A) 76
B) 80
C) 84
D) none of these
Correct Answer: A
Solution :
[a] Clearly, \[\frac{1}{{{x}_{1}}},\frac{1}{{{x}_{2}}},...\frac{1}{{{x}_{20}}}\] will be in A.P. Hence., \[\frac{1}{{{x}_{2}}}-\frac{1}{{{x}_{1}}}=\frac{1}{{{x}_{3}}}-\frac{1}{{{x}_{2}}}=...=\frac{1}{{{x}_{r+1}}}-\frac{1}{{{x}_{r}}}=...\lambda (say)\] \[\Rightarrow \frac{{{x}_{r}}-{{x}_{r+1}}}{{{x}_{1}}{{x}_{r+1}}}=\lambda \] Or \[{{x}_{r}}{{x}_{r+1}}=-\frac{1}{\lambda }({{x}_{r+1}}-{{x}_{r}})\] \[\Rightarrow \sum\limits_{r=1}^{19}{{{x}_{r}}{{x}_{r+1}}=-\frac{1}{\lambda }\sum\limits_{r=1}^{19}{({{x}_{r+1}}-{{x}_{r}})}}\] \[=-\frac{1}{\lambda }({{x}_{20}}-{{x}_{1}})\] Now, \[\frac{1}{{{x}_{20}}}=\frac{1}{{{x}_{1}}}+19\lambda \] Or \[\frac{{{x}_{1}}-{{x}_{20}}}{{{x}_{1}}{{x}_{20}}}=19\lambda \] \[\Rightarrow \sum\limits_{r=1}^{19}{{{x}_{r}}{{x}_{r+1}}=19{{x}_{1}}{{x}_{20}}=19\times 4=76}\] \[(\because {{x}_{1}},2,{{x}_{20}}are\,in\,G.P.,then\,{{x}_{1}}{{x}_{20}}=4)\]
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