Question

Suppose a, x, y, z and b are in A.P. where x+y+z = 15, and $a,\alpha ,\beta ,\gamma$ and b are in HP,
where $\frac{1}{\alpha }+\frac{1}{\beta }+\frac{1}{\gamma }=\frac{5}{3}$
Find a and b.

• A. 1 and 9
• B. 3 and 7
• C. 2 and 8
• D. 5 and 6

Answer 1

The correct option is A 1 and 9

Since a, x, y, z and b are in AP, we have a+b = x + z = 2y
$Alsox+y+z=15\Rightarrow 2y+y=15\Rightarrow 3y=15ory=5\therefore a+b=10$
Since $a,\alpha ,\beta ,\gamma$ and b are in HP, we have
$\frac{1}{a}+\frac{1}{b}=\frac{1}{\alpha }+\frac{1}{\gamma }=\frac{2}{\beta }$
$Also,\frac{1}{\alpha }+\frac{1}{\gamma }+\frac{1}{\beta }=\frac{5}{3}\Rightarrow \frac{1}{\beta }=\frac{5}{9}$
Thus $\frac{1}{a}+\frac{1}{b}=\frac{10}{9}\Rightarrow \frac{b+a}{ab}=\frac{10}{9}\Rightarrow ab=9$
Hence, either a = 1, b = 9 or a = 9, b = 1