Question

If the 6th term of a series in Geometric Progression (G.P.) is 32 and the 9th term is 256, find the: 

1. first term and the common ratio.
2. sum of its first 10 terms.

Answer 1

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

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

$$T_6 = ar^5 = 32 \quad ...(i)$$

$$T_9 = ar^8 = 256 \quad ...(ii)$$

i. Divide (ii) by (i):

$$r^3 = \frac{256}{32} = 8 = 2^3$$

$$\Rightarrow r = 2$$

Put $$r = 2$$ in (i):

$$a \times 2^5 = 32$$

$$\Rightarrow 32a = 32$$

$$\Rightarrow a = 1$$

So, first term $$a = 1$$ and common ratio $$r = 2$$.

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

$$S_{10} = \frac{1 \times (2^{10} - 1)}{(2 - 1)}$$

$$= (1024 - 1)$$

$$= 1023$$

Hence, the sum of the first 10 terms is 1023.