Question

The value of $x+y+zis15ifa,x,y,z,b$ are in A.P. while the value of $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}is\frac{5}{3}ifa,x,y,z,b$ are in H. P. Then the value of and b are

• A. 2 and 8
• B. 1 and 9
• C. 3 and 7
• D. None

Answer 1

The correct option is B

1 and 9

As x,y,z, are A.M. of a and b
$\therefore x+y+z=3\left(\frac{a+b}{2}\right)\therefore 15=\frac{3}{2}(a+b)\Rightarrow a+b=10...\left(i\right)Again\frac{1}{x},\frac{1}{y},\frac{1}{z}areA.M.of\frac{1}{a}and\frac{1}{b}\therefore \frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{3}{2}(\frac{1}{a}+\frac{1}{b})\therefore \frac{5}{3}=\frac{3}{2}.\frac{a+b}{ab}\Rightarrow \frac{10}{9}=\frac{10}{ab}\Rightarrow ab=9...\left(ii\right)$
Solving (i) and (ii), we get
a = 9, 1, b = 1, 9