Question

How many terms of the G.P. $$\frac{2}{9}, \frac{-1}{3}, \frac{1}{2}, ............$$ must be taken to make the sum equal to $$\frac{55}{72}$$?

Answer 1

Here, $$a = \dfrac{2}{9}$$ and $$r = \dfrac{-1/3}{2/9} = -\dfrac{3}{2}$$.

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

$$\frac{2}{9} \times \frac{\left[1 - \left(-\frac{3}{2}\right)^n\right]}{\left(1 + \frac{3}{2}\right)} = \frac{55}{72}$$

$$\frac{2}{9} \times \frac{2}{5} \left[1 - \left(-\frac{3}{2}\right)^n\right] = \frac{55}{72}$$

$$\frac{4}{45} \left[1 - \left(-\frac{3}{2}\right)^n\right] = \frac{55}{72}$$

$$\Rightarrow 1 - \left(-\frac{3}{2}\right)^n = \frac{55}{72} \times \frac{45}{4}$$

$$= \frac{275}{32}$$

$$\left(-\frac{3}{2}\right)^n = 1 - \frac{275}{32}$$

$$= -\frac{243}{32}$$

$$= \left(-\frac{3}{2}\right)^5$$

$$\Rightarrow n = 5$$

Check: $$\dfrac{2}{9} - \dfrac{1}{3} + \dfrac{1}{2} - \dfrac{3}{4} + \dfrac{9}{8}$$

$$= \dfrac{16 - 24 + 36 - 54 + 81}{72}$$

$$= \dfrac{55}{72}$$ 

Hence, 5 terms must be taken.