A) \[n({{a}_{1}}-{{a}_{n}})\]
B) \[(n-1)({{a}_{1}}-{{a}_{n}})\]
C) \[n{{a}_{1}}{{a}_{n}}\]
D) \[(n-1){{a}_{1}}{{a}_{n}}\]
Correct Answer: D
Solution :
[d] \[\frac{1}{{{a}_{2}}}-\frac{1}{{{a}_{1}}}=\frac{1}{{{a}_{3}}}-\frac{1}{{{a}_{2}}}=...=\frac{1}{{{a}_{n}}}-\frac{1}{{{a}_{n-1}}}=d(say)\] Then \[{{a}_{1}}{{a}_{2}}=\frac{{{a}_{1}}-{{a}_{2}}}{d},{{a}_{2}}{{a}_{3}}=\frac{{{a}_{2}}-{{a}_{3}}}{d},...,{{a}_{n-1}}{{a}_{n}}=\frac{{{a}_{n-1}}-{{a}_{n}}}{d}\] \[\therefore {{a}_{1}}{{a}_{2}}+{{a}_{2}}{{a}_{3}}+...+{{a}_{n-1}}{{a}_{n}}=\frac{{{a}_{1}}-{{a}_{n}}}{d}\] Also, \[\frac{1}{{{a}_{n}}}=\frac{1}{{{a}_{1}}}+(n-1)d\] \[\Rightarrow \frac{{{a}_{1}}-{{a}_{n}}}{d}=(n-1){{a}_{1}}{{a}_{n}}\]
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