Question
The value of $$x + y + z$$ is 15 if $$a, x, y, z, b$$ are in A.P. while the value of $$\frac{1}{x} + \frac{1}{y} + \frac{1}{z}$$ is $$\frac{5}{3}$$ if $$a, x, y, z, b$$ are in H.P. Then the value of $$a$$ and $$b$$ are
A.
2 and 8
B.
1 and 9
C.
3 and 7
D.
None
Answer :
1 and 9
Solution :
As $$x, y, z$$ are A.M. of $$a$$ and $$b$$
$$\eqalign{
& \therefore x + y + z = 3\left( {\frac{{a + b}}{2}} \right) \cr
& \therefore 15 = \frac{3}{2}\left( {a + b} \right) \cr
& \Rightarrow a + b = 10\,\,\,.....\left( 1 \right) \cr} $$
Again $$\frac{1}{x} , \frac{1}{y} , \frac{1}{z}$$ are A.M. of $$\frac{1}{a}$$ and $$\frac{1}{b}$$
$$\eqalign{
& \therefore \frac{1}{x} + \frac{1}{y} + \frac{1}{z} = \frac{3}{2}\left( {\frac{1}{a} + \frac{1}{b}} \right) \cr
& \therefore \frac{5}{3} = \frac{3}{2} \cdot \frac{{a + b}}{{ab}} \cr
& \Rightarrow \frac{{10}}{9} = \frac{{10}}{{ab}} \cr
& \Rightarrow ab = 9\,\,\,\,\,.....\left( 2 \right) \cr} $$
Solving (1) and (2), we get
$$a = 9, 1, b = 1, 9$$