A) \[\frac{xy}{x+y-1}\]
B) \[\frac{xy}{x-y-1}\]
C) \[\frac{xy}{x-y+1}\]
D) \[\frac{xy}{x+y+1}\]
Correct Answer: A
Solution :
| [a] If a, \[ar,\,a{{r}^{2}},a{{r}^{3}}..........\]are in GP, then |
| sum of infinite GP.\[=a+ar+.....+\infty =\frac{a}{1-r}\] |
| where 'a' is the first term and V is the common ratio of GP. |
| Given \[x=1+a+{{a}^{2}}+....\infty \] |
| This is a GP, with common ratio 'a'. |
| \[\Rightarrow \,\,\,\,x=\frac{1}{1-a}\Rightarrow x-ax=1\Rightarrow a=\frac{x-1}{x}\] |
| Again, \[y=1+b+{{b}^{2}}+....\infty \]This is also a GP, with common ratio 'b?. |
| \[\Rightarrow \,\,\,y=\frac{1}{1-b}\Rightarrow b=\frac{y-1}{y}\] |
| Now, consider \[1+ab+{{a}^{2}}{{b}^{2}}+....\infty \] |
| which is again a GP with common ratio 'ab?. |
| \[\therefore \] Sum \[=\frac{1}{1-ab}=\frac{1}{1-\frac{x-1}{x}.\frac{y-1}{y}}\] |
| \[=\frac{xy}{xy-xy+x+y-1}=\frac{xy}{x+y-1}\] |
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