Question

(ii) 2, -4, 8, -16, ... Find S₉ and S₁₂.

Answer 1

In the given G.P., we have t₁ = 2, t₂ = –4, t₃ = 8 and t₄ = –16.
Thus, we have the first term, i.e., a = t₁ = 2, and the common ratio, i.e., $r=\frac{t_2}{t_1}=\frac{-4}{2}=-2<1$.
We know that the sum of the first n^th term of the G.P. is $S_n=\frac{a(1-r^n)}{1-r}\mathrm{for}r<1$.
To find S₉, we will take a = 2, r = –2 and n = 9.
Thus, we have:
$$\begin{gathered} S_9=\frac{2\left(1-(-2{)}^9\right)}{1-(-2)} \\ =\frac{2\left(1+512\right)}{1+2} \\ =\frac{2\times 513}{3} \\ =342 \end{gathered}$$

Now,
To find S₁₂, we will take a = 2, r = –2 and n = 12.
Thus, we have:

$$\begin{gathered} S_{12}=\frac{2\left(1-(-2{)}^{12}\right)}{1-(-2)} \\ =\frac{2(1-4096)}{1+2} \\ =\frac{2\times (-4095)}{3} \\ =-2730 \end{gathered}$$