A) \[(n-1)({{a}_{1}}-{{a}_{n}})\]
B) \[n{{a}_{1}}{{a}_{n}}\]
C) \[(n-1){{a}_{1}}{{a}_{n}}\]
D) \[n({{a}_{1}}-{{a}_{n}})\]
Correct Answer: C
Solution :
First, we will find d by using formula\[{{T}_{n}}=a+(n-1)d\]and then determine the sum. Since, \[{{a}_{1}},{{a}_{2}},{{a}_{3}},....{{a}_{n}}\]are in HP. \[\therefore \] \[\frac{1}{{{a}_{1}}},\frac{1}{{{a}_{2}}},\frac{1}{{{a}_{3}}},....\frac{1}{{{a}_{n}}}\]are in AP. Let \[d\]be the common difference of AP. \[\therefore \]\[\frac{1}{{{a}_{2}}}-\frac{1}{{{a}_{1}}}=d\] \[\Rightarrow \]\[{{a}_{1}}-{{a}_{2}}={{a}_{1}}{{a}_{2}}d\] Similarly, \[{{a}_{2}}-{{a}_{3}}={{a}_{2}}{{a}_{3}}d\] ????????? ????????? \[{{a}_{n-1}}-{{a}_{n}}={{a}_{n-1}}{{a}_{n}}d\] On adding all the equations, we get \[{{a}_{1}}-{{a}_{n}}=d\{{{a}_{1}}{{a}_{2}}+{{a}_{2}}{{a}_{3}}+......+{{a}_{n-1}}{{a}_{n}}\}\] ...(i) Also, \[\frac{1}{{{a}_{n}}}=\frac{1}{{{a}_{1}}}+(n-1)d\] \[\Rightarrow \]\[d=\frac{{{a}_{1}}-{{a}_{n}}}{{{a}_{1}}{{a}_{n}}(n-1)}\] On putting this value of d in Eq. (i), we get \[{{a}_{1}}-{{a}_{n}}=\frac{{{a}_{1}}-{{a}_{n}}}{{{a}_{1}}{{a}_{n}}(n-1)}\{{{a}_{1}}{{a}_{2}}+{{a}_{2}}{{a}_{3}}+.....+{{a}_{n-1}}{{a}_{n}}\}\] \[\Rightarrow \]\[{{a}_{1}}{{a}_{2}}+{{a}_{2}}{{a}_{3}}+....+{{a}_{n-1}}{{a}_{n}}={{a}_{1}}{{a}_{n}}(n-1)\]
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