Question

If x, 3/2, Z are in AP; 2,3,z are in GP
then which of the following will be in HP ?

(a) x, 6, z
b) x, 4 , z
c) x, 2, z
d) x , 1 ,z​

Answer 1

Answer with explanation:

→It is given that, $x,\frac{3}{2},z$ are in A.P.

$2\times \frac{3}{2}=x+z\\\\ 3=x+z$

→2,3,z are in GP.

 3²=2 × z

9=2 z

$z=\frac{9}{2}\\\\3=x+\frac{9}{2}\\\\3-\frac{9}{2}=x\\\\x=\frac{-3}{2}$

→Three numbers , a,b and c are in H.P ,if

$\frac{2}{b}=\frac{1}{a}+\frac{1}{c}$

→We will start from option (1), that which three numbers are in G.P

$1.\frac{2}{6}=\frac{1}{x}+\frac{1}{z}\\\\LHS=\frac{1}{3}\\\\RHS=\frac{2}{9}+\frac{-2}{3}=\frac{-4}{9}\\\\2.\frac{2}{4}=\frac{1}{x}+\frac{1}{z}\\\\LHS=\frac{1}{2}\\\\RHS=\frac{2}{9}+\frac{-2}{3}=\frac{-4}{9}\\\\3.\frac{2}{2}=\frac{1}{x}+\frac{1}{z}\\\\LHS=1\\\\RHS=\frac{2}{9}+\frac{-2}{3}=\frac{-4}{9}\\\\4.\frac{2}{1}=\frac{1}{x}+\frac{1}{z}\\\\LHS=2\\\\RHS=\frac{2}{9}+\frac{-2}{3}=\frac{-4}{9}$

→None of the four given options are in H.P.