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Mathematics

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

(a) the value of 'n'

(b) hence find the sum of the 'n' terms of the G.P.

G.P.

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Answer

Given,

G.P. : 7, 21, 63,…….

First term (a) = 7

Common ratio (r) = 217=3\dfrac{21}{7} = 3

(a) By formula,

⇒ Tn = a × r(n-1)

⇒ 1701 = 7 × 3(n-1)

⇒3n−1=17017\Rightarrow 3^{n-1} = \dfrac{1701}{7}

⇒ 3(n-1) = 243

⇒ 3(n-1) = 35

⇒ n - 1 = 5

⇒ n = 6.

Hence, n = 6.

(b) By formula,

Sn=a×(r n−1r−1)S_{n} = a \times \Big(\dfrac{r^{\,n}-1}{r-1}\Big)

Substituting values we get :

⇒S6=7×(36−13−1)=7×(729−12)=7×7282=7×364=2548.\Rightarrow S_{6} = 7 \times \Big(\dfrac{3^{6}-1}{3-1}\Big) \\[1em] = 7 \times \Big(\dfrac{729-1}{2}\Big) \\[1em] = 7 \times \dfrac{728}{2} \\[1em] = 7 \times 364 \\[1em] = 2548.

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

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