Question

If the ratio of the sums of first n terms of two A.P.s is $\frac{5n+13}{7n+27}$, then the ratio of their 4^th terms is ________.

Answer 1

Given:
$$\begin{gathered} \frac{S_n}{S_n^{\text{'}}}=\frac{5n+13}{7n+27} \\ \mathrm{Let}\mathrm{the}\mathrm{first}\mathrm{term}\mathrm{of}{1}^{\mathrm{st}}\mathrm{A}.\mathrm{P}.\mathrm{be}a\mathrm{and}\mathrm{of}{2}^{\mathrm{nd}}\mathrm{A}.\mathrm{P}.\mathrm{be}{a}^{\text{'}}. \\ \mathrm{Let}\mathrm{the}\mathrm{common}\mathrm{difference}\mathrm{of}{1}^{\mathrm{st}}\mathrm{A}.\mathrm{P}.\mathrm{be}d\mathrm{and}\mathrm{of}{2}^{\mathrm{nd}}\mathrm{A}.\mathrm{P}.\mathrm{be}{d}^{\text{'}}. \\ \mathrm{Let}{\mathrm{S}}_{\mathrm{n}}\mathrm{be}\mathrm{the}\mathrm{sum}\mathrm{of}n\mathrm{terms}\mathrm{of}{1}^{\mathrm{st}}\mathrm{A}.\mathrm{P}.\mathrm{and}{\mathrm{S}}_{\mathrm{n}}^{\text{'}}\mathrm{be}\mathrm{the}\mathrm{sum}\mathrm{of}n\mathrm{terms}\mathrm{of}{2}^{\mathrm{nd}}\mathrm{A}.\mathrm{P}.. \\ \mathrm{Now}, \\ \frac{S_n}{S_n^{\text{'}}}=\frac{{\displaystyle \frac{n}{2}}\left[2a+\left(n-1\right)d\right]}{{\displaystyle \frac{n}{2}}\left[2a^{\text{'}}+\left(n-1\right)d^{\text{'}}\right]} \\ \Rightarrow \frac{5n+13}{7n+27}=\frac{2a+\left(n-1\right)d}{2a^{\text{'}}+\left(n-1\right)d^{\text{'}}}....\left(1\right) \\ \mathrm{Let}{a}_4\mathrm{be}\mathrm{the}{4}^{\mathrm{th}}\mathrm{term}\mathrm{of}{1}^{\mathrm{st}}\mathrm{A}.\mathrm{P}.\mathrm{and}{a}_4^{\text{'}}\mathrm{be}\mathrm{the}{4}^{\mathrm{th}}\mathrm{term}\mathrm{of}{2}^{\mathrm{nd}}\mathrm{A}.\mathrm{P}.. \\ \frac{a_4}{a_4^{\text{'}}}=\frac{a+\left(4-1\right)d}{a^{\text{'}}+\left(4-1\right)d^{\text{'}}} \\ =\frac{a+3d}{a^{\text{'}}+3d^{\text{'}}} \\ \mathrm{Multiplying}\mathrm{and}\mathrm{dividing}\mathrm{the}\mathrm{RHS}\mathrm{by}2,\mathrm{we}\mathrm{get} \\ \frac{a_4}{a_4^{\text{'}}}=\frac{2a+6d}{2a^{\text{'}}+6d^{\text{'}}}....\left(2\right) \\ \mathrm{Putting}\mathrm{n}=7\mathrm{in}\left(1\right),\mathrm{we}\mathrm{get} \\ \frac{5\left(7\right)+13}{7\left(7\right)+27}=\frac{2a+\left(7-1\right)d}{2a^{\text{'}}+\left(7-1\right)d^{\text{'}}} \\ \Rightarrow \frac{35+13}{49+27}=\frac{2a+6d}{2a^{\text{'}}+6d^{\text{'}}} \\ \Rightarrow \frac{2a+6d}{2a^{\text{'}}+6d^{\text{'}}}=\frac{48}{76}=\frac{12}{19}....\left(3\right) \\ \mathrm{From}\left(2\right)\mathrm{and}\left(3\right),\mathrm{we}\mathrm{get} \\ \frac{a_4}{a_4^{\text{'}}}=\frac{12}{19} \end{gathered}$$

Hence, the ratio of their 4^th terms is 12 : 19.