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The first term of a G.P. is 27 and its 8th term is $$\frac{1}{81}$$. Find the sum of first seven terms of the G.P.

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Question

The first term of a G.P. is 27 and its 8th term is $$\frac{1}{81}$$. Find the sum of first seven terms of the G.P.

Sum
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Solution

Let the first term be $$a = 27$$ and the common ratio be $$r$$.

Using $$T_n = ar^{n - 1}$$:

$$T_8 = 27 \times r^7$$

$$= \frac{1}{81}$$ 

$$\Rightarrow r^7 = \frac{1}{27 \times 81}$$

$$= \frac{1}{2187}$$

$$= \left(\frac{1}{3}\right)^7$$

$$\Rightarrow r = \frac{1}{3}$$

Since $$r < 1$$, use $$S_n = \dfrac{a(1 - r^n)}{1 - r}$$.

$$S_7 = \frac{27 \times \left[1 - \left(\frac{1}{3}\right)^7\right]}{\left(1 - \frac{1}{3}\right)}$$

$$= \frac{27 \times \frac{2186}{2187}}{\frac{2}{3}}$$

$$= \frac{27 \times 2186 \times 3}{2187 \times 2}$$

$$= \frac{1093}{27}$$

Hence, the required sum is $$\dfrac{1093}{27} \approx 40.48$$.

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Chapter 11: Geometric Progression - EXERCISE 11B [Page 180]

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R.S. Aggarwal Mathematics [English] Class 10 ICSE
Chapter 11 Geometric Progression
EXERCISE 11B | Q 6. | Page 180
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