Question

The ratio of the sum of n terms of two AP’s is $\left( 7n+1 \right):\left( 4n+27 \right)$. Find the ratio of ${{m}^{th}}$ terms.

Answer 1

From the given series of arithmetic sequences, we find the general term of the series. We assume two AP sequences and express it in its general form. From the given relation, we find an equation of ratios. We put the value of $n=2m-1$ in the equation.

Complete answer:
We assume two AP sequences and express it in its general form.
We take the first sequence. The first term and common difference is ${{a}_{1}},{{d}_{1}}$ respectively.
For the second sequence, the first term and common difference is ${{a}_{2}},{{d}_{2}}$ respectively.
In general form we express the terms as ${{t}_{n}}$, the ${{n}^{th}}$ term of the series.
The first term be ${{t}_{1}}$ and the common difference be $d$ where $d={{t}_{2}}-{{t}_{1}}={{t}_{3}}-{{t}_{2}}={{t}_{4}}-{{t}_{3}}$.
We can express the general term ${{t}_{n}}$ based on the first term be and the common difference.
The formula being ${{t}_{n}}={{t}_{1}}+\left( n-1 \right)d$.
The general formula for n terms is ${{S}_{n}}=\dfrac{n}{2}\left[ 2{{t}_{1}}+\left( n-1 \right)d \right]$.
It is given that the ratio of the sum of n terms of two AP’s is $\left( 7n+1 \right):\left( 4n+27 \right)$.
So, $\dfrac{\dfrac{n}{2}\left[ 2{{a}_{1}}+\left( n-1 \right){{d}_{1}} \right]}{\dfrac{n}{2}\left[ 2{{a}_{2}}+\left( n-1 \right){{d}_{2}} \right]}=\dfrac{2{{a}_{1}}+\left( n-1 \right){{d}_{1}}}{2{{a}_{2}}+\left( n-1 \right){{d}_{2}}}=\dfrac{7n+1}{4n+27}$.
We need to find the ratio of ${{m}^{th}}$ terms which is $\dfrac{{{a}_{1}}+\left( m-1 \right){{d}_{1}}}{{{a}_{2}}+\left( m-1 \right){{d}_{2}}}$.
We use the replacement of $n=2m-1$ in the equality of $\dfrac{2{{a}_{1}}+\left( n-1 \right){{d}_{1}}}{2{{a}_{2}}+\left( n-1 \right){{d}_{2}}}=\dfrac{7n+1}{4n+27}$.

& \dfrac{2{{a}_{1}}+\left( 2m-1-1 \right){{d}_{1}}}{2{{a}_{2}}+\left( 2m-1-1 \right){{d}_{2}}}=\dfrac{7\left( 2m-1 \right)+1}{4\left( 2m-1 \right)+27} \\\ & \Rightarrow \dfrac{2\left[ {{a}_{1}}+\left( m-1 \right){{d}_{1}} \right]}{2\left[ {{a}_{2}}+\left( m-1 \right){{d}_{2}} \right]}=\dfrac{14m-7+1}{8m-4+27} \\\ & \Rightarrow \dfrac{{{a}_{1}}+\left( m-1 \right){{d}_{1}}}{{{a}_{2}}+\left( m-1 \right){{d}_{2}}}=\dfrac{14m-6}{8m+23} \\\ \end{aligned}$$ **Therefore, the ratio of ${{m}^{th}}$ terms is $$\dfrac{14m-6}{8m+23}$$.** **Note:** The sequence is a decreasing or increasing sequence depending on the common difference being a negative or positive number respectively. The common difference will never be calculated according to the difference of greater number from the lesser number.