Question

If the $(m+1{)}^{th},(n+1{)}^{th}\&(r+1{)}^{th}$ terms of an AP are in GP & $m,n,r$ are in HP, then the ratio of the common difference to the first term of the AP is

• A. $\frac{1}{n}$
• B. $\frac{2}{n}$
• C. $-\frac{2}{n}$
• D. none of these

Answer 1

Answer: (C)

$-\frac{2}{n}$

Given $m,n,r$ are in H.P.

$\Rightarrow mr=\frac{n(m+r)}{2}$..........(i)

Given, $(m+1{)}^{th},(n+1{)}^{th}\&(r+1{)}^{th}$ are terms of an AP

$T_{m+1}=a+md$

$T_{n+1}=a+nd$

$T_{r+1}=a+rd$

Also, given these are in GP.

$\Rightarrow \frac{a+rd}{a+nd}=\frac{a+nd}{a+md}$
$\Rightarrow \frac{1+r\left(d/a\right)}{1+n\left(d/a\right)}=\frac{1+n\left(d/a\right)}{1+m\left(d/a\right)}$
Let $\frac{d}{a}=x$

$\Rightarrow (1+nx{)}^2=(1+rx\left)\right(1+mx)$

$\Rightarrow (n^2-mr)x^2+(2n-r-m)x=0$

$\Rightarrow x=0$ or $x=-\left(\frac{2n-r-m}{n^2-mr}\right)=\left(\frac{2n-r-m}{\frac{n(m+r)}{2}-n^2}\right)$ (by (i))

$\Rightarrow x=\frac{-2}{n}$