A) \[\frac{a-b}{a+b}\]
B) \[\frac{b-a}{a+b}\]
C) \[\frac{a+b}{a-b}\]
D) \[\frac{b+a}{b-a}\]
Correct Answer: A
Solution :
Let d be the common difference of the given AP then, \[{{a}_{2}}-{{a}_{1}}={{a}_{3}}-{{a}_{2}}={{a}_{4}}-{{a}_{3}}\] \[=...={{a}_{n}}-{{a}_{n-1}}=d\] ?(i) Now, \[\frac{1}{\sqrt{{{a}_{1}}}+\sqrt{{{a}_{2}}}}+\frac{1}{\sqrt{{{a}_{2}}}+\sqrt{{{a}_{3}}}}+\frac{1}{\sqrt{{{a}_{3}}}+\sqrt{{{a}_{4}}}}\] \[+....+\frac{1}{\sqrt{{{a}_{n-1}}}+\sqrt{{{a}_{n}}}}\] ?(ii) On multiplying by the rationalisation factor of every term in numerator and denominator, we get \[\frac{\sqrt{{{a}_{2}}}-\sqrt{{{a}_{1}}}}{{{a}_{2}}-{{a}_{1}}}+\frac{\sqrt{{{a}_{3}}}-\sqrt{{{a}_{2}}}}{{{a}_{3}}-{{a}_{2}}}+\frac{\sqrt{{{a}_{4}}}-\sqrt{{{a}_{3}}}}{{{a}_{4}}-{{a}_{3}}}\] \[+....+\frac{\sqrt{{{a}_{n}}}-\sqrt{{{a}_{n-1}}}}{{{a}_{n}}-{{a}_{n-1}}}\] [using Eq.(i)] \[=\frac{1}{d}[\sqrt{{{a}_{2}}}-\sqrt{{{a}_{1}}}+\sqrt{{{a}_{3}}}-\sqrt{{{a}_{2}}}+\sqrt{{{a}_{4}}}-\sqrt{{{a}_{3}}}\] \[+...+\sqrt{{{a}_{n}}}-\sqrt{{{a}_{n-1}}}]\] \[=\frac{1}{d}(\sqrt{{{a}_{n}}}-\sqrt{{{a}_{1}}})\] \[=\frac{1}{d}(\sqrt{{{a}_{n}}}-\sqrt{{{a}_{1}}}).\frac{\sqrt{{{a}_{n}}}+\sqrt{{{a}_{1}}}}{\sqrt{{{a}_{n}}}+\sqrt{{{a}_{1}}}}\] \[=\frac{{{a}_{n}}-{{a}_{1}}}{d(\sqrt{{{a}_{n}}}+\sqrt{{{a}_{1}}})}\] \[=\frac{n-1}{d(\sqrt{{{a}_{n}}}+\sqrt{{{a}_{1}}})}\] \[[\because {{a}_{n}}={{a}_{1}}+(n-1)d]\] \[=\frac{n-1}{\sqrt{{{a}_{n}}}+\sqrt{{{a}_{1}}}}\] \[=\frac{n-1}{\sqrt{{{a}_{1}}}+\sqrt{{{a}_{n}}}}\]
You need to login to perform this action.
You will be redirected in
3 sec
