Answer
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Hint: We start solving the problem by assigning variables for the first term and common difference for the two of given arithmetic progressions. We compare the given ratio of the sum of n terms using the sum of n terms of A.P \[{{S}_{n}}=\dfrac{n}{2}\left( 2a+(n-1)d \right)\]. We then make substitution for the value of n in order to get ${{m}^{th}}$ terms in both numerator and denominator by comparing with \[{{n}^{th}}\ term=a+(n-1)d\] to get the final result.
Complete step-by-step answer:
As mentioned in the question, the ratio of the sum of first n terms of two different arithmetic progressions is as follows
For 1st arithmetic progression, let the first term and common difference be ‘a’ and ‘d’ respectively and for the 2nd arithmetic progression, let the first term and common difference be ‘A’ and ‘D’.
Now, we can write as following using the formula given in the hint,
\[\dfrac{{{S}_{n}}}{{{S}_{n}}\grave{\ }}=\dfrac{\left( 7n+1 \right)}{\left( 4n+27 \right)}\]
Now, we can write this ratio as follows
\[\dfrac{\dfrac{n}{2}\left( 2a+(n-1)d \right)}{\dfrac{n}{2}\left( 2A+(n-1)D \right)}=\dfrac{\left( 7n+1 \right)}{\left( 4n+27 \right)}\].
\[\dfrac{\left( 2a+(n-1)d \right)}{\left( 2A+(n-1)D \right)}=\dfrac{\left( 7n+1 \right)}{\left( 4n+27 \right)}\].
\[\dfrac{\left( a+\dfrac{(n-1)}{2}d \right)}{\left( A+\dfrac{(n-1)}{2}D \right)}=\dfrac{\left( 7n+1 \right)}{\left( 4n+27 \right)}\] ---(a).
We know that the ${{n}^{th}}$ term of an A.P (Arithmetic progression) is defined as ${{T}_{n}}=a+\left( n-1 \right)d$---(1), where a is first term and d is the common difference.
Now, we are asked the ratio of the \[{{m}^{th}}\] term of these two series and let us assume it as ‘m’. Using equation (1) we get,
\[\dfrac{a+\left( m-1 \right)d}{A+\left( m-1 \right)D}=x\ \ \ \ \ ...(b)\]
We need to make a substitution in place of n of the equation (a) to get equation (b). So, we compare both numerators (or denominators) to find the value that needs to substitute in n.
So, we have \[\dfrac{(n-1)}{2}=\left( m-1 \right)\].
$n-1=2\left( m-1 \right)$.
$n-1=2m-2$.
$n=2m-2+1$.
$n=2m-1$ ---(c).
We now substitute the result obtained from equation (c) in equation (a).
\[\dfrac{\left( a+\dfrac{(2m-1-1)}{2}d \right)}{\left( A+\dfrac{(2m-1-1)}{2}D \right)}=\dfrac{\left( 7\times \left( 2m-1 \right)+1 \right)}{\left( 4\times \left( 2m-1 \right)+27 \right)}\].
\[\dfrac{\left( a+\dfrac{(2m-2)}{2}d \right)}{\left( A+\dfrac{(2m-2)}{2}D \right)}=\dfrac{\left( 14m-7+1 \right)}{\left( 8m-4+27 \right)}\].
\[\dfrac{\left( a+\left( m-1 \right)d \right)}{\left( A+\left( m-1 \right)D \right)}=\dfrac{\left( 14m-6 \right)}{\left( 8m+23 \right)}\].
On comparing numerator and denominator with equation (1), we get
\[\dfrac{{{T}_{m}}}{T_{m}^{'}}=\dfrac{\left( 14m-6 \right)}{\left( 8m+23 \right)}\].
We have found the ratio of the \[{{m}^{th}}\] terms of two progressions as \[\dfrac{14m-6}{8m+23}\].
Note: We should not take the same first term and common difference for both progressions as it will make us confused and give wrong results. Whenever we get this type of problem, we need to first take the ratio and make a substitution that will be fit in order to get the ratio of required terms. We should not confuse the general terms of geometric, arithmetic and harmonic progressions.
Complete step-by-step answer:
As mentioned in the question, the ratio of the sum of first n terms of two different arithmetic progressions is as follows
For 1st arithmetic progression, let the first term and common difference be ‘a’ and ‘d’ respectively and for the 2nd arithmetic progression, let the first term and common difference be ‘A’ and ‘D’.
Now, we can write as following using the formula given in the hint,
\[\dfrac{{{S}_{n}}}{{{S}_{n}}\grave{\ }}=\dfrac{\left( 7n+1 \right)}{\left( 4n+27 \right)}\]
Now, we can write this ratio as follows
\[\dfrac{\dfrac{n}{2}\left( 2a+(n-1)d \right)}{\dfrac{n}{2}\left( 2A+(n-1)D \right)}=\dfrac{\left( 7n+1 \right)}{\left( 4n+27 \right)}\].
\[\dfrac{\left( 2a+(n-1)d \right)}{\left( 2A+(n-1)D \right)}=\dfrac{\left( 7n+1 \right)}{\left( 4n+27 \right)}\].
\[\dfrac{\left( a+\dfrac{(n-1)}{2}d \right)}{\left( A+\dfrac{(n-1)}{2}D \right)}=\dfrac{\left( 7n+1 \right)}{\left( 4n+27 \right)}\] ---(a).
We know that the ${{n}^{th}}$ term of an A.P (Arithmetic progression) is defined as ${{T}_{n}}=a+\left( n-1 \right)d$---(1), where a is first term and d is the common difference.
Now, we are asked the ratio of the \[{{m}^{th}}\] term of these two series and let us assume it as ‘m’. Using equation (1) we get,
\[\dfrac{a+\left( m-1 \right)d}{A+\left( m-1 \right)D}=x\ \ \ \ \ ...(b)\]
We need to make a substitution in place of n of the equation (a) to get equation (b). So, we compare both numerators (or denominators) to find the value that needs to substitute in n.
So, we have \[\dfrac{(n-1)}{2}=\left( m-1 \right)\].
$n-1=2\left( m-1 \right)$.
$n-1=2m-2$.
$n=2m-2+1$.
$n=2m-1$ ---(c).
We now substitute the result obtained from equation (c) in equation (a).
\[\dfrac{\left( a+\dfrac{(2m-1-1)}{2}d \right)}{\left( A+\dfrac{(2m-1-1)}{2}D \right)}=\dfrac{\left( 7\times \left( 2m-1 \right)+1 \right)}{\left( 4\times \left( 2m-1 \right)+27 \right)}\].
\[\dfrac{\left( a+\dfrac{(2m-2)}{2}d \right)}{\left( A+\dfrac{(2m-2)}{2}D \right)}=\dfrac{\left( 14m-7+1 \right)}{\left( 8m-4+27 \right)}\].
\[\dfrac{\left( a+\left( m-1 \right)d \right)}{\left( A+\left( m-1 \right)D \right)}=\dfrac{\left( 14m-6 \right)}{\left( 8m+23 \right)}\].
On comparing numerator and denominator with equation (1), we get
\[\dfrac{{{T}_{m}}}{T_{m}^{'}}=\dfrac{\left( 14m-6 \right)}{\left( 8m+23 \right)}\].
We have found the ratio of the \[{{m}^{th}}\] terms of two progressions as \[\dfrac{14m-6}{8m+23}\].
Note: We should not take the same first term and common difference for both progressions as it will make us confused and give wrong results. Whenever we get this type of problem, we need to first take the ratio and make a substitution that will be fit in order to get the ratio of required terms. We should not confuse the general terms of geometric, arithmetic and harmonic progressions.