A) \[{{a}_{1}}{{a}_{n}}\]
B) \[n{{a}_{1}}{{a}_{n}}\]
C) \[(n-1){{a}_{1}}{{a}_{n}}\]
D) None of these
Correct Answer: C
Solution :
Since \[{{a}_{1}},\ {{a}_{2}},\ {{a}_{3}},\ .........{{a}_{n}}\] are in H.P. Therefore \[\frac{1}{{{a}_{1}}},\ \frac{1}{{{a}_{2}}},\ \frac{1}{{{a}_{3}}}.......\frac{1}{{{a}_{n}}}\] will be in A.P. Which gives \[\frac{1}{{{a}_{2}}}-\frac{1}{{{a}_{1}}}=\frac{1}{{{a}_{3}}}-\frac{1}{{{a}_{2}}}=.......=\frac{1}{{{a}_{n}}}-\frac{1}{{{a}_{n-1}}}=d\] \[\Rightarrow \]\[\frac{{{a}_{1}}-{{a}_{2}}}{{{a}_{1}}{{a}_{2}}}=\frac{{{a}_{3}}-{{a}_{2}}}{{{a}_{2}}{{a}_{3}}}=.......=\frac{{{a}_{n-1}}-{{a}_{n}}}{{{a}_{n-1}}{{a}_{n}}}=d\] \[\Rightarrow \]\[{{a}_{1}}-{{a}_{2}}=d{{a}_{1}}{{a}_{2}}\] \[{{a}_{2}}-{{a}_{3}}=d{{a}_{2}}{{a}_{3}}\] ............................. ............................. and \[{{a}_{n-1}}-{{a}_{n}}=d{{a}_{n}}{{a}_{n-1}}\] Adding these, we get \[d({{a}_{1}}{{a}_{2}}+{{a}_{2}}{{a}_{3}}+......+{{a}_{n}}{{a}_{n-1}})\] \[=({{a}_{1}}+{{a}_{2}}+......+{{a}_{n-1}})-({{a}_{2}}+{{a}_{3}}+.....+{{a}_{n}})\] \[={{a}_{1}}-{{a}_{n}}\] ?..(i) Also \[{{n}^{th}}\] term of this A.P. is given by \[\frac{1}{{{a}_{n}}}=\frac{1}{{{a}_{1}}}+(n-1)d\Rightarrow d=\frac{{{a}_{1}}-{{a}_{n}}}{{{a}_{1}}{{a}_{n}}(n-1)}\] Substituting this value of \[d\] in (i) \[({{a}_{1}}-{{a}_{n}})=\frac{{{a}_{1}}-{{a}_{n}}}{{{a}_{1}}{{a}_{n}}(n-1)}({{a}_{1}}{{a}_{2}}+{{a}_{2}}{{a}_{3}}+.......+{{a}_{n}}{{a}_{n-1}})\] \[({{a}_{1}}{{a}_{2}}+{{a}_{2}}{{a}_{3}}+.........+{{a}_{n}}{{a}_{n-1}})={{a}_{1}}{{a}_{n}}(n-1)\].
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