QUESTION

# Find the relation between x and y in order that the ‘$r^{th}$’ mean between x and 2y may be the same as the ‘$r^{th}$’ mean between 2x and y, n means being inserted in each case.

Hint: Firstly assume n means between x and 2y and then consider whole series as an AP and find out the common difference using the formula ${{t}_{n+2}}={{t}_{1}}+\left[ \left( n+2 \right)-1 \right]d$ as there are two extra terms in your AP. Then find the $r^{th}$ term by using the same formula. Do the same process in second condition and then equate both the $r^{th}$ terms to get the relation between x and y.

As we have given in the question that we have to insert ‘n’ means in both the conditions therefore,
Let us assume ‘n’ means inserted namely “$a_1$, $a_2$, $a_3$, ……………………$a_n$” between x and 2y and therefore we can say that,
x, $a_1$, $a_2$, $a_3$, …………………………, $a_n$, 2y are in Arithmetic Progression. ………………….. (1)
As we have inserted ‘n’ means between x and 2y therefore number of terms will become (n + 2)
Therefore in the above AP,
First term= ${{t}_{1}}=x$
Last term = ${{t}_{n+2}}=2y$
Now to find the $r^{th}$ term we should know the common difference and for that we should know the formula given below,
Formula:
${{t}_{n}}={{t}_{1}}+(n-1)d$ , where d is the common difference. ……………………………………………………. (2)
As our nth term is (n+2)th term therefore the above formula can be replaced as,
${{t}_{n+2}}={{t}_{1}}+\left[ \left( n+2 \right)-1 \right]d$
If we put the above values in the formula we will get,
$2y=x+\left( n+1 \right)d$
Further simplification in the above equation will give,
$\Rightarrow \dfrac{2y-x}{\left( n+1 \right)}=d$
By rearranging the above equation we will get,
$\Rightarrow d=\dfrac{2y-x}{\left( n+1 \right)}$ ……………………………………………………. (3)
As we have to find the $r^{th}$ mean but in our AP the $r^{th}$ mean is the (r+1)th term and therefore we can write the formula for (r+1)th term by using formula (2) as follows,
$\Rightarrow {{t}_{r+1}}={{t}_{1}}+\left[ \left( r+1 \right)-1 \right]d$
As our (r+1)th term is ${{a}_{r}}^{th}$ mean therefore the above equation can be written as,
$\Rightarrow {{a}_{r}}={{t}_{1}}+\left[ \left( r+1 \right)-1 \right]d$
If we put the values of AP and equation (3) in the above formula we will get,
$\Rightarrow {{a}_{r}}=x+r\left[ \dfrac{2y-x}{\left( n+1 \right)} \right]$ ……………………………………………… (4)
Now Let us assume ‘n’ means inserted namely “b1, b2, b3, ……………………bn” between 2x and y and therefore we can say that,
2x, $a_1$, $a_2$, $a_3$, …………………………, $a_n$, y are in Arithmetic Progression. ………………………….. (5)
As we have inserted ‘n’ means between 2x and y therefore number of terms will become (n + 2)
Therefore in the above AP,
First term= ${{t}_{1}}=2x$
Last term = ${{t}_{n+2}}=y$
Now to find the $r^{th}$ term we should know the common difference and for that we should know the formula given below,
Formula:
${{t}_{n}}={{t}_{1}}+(n-1)D$ , where D is the common difference. ……………………………………………………. (6)
As our nth term is (n+2)th term therefore the above formula can be replaced as,
${{t}_{n+2}}={{t}_{1}}+\left[ \left( n+2 \right)-1 \right]D$
If we put the above values in the formula we will get,
$y=2x+\left( n+1 \right)D$
Further simplification in the above equation will give,
$\Rightarrow \dfrac{y-2x}{\left( n+1 \right)}=D$
By rearranging the above equation we will get,
$\Rightarrow D=\dfrac{y-2x}{\left( n+1 \right)}$ ……………………………………………………. (7)
As we have to find the $r^{th}$ mean but in our AP the $r^{th}$ mean is the $(r+1)^{th}$ term and therefore we can write the formula for (r+1)th term by using formula (2) as follows,
$\Rightarrow {{t}_{r+1}}={{t}_{1}}+\left[ \left( r+1 \right)-1 \right]D$
As our (r+1)th term is ${{b}_{r}}^{th}$ mean therefore the above equation can be written as,
$\Rightarrow {{b}_{r}}={{t}_{1}}+\left[ \left( r+1 \right)-1 \right]D$
If we put the values of AP and equation (7) in the above formula we will get,
$\Rightarrow {{b}_{r}}=2x+r\left[ \dfrac{y-2x}{\left( n+1 \right)} \right]$ ……………………………………………… (8)
As per the condition given in question, the $r^{th}$ means of both the cases are equal, therefore
$\Rightarrow {{a}_{r}}={{b}_{r}}$
If we put the values of equation (4) and equation (8) in the above equation we will get,
$\Rightarrow 2x+r\left[ \dfrac{y-2x}{\left( n+1 \right)} \right]=x+r\left[ \dfrac{2y-x}{\left( n+1 \right)} \right]$
If we multiply the whole equation by (n + 1) we will get,
$\Rightarrow \left( n+1 \right)\left\{ 2x+r\left[ \dfrac{y-2x}{\left( n+1 \right)} \right] \right\}=\left( n+1 \right)\left\{ x+r\left[ \dfrac{2y-x}{\left( n+1 \right)} \right] \right\}$
Further simplification in the above equation will give,
$\Rightarrow 2x\left( n+1 \right)+r\left[ \dfrac{y-2x}{\left( n+1 \right)} \right]\left( n+1 \right)=x\left( n+1 \right)+r\left[ \dfrac{2y-x}{\left( n+1 \right)} \right]\left( n+1 \right)$
$\Rightarrow 2x\left( n+1 \right)+r\left( y-2x \right)=x\left( n+1 \right)+r\left( 2y-x \right)$
If we shift the term $2x\left( n+1 \right)$ on the right hand side of the equation and the term $r\left( 2y-x \right)$ on the left hand side of the equation we will get,
$\Rightarrow r\left( y-2x \right)-r\left( 2y-x \right)=x\left( n+1 \right)-2x\left( n+1 \right)$
$\Rightarrow r\left( y-2x-2y+x \right)=x\left( n+1 \right)-x\left( 2n+2 \right)$
$\Rightarrow r\left( y-2x-2y+x \right)=x\left( n+1-2n-2 \right)$
$\Rightarrow r\left( -x-y \right)=x\left( -n-1 \right)$
$\Rightarrow -r\left( x+y \right)=-x\left( n+1 \right)$
$\Rightarrow r\left( x+y \right)=x\left( n+1 \right)$
As we have to find the relation between x and y therefore we will convert the equation in the form y = f(x), therefore,
$\Rightarrow x+y=\dfrac{x\left( n+1 \right)}{r}$
$\Rightarrow y=\dfrac{x\left( n+1 \right)}{r}-x$
$\Rightarrow y=x\left[ \dfrac{\left( n+1 \right)}{r}-1 \right]$
Therefore the relation between x and y is given by $y=x\left[ \dfrac{\left( n+1 \right)}{r}-1 \right]$.

Note: Students generally use the formula ${{t}_{n}}={{t}_{1}}+(n-1)d$ as it is and they forgot that there is first and last term are available which we have to add in n means while considering the whole as an AP.