A) 1
B) 2
C) 3
D) 9
Correct Answer: A
Solution :
[a] 9, x, y, z, a are in A.P. \[x+y+z=15\Rightarrow 3\left[ \frac{9+a}{2} \right]=15\] \[a=1\] 9, X, Y, Z, a are in H.P. \[\frac{1}{9},\frac{1}{X},\frac{1}{Y},\frac{1}{Z},\frac{1}{a}\] are in A.P. \[\frac{1}{X}+\frac{1}{Y}+\frac{1}{Z}=\frac{5}{3}\Rightarrow 3\frac{\left[ \frac{1}{9}+\frac{1}{a} \right]}{2}=\frac{5}{3}\Rightarrow a=1\]You need to login to perform this action.
You will be redirected in
3 sec