A) \[(n-2)+\frac{1}{(n-2)}\]
B) \[\frac{1}{(n-2)}\]
C) \[(n-2)\]
D) \[(n-1)\]
E) \[(n+2)\]
Correct Answer: A
Solution :
Given;\[{{a}_{1}},{{a}_{2}},{{a}_{3}}......{{a}_{n}}\]are in AP and\[{{a}_{1}}=0\] Then, \[{{a}_{2}}={{a}_{1}}+d=0+d=d\] \[{{a}_{3}}={{a}_{1}}+2d=0+2d=2d\] ???????????????. ???????????????. \[{{a}_{n}}={{a}_{1}}+(n-1)d\] \[=(n-1)d\] Now, \[\left( \frac{{{a}_{3}}}{{{a}_{2}}}+\frac{{{a}_{4}}}{{{a}_{3}}}+....+\frac{{{a}_{n}}}{{{a}_{n-1}}} \right)\] \[-{{a}_{2}}\left( \frac{1}{{{a}_{2}}}+\frac{1}{{{a}_{3}}}+....+\frac{1}{{{a}_{n-2}}} \right)\] \[=\left( \frac{2d}{{{a}_{2}}}+\frac{3d}{2d}+....+\frac{(n-1)d}{(n-2)d} \right)\] \[-d\left( \frac{1}{d}+\frac{1}{2d}+....+\frac{1}{(n-3)d} \right)\] \[=\left( \frac{2}{1}+\frac{3}{2}+....+\frac{(n-1)}{(n-2)} \right)\] \[-\left( 1+\frac{1}{2}+....+\frac{1}{n-3} \right)\] \[=\left\{ (1+1)+\left( 1+\frac{1}{2} \right)+....+\left( 1+\frac{1}{n-2} \right) \right\}\] \[-\left\{ 1+\frac{1}{2}+...+\frac{1}{n-3} \right\}\] \[=\left\{ (1+1+1+.....+(n-2)terms) \right.\] \[\left. +\left( 1+\frac{1}{2}+\frac{1}{3}+.....+\frac{1}{n-3} \right)+\frac{1}{n-2} \right\}\] \[-\left\{ 1+\frac{1}{2}+....+\frac{1}{n-3} \right\}\] \[=\left\{ (1+1+....+(n-2)terms)+\frac{1}{n-2} \right\}\] \[=(n-2)+\frac{1}{(n-2)}\]
You need to login to perform this action.
You will be redirected in
3 sec
