Question

The 4th and the 7th terms of a G.P. are $$\frac{1}{27}$$ and $$\frac{1}{729}$$ respectively. Find the sum of first 6 terms of the G.P.

Answer 1

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

$$T_4 = ar^3 = \frac{1}{27} \quad ...(i)$$

$$T_7 = ar^6 = \frac{1}{729} \quad ...(ii)$$

Divide (ii) by (i):

$$r^3 = \frac{1}{729} \times 27$$

$$= \frac{1}{27}$$

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

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

Put $$r = \dfrac{1}{3}$$ in (i):

$$a \times \frac{1}{27} = \frac{1}{27}$$

$$\Rightarrow a = 1$$

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

$$S_6 = \frac{1 \times \left[1 - \left(\frac{1}{3}\right)^6\right]}{\left(1 - \frac{1}{3}\right)}$$

$$= \frac{1 - \frac{1}{729}}{\frac{2}{3}}$$

$$= \frac{728}{729} \times \frac{3}{2}$$

$$= \frac{364}{243}$$

Hence, the required sum is $$\dfrac{364}{243} \approx 1.498$$.