Sequence And Series Question 100
Question: If $ x_1,x_2,…..x_{20} $ are in H.P. and $ x_1,2,x_{20} $ are in G.P., then $ \sum\limits_{r=1}^{19}{x_{r}{x_{r+1}}} $ =
Options:
A) 76
B) 80
C) 84
D) none of these
Show Answer
Answer:
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_1x_{20}}=19\lambda $
$ \Rightarrow \sum\limits_{r=1}^{19}{x_{r}{x_{r+1}}=19x_1x_{20}=19\times 4=76} $ $ (\because x_1,2,x_{20}are,in,G.P.,then,x_1x_{20}=4) $
BETA
