Question

Given that n numbers of AMs are inserted between two sets of numbers a, 2b and 2a, b where $a,b\in R.$ Suppose further that the ${{m}^{th}}$ mean between these sets of numbers are the same, then the ratio a:b equals
(a) n – m + 1: m
(b) n – m + 1: n
(c) m: n – m + 1
(d) n: n – m + 1

Answer 1

To solve the given question, we will first find out the common difference of both the APs formed when n numbers of AM are inserted between them. After finding the common difference of both the APs in terms of a, b, m, and n, we will find the ${{m}^{th}}$ mean of both the APs which will be equal to the ${{\left( m+1 \right)}^{th}}$ term of these APs. For this, we will use the formula: ${{a}_{r}}=a+\left( r-1 \right)d$ where ${{a}_{r}}$ is the ${{r}^{th}}$ term and d is the common difference of the AP. Then we will equate these ${{\left( m+1 \right)}^{th}}$ terms and find the relation between a and b in terms of m and n.

Complete step-by-step answer:
To start with, we must know what an AP is. An AP or an arithmetic progression is a series of numbers in which the difference between any two consecutive terms is the same. In the question, we are given that we have put n numbers of AMs between (a, 2b) and (2a, b). Thus, we can say that when n AMs are inserted between a and 2b, the total number of terms becomes (n + 2) and we have got an AP with the first term as a and ${{\left( n+2 \right)}^{th}}$ term as 2b.
Now, the formula to find the ${{r}^{th}}$ term of any AP with the first term P and common difference Q is shown below.
${{r}^{th}}\text{ term}=P+\left( r-1 \right)Q$
In our case, r = n + 2, ${{r}^{th}}$ term is 2b and P = a. We have to find the value of Q = d which is the common difference. Thus, we have,
$2b=a+\left( n+2-1 \right)d$
$\Rightarrow 2b-a=\left( n+1 \right)d$
$\Rightarrow d=\dfrac{2b-a}{n+1}......\left( i \right)$
Similarly, n AMs are inserted between 2a and b. Thus, we can say that the total terms become (n + 2) and b is the ${{\left( n+2 \right)}^{th}}$ term. Here, d’ will be the common difference. Thus, we have,
$b=2a+\left( n+1 \right){{d}^{'}}$
$\Rightarrow b-a=\left( n+1 \right){{d}^{'}}$
$\Rightarrow {{d}^{'}}=\dfrac{b-2a}{n+1}......\left( ii \right)$
Now, the ${{m}^{th}}$ mean of these sets will be equal to the ${{\left( m+1 \right)}^{th}}$ term of these APs. The ${{\left( m+1 \right)}^{th}}$ term of the first set will be
${{\left( m+1 \right)}^{th}}\text{ term}=a+\left( m+1-1 \right)d$
$\Rightarrow {{\left( m+1 \right)}^{th}}\text{ term}=a+md.......\left( iii \right)$
Now, we will substitute the value of d from (i) to (iii). Thus, we will get,
$\Rightarrow {{\left( m+1 \right)}^{th}}\text{ term of the first set}=a+m\left( \dfrac{2b-a}{n+1} \right).......\left( iv \right)$
Now, the ${{\left( m+1 \right)}^{th}}$ term of the second set will be
${{\left( m+1 \right)}^{th}}\text{ term of the second set}=2a+\left( m+1-1 \right){{d}^{'}}$
${{\left( m+1 \right)}^{th}}\text{ term of the second set}=2a+m{{d}^{'}}........\left( v \right)$
Now, we will substitute the value of d’ from (ii) to (v). Thus, we will get,
${{\left( m+1 \right)}^{th}}\text{ term of the second set}=2a+m\left( \dfrac{b-2a}{n+1} \right)........\left( vi \right)$
Now, the ${{m}^{th}}$ mean of both the sets are equal, i.e. (iv) = (vi). Hence, we will get,
$a+m\left( \dfrac{2b-a}{n+1} \right)=2a+m\left( \dfrac{b-2a}{n+1} \right)$
$\Rightarrow \dfrac{a\left( n+1 \right)+m\left( 2b-a \right)}{n+1}=\dfrac{2a\left( n+1 \right)+m\left( b-2a \right)}{n+1}$
$\Rightarrow an+a+2mb-ma=2an+2a+mb-2am$
$\Rightarrow mb+am=a+an$
$\Rightarrow mb=a+an-am$
On dividing the above equation by b, we will get,
m=\dfrac{a}{b}\left( 1+n-m \right)$$$$\Rightarrow \dfrac{a}{b}=\dfrac{m}{1+n-m}
$\Rightarrow a:b=m:n-n+1$
Hence, the option (c) is the right answer.

Note: Here, we cannot take both a and b as zero because in that case, the ratio will become indeterminant form. Also, a and b should have the same sign i.e. both a and b should be either positive or both should be negative, because if one of them is positive and the other one is negative, then the ratio will become negative.