An A.P. whose first term is unity and in which the sum of first half of any even number of terms to that of second half of the same number of terms is a constant ratio, then the common difference is:
A. 2
B. 1
C. 3
D. None of these

Hint: In this case, we have to discover the common distinction between the Arithmetic progression and the series. Then we take a few terms from each series to locate two common terms that will form our new common term series. Using the two terms in the new series, we determine the common variance of the new series and its total using the sum of n terms formula.

Formula Used: N term’s sum can be determined by
\[{S_n} = \dfrac{n}{2}(2a + (n - 1)d)\]

Complete step by step solution: We have been provided in the question that,
An A.P. with a constant ratio of the summation of the first half of any equal number of terms to the second half of the same number of terms
That implies,
\[{S_n} = \dfrac{n}{2}(2a + (n - 1)d)\]
Now we have to rewrite the above formula in terms of \[2n\] we get
\[{S_{2n}} = \dfrac{n}{2}(2a + (n - 1)d)\]
As per the given statement, we can write as
\[\dfrac{{{{\rm{S}}_{\rm{n}}}}}{{{{\rm{S}}_{2{\rm{n}}}} - {{\rm{S}}_{\rm{n}}}}} = {\rm{ constant }} = {\rm{k}}\]
Or we can rewrite the above as,
\[\dfrac{{{{\rm{S}}_{2{\rm{n}}}} - {{\rm{S}}_{\rm{n}}}}}{{{{\rm{S}}_{\rm{n}}}}} = {\rm{k}}\]
On replacing the value for \[{{\rm{S}}_{\rm{n}}}\] we get
\[ \Rightarrow \dfrac{{{{\rm{S}}_{2{\rm{n}}}}}}{{{{\rm{S}}_{\rm{n}}}}} - 1 = {\rm{k}}\]
Now, we have to move the constant 1 to the other side of the equation, we get
\[\therefore \dfrac{{{{\rm{S}}_{2{\rm{n}}}}}}{{{{\rm{S}}_{\rm{n}}}}} = {\rm{k}} + 1\]
Now on applying the values of equation (1) and equation (2), we get
\[\dfrac{{\dfrac{{2n}}{2}[2a + (2n - 1)d]}}{{\dfrac{n}{2}[2a + (n - 1)d]}} = k + 1\]
Now, we have to reduce the above expression by replacing with k, therefore we get
\[d = \dfrac{{2(k - 1)}}{{n(3 - k) + k - 1}}\]
Because the ratio is constant, the factor of n must be 0
Therefore, we have
\[3 - {\rm{k}} = 0\]
On simplification, we get
\[k = 3\]
Now, we have to substitute the value of k, so that we obtain
\[d = \dfrac{{2(3 - 1)}}{{n(3 - 3) + 3 - 1}}\]
On simplifying the above expression, we get
\[{\rm{d}} = 2\]
Therefore, the common difference is \[{\rm{d}} = 2\]

Option ‘A’ is correct

Note: It is essential for students to note that forming a general sequence for the nth terms of all the A.P.’s are critical. If we do not create any sequence, we will become perplexed by the terms and will be unable to use the method for the sum of 'n' terms of an A.P.