Question

If the ${{\left( m+1 \right)}^{th}}$ , ${{\left( n+1 \right)}^{th}}$ and ${{\left( r+1 \right)}^{th}}$ terms of an A.P. are in G.P. and $m,n,r$ are in H.P. then the ratio of the first term to the common difference of the A.P. is:
(1) $\dfrac{n}{2}$
(2) $\dfrac{-n}{2}$
(3) $\dfrac{n}{3}$
(4) $\dfrac{-n}{3}$

Answer 1

Hint:Here in this question we have been asked to find the ratio of the first term and the common difference of the A.P. given that ${{\left( m+1 \right)}^{th}}$ , ${{\left( n+1 \right)}^{th}}$ and ${{\left( r+1 \right)}^{th}}$ terms of an A.P. are in G.P. and $m,n,r$ are in H.P. We know that the ${{n}^{th}}$ term of an A.P. and G.P. are $a+\left( n-1 \right)d$ and $a{{r}^{n-1}}$ respectively.

Complete step-by-step solution:
Now considering from the question we have been asked to find the ratio of the first term and the common difference of the A.P. given that ${{\left( m+1 \right)}^{th}}$ , ${{\left( n+1 \right)}^{th}}$ and ${{\left( r+1 \right)}^{th}}$ terms of an A.P. are in G.P. and $m,n,r$ are in H.P.
From the basic concepts of progression, we know that the ${{n}^{th}}$ term of an A.P. and G.P. are $a+\left( n-1 \right)d$ and $a{{r}^{n-1}}$ respectively.
H.P. implies that the reciprocals of the terms are in A.P.
Here given that $m,n,r$ are in H.P. it implies that $\dfrac{1}{m},\dfrac{1}{n},\dfrac{1}{r}$ are in A.P.
 Terms in A.P. have a common difference it means that $\dfrac{1}{n}-\dfrac{1}{m}=\dfrac{1}{r}-\dfrac{1}{n}\Rightarrow \dfrac{2}{n}=\dfrac{1}{m}+\dfrac{1}{r}$ .
Let us suppose that the first term of the given A.P. is $a$ and the common difference is $d$ then ${{a}_{m+1}}=a+md$ , ${{a}_{n+1}}=a+nd$ and ${{a}_{r+1}}=a+rd$ .
As it is given that ${{\left( m+1 \right)}^{th}}$ , ${{\left( n+1 \right)}^{th}}$ and ${{\left( r+1 \right)}^{th}}$ terms of an A.P. are in G.P. We can say that $\dfrac{a+nd}{a+md}=\dfrac{a+rd}{a+nd}\Rightarrow {{\left( a+nd \right)}^{2}}=\left( a+md \right)\left( a+rd \right)$ .
Now we have two expressions they are $2mr=n\left( r+m \right)$ and ${{\left( a+nd \right)}^{2}}=\left( a+md \right)\left( a+rd \right)$ .
Now we need to solve the above expressions.
By simplifying them we will have ${{\left( 1+n\left( \dfrac{d}{a} \right) \right)}^{2}}=\left( 1+m\left( \dfrac{d}{a} \right) \right)\left( 1+r\left( \dfrac{d}{a} \right) \right)$ . Let us suppose that $\left( \dfrac{d}{a} \right)=x$ then we will have ${{\left( 1+nx \right)}^{2}}=\left( 1+mx \right)\left( 1+rx \right)$ .
Now by expanding the expression we have we will get
$\begin{align}
  & 1+{{n}^{2}}{{x}^{2}}+2nx=1+mx+rx+mr{{x}^{2}} \\
 & \Rightarrow {{x}^{2}}\left( mr-{{n}^{2}} \right)+x\left( m+r-2n \right)=0 \\
 & \Rightarrow x=0\text{ }or\text{ }x=\dfrac{-\left( m+r-2n \right)}{\left( mr-{{n}^{2}} \right)} \\
\end{align}$
Now by using first expression $2mr=n\left( r+m \right)$ in $x=\dfrac{-\left( m+r-2n \right)}{\left( mr-{{n}^{2}} \right)}$ we will get
$\begin{align}
  & x=\dfrac{-\left( m+r-2n \right)}{\left( \dfrac{n\left( m+r \right)}{2}-{{n}^{2}} \right)} \\
 & \Rightarrow x=\dfrac{-2}{n} \\
\end{align}$
Therefore we can conclude that the ratio of first term to the common difference in the given A.P is $\dfrac{-n}{2}$ .
Hence we will mark the option “2” as correct.

Note:While answering questions of this type we should be sure with the concepts that we are going to apply in between the steps. If someone had not carefully read the question and taught that we have been asked to find the ratio of common difference to the first term then they will end up having the answer as $\dfrac{-2}{n}$ which is a wrong answer.