Question

Find the sum of the following geometric progressions:
(i) 2, 6, 18, ... to 7 terms;
(ii) 1, 3, 9, 27, ... to 8 terms;
(iii) 1, −1/2, 1/4, −1/8, ... to 9 terms;
(iv) (a² − b²), (a − b), $\left(\frac{a-b}{a+b}\right)$, ... to n terms;
(v) 4, 2, 1, 1/2 ... to 10 terms.

Answer 1

(i) Here, a = 2 and r = 3.
$$\begin{gathered} \therefore S_7=a\left(\frac{r^7-1}{r-1}\right) \\ =2\left(\frac{3^7-1}{3-1}\right) \\ =2187-1 \\ =2186 \end{gathered}$$

(ii) Here, a = 1 and r = 3.
$$\begin{gathered} \therefore S_8=a\left(\frac{r^8-1}{r-1}\right) \\ =1\left(\frac{3^8-1}{3-1}\right) \\ =\frac{6561-1}{2} \\ =3280 \end{gathered}$$

(iii) Here, a = 1 and r = −$\frac{1}{2}$.
$$\begin{gathered} \therefore S_9=a\left(\frac{1-r^9}{1-r}\right) \\ =1\left(\frac{1-{\left(-{\displaystyle \frac{1}{2}}\right)}^9}{1-\left(-{\displaystyle \frac{1}{2}}\right)}\right) \\ =\frac{1-\left(-{\displaystyle \frac{1}{512}}\right)}{{\displaystyle \frac{3}{2}}} \\ =\frac{{\displaystyle \frac{513}{512}}}{{\displaystyle \frac{3}{2}}} \\ =\frac{513\times 2}{512\times 3} \\ =\frac{171}{256} \end{gathered}$$

(iv) Here, a = a² − b² and r = $\frac{1}{a+b}$.
$$\begin{gathered} \therefore S_n=a\left(\frac{1-r^n}{1-r}\right) \\ =\left(a^2-b^2\right)\left(\frac{1-{\left({\displaystyle \frac{1}{a+b}}\right)}^n}{1-\left(\frac{1}{a+b}\right)}\right) \\ =\left(a^2-b^2\right)\left(\frac{\left({\displaystyle \frac{{\left(a+b\right)}^n-1}{{\left(a+b\right)}^n}}\right)}{{\displaystyle \frac{\left(a+b\right)-1}{a+b}}}\right) \\ \Rightarrow S_n=\frac{\left(a+b\right)\left(a-b\right)}{{\left(a+b\right)}^{n-1}}\left(\frac{{\left(a+b\right)}^n-1}{\left(a+b\right)-1}\right) \\ =\frac{\left(a-b\right)}{{\left(a+b\right)}^{n-2}}\left(\frac{{\left(a+b\right)}^n-1}{\left(a+b\right)-1}\right) \end{gathered}$$

(v) Here, a = 4 and r = $\frac{1}{2}$
$$\begin{gathered} \therefore S_n=a\left(\frac{1-r^{10}}{1-r}\right) \\ =4\left(\frac{1-{\left({\displaystyle \frac{1}{2}}\right)}^{10}}{1-\left({\displaystyle \frac{1}{2}}\right)}\right) \\ =4\left(\frac{1-\left({\displaystyle \frac{1}{1024}}\right)}{{\displaystyle \frac{1}{2}}}\right) \\ =8\left(1-\frac{1}{1024}\right) \\ =\frac{1023}{128} \end{gathered}$$