Question

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

Answer 1

Let $a=A-2d,x=A-d,y=A,z=A+d,b=A+2d$
Given $x+y+z=15\Rightarrow A=5$
$\Rightarrow a=5-2d,b=5+2d$
Also $\frac{1}{a},\frac{1}{\alpha }.\frac{1}{\beta },\frac{1}{\gamma },\frac{1}{b}$ in A.P.
$\Rightarrow \frac{1}{\alpha }+\frac{1}{\beta }+\frac{1}{\gamma }=\frac{3}{2}[\frac{1}{a}+\frac{1}{b}]=\frac{5}{3}$

$\Rightarrow \frac{3}{2}(\frac{1}{5-2d}+\frac{1}{5+2d})=\frac{5}{3}\Rightarrow d=\pm 2$
Take $d=-2$ since $a>b$
Hence $a=9$