# $a,x,y,z,b$ are in AP such that $x+y+z=15$ and $a$, $\alpha$, $\beta$, $y$, $b$ are in HP such that $\large\frac{1}{\alpha}+\large\frac{1}{\beta}+\large\frac{1}{\gamma}=\large\frac{5}{3}$. Find $a$, $(a > b)$.

$(a)\;9\qquad(b)\;1\qquad(c)\;\large\frac{1}{9}\qquad(d)\;5$

Answer : (a) $\; 9$
$x+y+z=3y=15$
$y=5$
$a+b=2y=10$
$\large\frac{1}{\alpha}+\large\frac{1}{\beta}+\large\frac{1}{\gamma}=\large\frac{3}{\beta}=\large\frac{5}{3}$
$\large\frac{1}{\beta}=\large\frac{5}{9}$
$\large\frac{1}{a}+\large\frac{1}{b}=\large\frac{2}{\beta}=\large\frac{10}{9}$
$\large\frac{a+b}{ab}=\large\frac{10}{9}$
$ab=9$
$a > b$
$a=9\quad\;b=1\;.$