A) \[\frac{n}{2n-1}(a_{1}^{2}-a_{2n}^{2})\]
B) \[\frac{2n}{n-1}(a_{2n}^{2}-a_{1}^{2})\]
C) \[\frac{n}{n+1}(a_{1}^{2}+a_{2n}^{2})\]
D) None of these
Correct Answer: A
Solution :
Since \[{{a}_{1}},\ {{a}_{2}},\,{{a}_{3}},...........,{{a}_{n}}\] form an A.P. therefore, \[{{a}_{2}}-{{a}_{1}}={{a}_{4}}-{{a}_{3}}=.......={{a}_{2n}}-{{a}_{2n-1}}=d\] Here \[a_{1}^{2}-a_{2}^{2}+a_{3}^{2}-a_{4}^{2}+.......+a_{2n-1}^{2}-a_{2n}^{2}\] \[=({{a}_{1}}-{{a}_{2}})({{a}_{1}}+{{a}_{2}})+({{a}_{3}}-{{a}_{4}})({{a}_{3}}+{{a}_{4}})+........\] \[......+({{a}_{2n-1}}-{{a}_{2n}})({{a}_{2n-1}}+{{a}_{2n}})\] \[=-d({{a}_{1}}+{{a}_{2}}+.......+{{a}_{2n}})=-d\left\{ \frac{2n}{2}({{a}_{1}}+{{a}_{2n}}) \right\}\] Also we know \[{{a}_{2n}}={{a}_{1}}+(2n-1)d\]\[\Rightarrow \]\[d=\frac{{{a}_{2n}}-{{a}_{1}}}{2n-1}\] \[\Rightarrow \]\[-d=\frac{{{a}_{1}}-{{a}_{2n}}}{2n-1}\]. \[\therefore \] Therefore the sum is = \[\frac{n({{a}_{1}}-{{a}_{2n}}).({{a}_{1}}+{{a}_{2n}})}{2n-1}=\frac{n}{2n-1}(a_{1}^{2}-a_{2n}^{2})\].
You need to login to perform this action.
You will be redirected in
3 sec
