Question

If $m\ne n$ and the sequences $p,a,b,q$ and $p,m,n,q$ each are in AP, then $\frac{b-a}{n-m}$ is ___.

• A. $\frac{2}{3}$
• B. $\frac{3}{2}$
• C. 1
• D. $\frac{3}{4}$

Answer 1

The correct option is C 1

Given, $p,a,b,q$ are in AP.
$\therefore 2a=p+b$
$\Rightarrow 2a-b=p....\left(i\right)$
$\text{Also,}2b=a+q.$
$\Rightarrow 2b-a=q....\left(ii\right)$

Also given that $p,m,n,q$ are in AP.
$\therefore 2m=p+n$
$\Rightarrow 2m-n=p...\left(iii\right)$
$\text{Also,}2n=m+q.$
$\Rightarrow 2n-m=q...\left(iv\right)$

From eqns. $\left(i\right)$ & $\left(iii\right),$ we get
$2a-b=2m-n....\left(v\right)$
From eqns. $\left(ii\right)$ & $\left(iv\right),$ we get
$2b-a=2n-m....\left(vi\right)$

Subtracting $\left(vi\right)$ from $\left(v\right),$ we get

$2a-b-(2b-a)=2m-n-(2n-m).$
$\Rightarrow 2a-b-2b+a=2m-n-2n+m$
$3a-3b=3m-3n$
$b-a=n-m$
$\Rightarrow \frac{b-a}{n-m}=1$