Question

Find the $5$ numbers of an AP whose sum is $12\dfrac{1}{2}$ and the ratio of the first term to last term is $2:3$.

Answer 1

Hint: Arithmetic Progression, also known as AP, is a sequence of numbers such that the difference between the consecutive numbers is constant. This difference is commonly termed as the common difference and is denoted by $d$. The first term of an AP is denoted by $a$. The last term of an AP is denoted by $l$. The formula to find the ${{n}^{th}}$ term of an AP is ${{a}_{n}}=a+\left( n-1 \right)d$ where $n$ is the number of the term which we desire to find out.
Complete step-by-step solution:
We saw that the formula to find the ${{n}^{th}}$term of an AP is ${{a}_{n}}=a+\left( n-1 \right)d$.
So to find the second term of an AP, we have to plug-in $n=2$.
Upon doing so, we get the following :
$\begin{align}
  & \Rightarrow {{a}_{n}}=a+\left( n-1 \right)d \\
 & \Rightarrow {{a}_{2}}=a+\left( 2-1 \right)d \\
 & \Rightarrow {{a}_{2}}=a+d \\
\end{align}$
In the same way to find out the third term of the AP, we plug-in $n=3$ to find out.
So in the question, we are given the sum of the first five numbers of an AP.
We can do it using the formula for sum of $n$ numbers of an AP which is ${{S}_{n}}=\dfrac{n}{2}\left( 2a+\left( n-1 \right)d \right)$ where ${{S}_{n}}$ denotes the sum.
But now let us solve the sum by taking some terms to be in AP and then solving them. Either way is correct.
Let us take the terms $a-2d,a-d,a,a+d,a+2d$. We can clearly see that these terms are in AP and their common difference is +d.
In the question, we are given the sum of the first $5$ numbers to be $12\dfrac{1}{2}$ which is equal to $\dfrac{25}{2}$.
So let us add the terms which we took and equal them to $\dfrac{25}{2}$.
Upon doing so, we get the following :
$\begin{align}
  & \Rightarrow a-2d+a-d+a+a+d+a+2d=\dfrac{25}{2} \\
 & \Rightarrow 5a=\dfrac{25}{2} \\
 & \Rightarrow a=\dfrac{5}{2} \\
\end{align}$
We got the value of $a$ and it is $\dfrac{5}{2}$.
Now we have to find out the common difference. Let us look at the other piece of information we are provided with. We have the ratio of the first to last term and there are $5$numbers in this AP. It means that ${{a}_{5}}$ is the last term.
And $a+2d={{a}_{5}}$. This is the last term.
We are given the ratio of the first term to last term is $2:3$.
So $\dfrac{a-2d}{a+2d}=\dfrac{2}{3}$ .
Let us cross-multiple.
Upon doing so, we get the following :
$\begin{align}
  & \Rightarrow \dfrac{a-2d}{a+2d}=\dfrac{2}{3} \\
 & \Rightarrow 3\left( a-2d \right)=2\left( a+2d \right) \\
 & \Rightarrow 3a-6d=2a+4d \\
 & \Rightarrow a=10d \\
\end{align}$
Now let us substitute the value of $a=\dfrac{5}{2}$.
Upon doing so, we get the following :
$\begin{align}
  & \Rightarrow \dfrac{a-2d}{a+2d}=\dfrac{2}{3} \\
 & \Rightarrow 3\left( a-2d \right)=2\left( a+2d \right) \\
 & \Rightarrow 3a-6d=2a+4d \\
 & \Rightarrow a=10d \\
 & \Rightarrow \dfrac{5}{2}=10d \\
 & \Rightarrow d=\dfrac{1}{4} \\
\end{align}$
So the common difference of our AP is $\dfrac{1}{4}$.
So our 5 terms are:
 \[\begin{align}
  & \Rightarrow a-2d=\dfrac{5}{2}-2\left( \dfrac{1}{4} \right)=\dfrac{5}{2}-\dfrac{1}{2}=2. \\
 & \Rightarrow a-d=\dfrac{5}{2}-\left( \dfrac{1}{4} \right)=\dfrac{10}{4}-\dfrac{1}{4}=\dfrac{9}{4} \\
 & \Rightarrow a=\dfrac{5}{2} \\
 & \Rightarrow a+d=\dfrac{5}{2}+\left( \dfrac{1}{4} \right)=\dfrac{10}{4}+\dfrac{1}{4}=\dfrac{11}{4} \\
 & \Rightarrow a+2d=\dfrac{5}{2}+2\left( \dfrac{1}{4} \right)=\dfrac{5}{2}+\dfrac{1}{2}=3 \\
\end{align}\]
\[\therefore \] Hence, the $5$numbers of an AP whose sum is $12\dfrac{1}{2}$ and the ratio of the first term to last term is $2:3$ are $2,\dfrac{9}{4},\dfrac{5}{2},\dfrac{11}{4},3$.

Note: We can also do this using the formula. We just have to plug-in $n=5$ to get the sum of the first $5$ numbers of the AP and then equate it to $\dfrac{25}{2}$ to get a relation between $a$ and $d$ or else we can also take numbers which are already in AP and then solve for them. We should be careful of our first term as in this sum, the first term is not $a$ rather $a-2d$. We can also take terms as $a,a+d,a+2d....$ but the common difference does not cancel. So for the sake of simplicity we took it that way.