Question & Answer

QUESTION

ANSWER
Verified

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.

Complete step-by-step answer:

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.

Complete step-by-step answer:

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.