Question

The ratio of the sum of first n terms of two AP’s is given as (3n – 13) : (5n – 1) $\forall n\in N$. Find the ratio of their 13th terms.

Answer 1

Hint: We are given the ratio of the first n terms of two AP’s. We will use the formula of sum of n terms of an AP given by the relation ${{S}_{n}}=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]$, where a is the first term and d is the common difference. We will assume variables for the first term and the common differences of the AP’s. We will then compare the ratio of the formula to the given ratio. To find the ratio of the 13th term of the respective AP’s, we will first find the 13th term of each AP, given by the relation ${{T}_{n}}=a+\left( n-1 \right)d$. Then we will compare the two ratios to get a value of n such that the ratio of the 13th term is equal to ratio of sum of AP up to n terms. Once, we get that value of n, we can find the ratio of the 13th term equating the two ratios.

Complete step-by-step answer:
The ratio of the first n terms of two AP’s is given to us as (3n – 13) : (5n – 1). Let the first term of the one of the AP’s be ${{a}_{1}}$ and the common difference be ${{d}_{1}}$. Similarly, let the first term of the second AP be ${{a}_{2}}$ and the common difference be ${{d}_{2}}$.
We know the sum of the first n terms of an AP is given by the relation ${{S}_{n}}=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]$, where a is the first term and d is the common difference.
Therefore, sum of first n terms of first AP is given as ${{S}_{n1}}=\dfrac{n}{2}\left[ 2{{a}_{1}}+\left( n-1 \right){{d}_{1}} \right]$.
Similarly, sum of the first n terms of second AP is given as ${{S}_{n2}}=\dfrac{n}{2}\left[ 2{{a}_{2}}+\left( n-1 \right){{d}_{2}} \right]$.
The ratio of the sum of n terms of two AP’s will be $\dfrac{{{S}_{n1}}}{{{S}_{n2}}}=\dfrac{\dfrac{n}{2}\left[ 2{{a}_{1}}+\left( n-1 \right){{d}_{1}} \right]}{\dfrac{n}{2}\left[ 2{{a}_{2}}+\left( n-1 \right){{d}_{2}} \right]}$ .
\[\begin{align}
  & \Rightarrow \dfrac{\left[ 2{{a}_{1}}+\left( n-1 \right){{d}_{1}} \right]}{\left[ 2{{a}_{2}}+\left( n-1 \right){{d}_{2}} \right]}=\dfrac{\left( 3n-13 \right)}{\left( 5n-1 \right)} \\
 & \Rightarrow \dfrac{{{a}_{1}}+\left( \dfrac{n-1}{2} \right){{d}_{1}}}{{{a}_{1}}+\left( \dfrac{n-1}{2} \right){{d}_{1}}}=\dfrac{3n-13}{5n-1}......\left( 1 \right) \\
\end{align}\]
Now, we are supposed to find the ratio of the 13th term of the AP’s.
The nth term of an AP is given by the relation ${{T}_{n}}=a+\left( n-1 \right)d$, where a is the first term and d is the common difference.
Therefore, 13th term of the first AP will be ${{T}_{13\left( 1 \right)}}={{a}_{1}}+12{{d}_{1}}$.
Similarly, 13th term of the second AP will be ${{T}_{13\left( 2 \right)}}={{a}_{2}}+12{{d}_{2}}$.
The ratio of the 13th term of the two AP’s is given as follows:
\[\Rightarrow \dfrac{{{T}_{13\left( 1 \right)}}}{{{T}_{13\left( 2 \right)}}}=\dfrac{{{a}_{1}}+12{{d}_{1}}}{{{a}_{2}}+12{{d}_{2}}}......\left( 2 \right)\]
Now, we will compare (1) and (2) for the value of n.
$\begin{align}
  & \Rightarrow 12=\dfrac{n-1}{2} \\
 & \Rightarrow n=25 \\
\end{align}$
This means, the ratio of the sum of the first 25 terms of AP’s is equal to the ratio of 13th term of the AP’s.
$\begin{align}
  & \Rightarrow \dfrac{{{T}_{13\left( 1 \right)}}}{{{T}_{13\left( 2 \right)}}}=\dfrac{3\left( 25 \right)-13}{5\left( 25 \right)-1} \\
 & \Rightarrow \dfrac{{{T}_{13\left( 1 \right)}}}{{{T}_{13\left( 2 \right)}}}=\dfrac{75-13}{125-1} \\
 & \Rightarrow \dfrac{{{T}_{13\left( 1 \right)}}}{{{T}_{13\left( 2 \right)}}}=\dfrac{62}{124} \\
 & \Rightarrow \dfrac{{{T}_{13\left( 1 \right)}}}{{{T}_{13\left( 2 \right)}}}=\dfrac{1}{2} \\
\end{align}$
Therefore, the ratio of the 13th term of the two AP’s is 1 : 2.

Note: The trick for solving problems based on AP is to use the formula correctly. Students are advised to by heart the formulas of the nth term of AP and sum of first n terms of AP.