Question

First two terms of an, A.P. as well as an H.P. are $a$ and $b$. If $x$ be any term of the A.P. and $y$ the corresponding term of H.P., then will $\frac{x-a}{y-a}=\frac{a}{y}?$.
If Yes Write 1 otherwise write 0

Answer 1

We can not write

$x=a+(n-1)(b-a)$ and $\frac{1}{y}=\frac{1}{a}+(n-1)[\frac{1}{b}-\frac{1}{a}]$

$x-a=(n-1)(b-a)$ ...(1)
$\frac{b}{y}=\frac{b+(n-1)(a-b)}{a}$...(2)
Also $y=\frac{ab}{b+(n-1)(a-b)}$

So $y-a=\frac{ab}{b+(n-1)(a-b)}-a=\frac{(b-a)a(n-1)}{b+(n-1)(a-b)}$.......(3)
Hence $\frac{x-a}{y-a}=\frac{b+(n-1)(a-b)}{a}=\frac{b}{y}$ by Using (1),(3).