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Find the sum of the following geometric progressions:
(i) 1,-1 / 2,1 / 4,-1 / 8, \ldots
(ii) \left(a^{2}-b^{2}\right),(a-b),(a-b) /(a+b), \ldots to n
CBSE | Class 11 | Exercise 20.3 | Geometric Progressions | Maths | RD Sharma
Find the sum of the following geometric progressions:
(i) 1,-1 / 2,1 / 4,-1 / 8, \ldots
(ii) \left(a^{2}-b^{2}\right),(a-b),(a-b) /(a+b), \ldots to n

Solution:

(i) 1,-1 / 2,1 / 4,-1 / 8, \ldots
It is known that, the sum of \mathrm{GP} for infinity =\mathrm{a} /(1-\mathrm{r})
Given that,
\mathrm{a}=1, \mathrm{r}=\mathrm{t}_{2} / \mathrm{t}_{1}=(-1 / 2) / 1=-1 / 2
Substitute the values in
\begin{array}{l} \mathrm{a} /(1-\mathrm{r})=1 /(1-(-1 / 2)) \\ =1 /(1+1 / 2) \\ =1 /((2+1) / 2) \\ =1 /(3 / 2) \\ =2 / 3 \end{array}

(ii) \left(a^{2}-b^{2}\right),(a-b),(a-b) /(a+b), \ldots to n terms
It is known that, the sum of GP for n terms =a\left(r^{n}-1\right) /(r-1)
Given that,
a=\left(a^{2}-b^{2}\right), r=t_{2} / t_{1}=(a-b) /\left(a^{2}-b^{2}\right)=(a-b) /(a-b)(a+b)=1 /(a+b), n=n
Substitute the values in
\begin{array}{l} a\left(r^{n}-1\right) /(r-1)= \\ \quad=\left(a^{2}-b^{2}\right)\left(\frac{1-\left(\frac{1}{a+b}\right)^{n}}{1-\left(\frac{1}{a+b}\right)}\right) \\ =\left(a^{2}-b^{2}\right)\left(\frac{\left(\frac{(a+b)^{n}-1}{(a+b)^{n}}\right)}{\frac{(a+b)-1}{a+b}}\right) \\ =\frac{(a+b)(a-b)}{(a+b)^{n-1}}\left(\frac{(a+b)^{n}-1}{(a+b)-1}\right) \\ =\frac{(a-b)}{(a+b)^{n-2}}\left(\frac{(a+b)^{n}-1}{(a+b)-1}\right) \end{array}

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