Question

Find the sum of first 10 terms of the G.P. $$1, \sqrt{3}, 3, 3\sqrt{3}, .......$$

Answer 1

Here, $$a = 1$$, $$r = \sqrt{3}$$ and $$n = 10$$.

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

$$S_{10} = \frac{1 \times \left[(\sqrt{3})^{10} - 1\right]}{(\sqrt{3} - 1)}$$

$$= \frac{(3^5 - 1)}{(\sqrt{3} - 1)}$$

$$= \frac{242}{(\sqrt{3} - 1)}$$

Multiply numerator and denominator by $$(\sqrt{3} + 1)$$:

$$S_{10} = \frac{242(\sqrt{3} + 1)}{(\sqrt{3} - 1)(\sqrt{3} + 1)}$$

$$= \frac{242(\sqrt{3} + 1)}{(3 - 1)}$$

$$= 121(\sqrt{3} + 1)$$

Hence, the required sum is $$121(\sqrt{3} + 1)$$, i.e. $$121 + 121\sqrt{3}$$.