$.

Hence, the sequence is a G.P. with common ratio = $\sqrt{3}$.

(b) By formula,

Sum of G.P. (S) = $\dfrac{a(r^n - 1)}{(r - 1)}$

Substituting values we get :

$\Rightarrow S_{10} = \dfrac{\sqrt{3}[(\sqrt{3})^{10} - 1]}{\sqrt{3} - 1} \\[1em] \Rightarrow p(3 + \sqrt{3})= \dfrac{\sqrt{3}(243 - 1)}{\sqrt{3} - 1} \\[1em] \Rightarrow p = \dfrac{\sqrt{3}(243 - 1)}{(\sqrt{3} - 1)(3 + \sqrt{3})} \\[1em] \Rightarrow p = \dfrac{\sqrt{3} \times 242}{(\sqrt{3} - 1)\sqrt{3}(\sqrt{3} + 1)} \\[1em] \Rightarrow p = \dfrac{242}{3 - 1} \\[1em] \Rightarrow p = \dfrac{242}{2} = 121.$

Hence, p = 121.