Geometric Progression — Analytical & Application Based Questions

Write the first five terms of sequence given by $(\sqrt3)^n$, n ∈ N.

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

(b) If the sum of its first ten terms is p(3 + $\sqrt3$), find the value of p.

Answer

Terms of the sequence given by $(\sqrt3)^n$ are :

$(\sqrt3)^1, (\sqrt3)^2, (\sqrt3)^3, (\sqrt3)^4, (\sqrt3)^5, .......$

$\sqrt3, 9, 3\sqrt3, 9, 9\sqrt3, .......$

(a) Ratio between terms = $\dfrac{3}{\sqrt3} = \sqrt3$

Hence, the sequence is a G.P. with common ratio = $\sqrt3$.

(b) By formula,

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

Substituting values we get :

$\Rightarrow S_{10} = \dfrac{\sqrt3[(\sqrt3)^10 - 1)}{(\sqrt3 - 1)} \\[1em] \Rightarrow p(3 + \sqrt3) = \dfrac{\sqrt3[(\sqrt3)^10 - 1]}{(\sqrt3 - 1)} \\[1em] \Rightarrow p = \dfrac{\sqrt3(243 - 1)}{(\sqrt3 - 1)(3 + \sqrt3)} \\[1em] \Rightarrow p = \dfrac{\sqrt3(242)}{(\sqrt3 - 1)(3 + \sqrt3)} \\[1em] \Rightarrow p = \dfrac{\sqrt3(242)}{(\sqrt3 - 1)\sqrt3(\sqrt3 + 1)} \\[1em] \Rightarrow p = \dfrac{242}{(\sqrt3)^2 - (1)^2} \\[1em] \Rightarrow p = \dfrac{242}{3 - 1} \\[1em] \Rightarrow p = \dfrac{242}{2} = 121$

Hence, p = 121.