A) 1
B) 2
C) 3
D) 9
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 =\[\frac{1}{9}+\frac{5}{3}+\frac{1}{a}=\frac{5}{2}\left[ \frac{1}{9}+\frac{1}{a} \right]\]\[\Rightarrow \]\[a=1\].You need to login to perform this action.
You will be redirected in
3 sec