Two arithmetic progressions (A.Ps) contain the same number of terms. The ratio of the last term of the first progression to the first term of the second is equal to the ratio of the last term of the second progression to the first term of the first progression and is equal to $4$. The ratio of the sum of the first progression to that of the second is $2$. Find the ratio of the differences of the progressions.
Let the number of terms in the two A.Ps be $= n$
Let the $n^{th}$ term be the last term of both the progressions.
For first A.P:
$1^{st}$ term $= t_{11} = a_{11}$, $\;$ common difference $= d_1$
Sum of all terms $= S_{1n} = \dfrac{n}{2} \left[2a_{11} + \left(n - 1\right) d_1\right]$
For second A.P:
$1^{st}$ term $= t_{21} = a_{21}$, $\;$ common difference $= d_2$
Sum of all terms $= S_{2n} = \dfrac{n}{2} \left[2a_{21} + \left(n - 1\right) d_2\right]$
To find: $\;$ $\dfrac{d_1}{d_2}$
Given: $\;$ $\dfrac{t_{1n}}{t_{21}} = \dfrac{t_{2n}}{t_{11}} = 4$
i.e. $\;$ $\dfrac{a_{11} + \left(n - 1\right) d_1}{a_{21}} = \dfrac{a_{21} + \left(n - 1\right) d_2}{a_{11}} = 4$ $\;\;\; \cdots \; (1)$
$\implies$ $a_{11} + \left(n - 1\right) d_1 = 4 a_{21}$ $\;\;\; \cdots \; (2)$
and $\;$ $a_{21} + \left(n - 1\right)d_2 = 4 a_{11}$ $\;\;\; \cdots \; (3)$
Also: $\;$ $\dfrac{S_{1n}}{S_{2n}} = 2$
i.e. $\;$ $\dfrac{\dfrac{n}{2} \left[2a_{11} + \left(n - 1\right) d_1\right]}{\dfrac{n}{2} \left[2a_{21} + \left(n - 1\right) d_2\right]} = 2$
i.e. $\;$ $\dfrac{2a_{11} + \left(n - 1\right) d_1}{2 a_{21} + \left(n - 1\right) d_2} = 2$
i.e. $\;$ $\dfrac{a_{11} + a_{11} + \left(n - 1\right) d_1}{a_{21} + a_{21} + \left(n -1\right)d_2} = 2$
i.e. $\;$ $\dfrac{a_{11} + 4a_{21}}{a_{21} + 4 a_{11}} = 2$ $\;\;$ [By equations $(2)$ and $(3)$]
i.e. $\;$ $a_{11} + 4 a_{21} = 2 a_{21} + 8 a_{11}$
i.e. $\;$ $2 a_{21} = 7 a_{11}$
i.e. $\;$ $a_{21} = \dfrac{7}{2} a_{11}$ $\;\;\; \cdots \; (4)$
In view of equation $(4)$, equation $(3)$ becomes
$\dfrac{7}{2} a_{11} + \left(n - 1\right) d_2 = 4 a_{11}$
i.e. $\;$ $\left(n - 1\right) d_2 = 4a_{11} - \dfrac{7}{2} a_{11} = \dfrac{a_{11}}{2}$ $\;\;\; \cdots \; (5)$
In view of equation $(4)$, equation $(2)$ becomes
$a_{11} + \left(n - 1\right) d_1 = 4 \times \dfrac{7}{2} a_{11} = 14 a_{11}$
i.e. $\;$ $\left(n - 1\right) d_1 = 13 a_{11}$ $\;\;\; \cdots \; (6)$
$\therefore \;$ We have from equations $(5)$ and $(6)$
$\dfrac{\left(n - 1\right) d_1}{\left(n - 1\right) d_2} = \dfrac{13 a_{11}}{a_{11} / 2}$
$\implies$ $\dfrac{d_1}{d_2} = 26$