Question

Let $a(a\ne 0)$ is a fixed real number and $\frac{a-x}{px}=\frac{a-y}{qy}=\frac{a-z}{rz}$. if$b,q,r$ are in A.P., show that $\frac{1}{x},\frac{1}{y},\frac{1}{z}$ are in A.P.

Answer 1

$p,q.r$ are in A.P, So: $p+r=2q$-----(1)

Now, $\frac{a-x}{px}=\frac{a-y}{qy}=\frac{a-z}{rz}=k\left(say\right)$

$\frac{a-x}{px}=k\Rightarrow a-x=kpx$

$\Rightarrow a=x(kp+1)$

$\Rightarrow \frac{1}{x}=\frac{kp+1}{a}$

Similarly, $\frac{1}{y}=\frac{kq+1}{a},\frac{1}{z}=\frac{kr+1}{a}$

Now, $\frac{1}{x}+\frac{1}{z}=\frac{kp+1}{a}+\frac{kr+1}{a}$

$\Rightarrow \frac{k(p+r)+2}{a}$

From 1, $p+r=2q$

$\therefore \frac{1}{x}+\frac{1}{z}=\frac{k\left(2q\right)+2}{a}$

$\Rightarrow \frac{2(kq+1)}{a}$

$\therefore \frac{1}{x}+\frac{1}{z}=\frac{2}{y}$

Thus, $\frac{1}{x},\frac{1}{y},\frac{1}{z}$ are in A.P