Sequence And Series Question 233
Question: $ x+y+z=15 $ if $ 9,\ x,\ y,\ z,\ a $ are in A.P.; while $ \frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{5}{3} $ if $ 9,\ x,\ y,\ z,\ a $ are in H.P., then the value of $ a $ will be
[IIT 1978]
Options:
A) 1
B) 2
C) 3
D) 9
Show Answer
Answer:
Correct Answer: A
Solution:
$ x+y+z=15 $ , if $ x={{({z^{-3}})}^{-1}}=z^{3} $ are in A.P. Sum $ =9+15+a=\frac{5}{2}(9+a) $
$ \Rightarrow $ $ 24+a=\frac{5}{2}(9+a) $
$ \Rightarrow $ $ 48+2a=45+5a $
$ \Rightarrow $ $ 3a=3 $
$ \Rightarrow $ $ a=1 $ ?..(i) and $ \frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{5}{3} $ , if $ 9,\ x,\ y,\ z,\ a $ are in H.P. Sum = $ \frac{1}{9}+\frac{5}{3}+\frac{1}{a}=\frac{5}{2}[ \frac{1}{9}+\frac{1}{a} ] $
$ \Rightarrow $ $ a=1 $ .
BETA
