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Home » Find the sum of the following geometric series: (i) to 5 terms; (ii) to terms
Find the sum of the following geometric series:
(i) 2 / 9-1 / 3+1 / 2-3 / 4+\ldots to 5 terms;
(ii) (x+y)+\left(x^{2}+x y+y^{2}\right)+\left(x^{3}+x^{2} y+x y^{2}+y^{3}\right)+\ldots to n terms
CBSE | Class 11 | Exercise 20.3 | Geometric Progressions | Maths | RD Sharma
Find the sum of the following geometric series:
(i) 2 / 9-1 / 3+1 / 2-3 / 4+\ldots to 5 terms;
(ii) (x+y)+\left(x^{2}+x y+y^{2}\right)+\left(x^{3}+x^{2} y+x y^{2}+y^{3}\right)+\ldots to n terms

Solution:

(i) 2 / 9-1 / 3+1 / 2-3 / 4+\ldots to 5 terms;
Given that
\begin{array}{l} a=2 / 9 \\ r=t_{2} / t_{1}=(-1 / 3) /(2 / 9)=-3 / 2 \\ n=5 \end{array}
Using the formula,
The sum of GP for n terms =a\left(1-r^{n}\right) /(1-r)
\begin{array}{l} \mathrm{a}\left(1-\mathrm{r}^{\mathrm{n}}\right) /(1-\mathrm{r})=(2 / 9)\left(1-(-3 / 2)^{5}\right) /(1-(-3 / 2)) \\ =(2 / 9)\left(1+(3 / 2)^{5}\right) /(1+3 / 2) \\ =(2 / 9)\left(1+(3 / 2)^{5}\right) /(5 / 2) \\ =(2 / 9)(1+243 / 32) /(5 / 2) \\ =(2 / 9)((32+243) / 32) /(5 / 2) \\ =(2 / 9)(275 / 32) \times 2 / 5 \\ =55 / 72 \end{array}

(ii) (x+y)+\left(x^{2}+x y+y^{2}\right)+\left(x^{3}+x^{2} y+x y^{2}+y^{3}\right)+\ldots to n terms;
Let S_{n}=(x+y)+\left(x^{2}+x y+y^{2}\right)+\left(x^{3}+x^{2} y+x y^{2}+y^{3}\right)+\ldots . to n terms
Multiply and divide by (x-y) we obtain,
\mathrm{S}_{n}=1 /(\mathrm{x}-\mathrm{y})\left[(\mathrm{x}+\mathrm{y})(\mathrm{x}-\mathrm{y})+\left(\mathrm{x}^{2}+\mathrm{xy}+\mathrm{y}^{2}\right)(\mathrm{x}-\mathrm{y}) \ldots\right. upto \mathrm{n} terms ] (x-y) S_{n}=\left(x^{2}-y^{2}\right)+x^{3}+x^{2} y+x y^{2}-x^{2} y-x y^{2}-y^{3}..upto n terms (x-y) S_{n}=\left(x^{2}+x^{3}+x^{4}+\ldots n\right. terms )-\left(y^{2}+y^{3}+y^{4}+\ldots n\right. terms )
Using the formula,
The sum of GP for n terms =a\left(1-r^{n}\right) /(1-r)
In the above sum we have two G.Ps, so,
\begin{array}{l} (x-y) S_{n}=x^{2}\left[\left(x^{n}-1\right) /(x-1)\right]-y^{2}\left[\left(y^{n}-1\right) /(y-1)\right] \\ S_{n}=1 /(x-y)\left\{x^{2}\left[\left(x^{n}-1\right) /(x-1)\right]-y^{2}\left[\left(y^{n}-1\right) /(y-1)\right]\right\} \end{array}

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