If 1701 is the n th term of the Geometric Progression (G.P.) 7, 21, 63......, find :

Given,

G.P. : 7, 21, 63,.......

First term (a) = 7

Common ratio (r) = $\frac{21}{7} = 3$

(a) By formula,

⇒ T_n = a × r^(n-1)

⇒ 1701 = 7 × 3^(n-1)

$\Rightarrow 3^{n-1} = \frac{1701}{7}$

⇒ 3^(n-1) = 243

⇒ 3^(n-1) = 3⁵

⇒ n - 1 = 5

⇒ n = 6.

Hence, n = 6.

(b) By formula,

$S_{n} = a \times \Big(\frac{r^{,n}-1}{r-1}\Big)$

Substituting values we get :

$$\begin{aligned}\Rightarrow S_{6} = 7 \times \Big(\frac{3^{6}-1}{3-1}\Big) \\[1em] = 7 \times \Big(\frac{729-1}{2}\Big) \\[1em] = 7 \times \frac{728}{2} \\[1em] = 7 \times 364 \\[1em] = 2548.\end{aligned}$$

Hence, the sum of the n terms (here n = 6) of the G.P. is 2548.