Question

If $8,-4$ and $13$ be three (not necessarily consecutive terms) of an A.P., how many such A.P.s are possible?
A. $1$
B. $2$
C. Infinitely many
D. No such A.P. is possible

Answer 1

Hint: In this question, we are to find the possibilities of arithmetic progression for the given three terms. Since it is given that, these three terms of an arithmetic series are not necessarily consecutive terms and they are of either increasing or decreasing order. So, we can assume any one of them as the first term and the other two as $m^{th}$ and $n^{th}$ terms. In this way, we can conclude the series with one of the given options.

Formula used: If the series is an Arithmetic series, then the sum of the $n$ terms of the arithmetic series is calculated by
${{S}_{n}}=\dfrac{n}{2}\left[ 2a+(n-1)d \right]$
Where the common difference $d={{a}_{n}}-{{a}_{n-1}}$
The general or the $n^{th}$ term of the arithmetic series is
${{t}_{n}}=a+(n-1)d$
Here $a$ - The first term; ${{t}_{n}}$ - $n^{th}$ term;

Complete step by step solution: Given that, $8,-4$, and $13$ be three terms of an arithmetic series. They are not consecutive terms. So, we can assume the first term as one of them and the other two as $m^{th}$ and $n^{th}$ terms.
Consider if the series is in increasing order, then the first term will be $a=-4$. Then, the $m^{th}$ and $n^{th}$ terms are $8$ and $13$ respectively.
I.e.,
$a=-4;{{t}_{m}}=8;{{t}_{n}}=13$
From the general term formula, we can write the above terms as
${{t}_{n}}=a+(n-1)d$
$\begin{align}
  & {{t}_{m}}=8=-4+(m-1)d \\
 & {{t}_{n}}=13=-4+(n-1)d \\
\end{align}$
So, on simplifying, we get
$\begin{align}
  & 8+4=(m-1)d \\
 & \Rightarrow d=\dfrac{12}{(m-1)}\text{ }...(1) \\
\end{align}$
And
$\begin{align}
  & 13+4=(n-1)d \\
 & \Rightarrow d=\dfrac{17}{(n-1)}\text{ }...(2) \\
\end{align}$
From (1) and (2), we get
$\begin{align}
  & \dfrac{12}{m-1}=\dfrac{17}{n-1} \\
 & \Rightarrow \dfrac{m-1}{12}=\dfrac{n-1}{17}=k \\
 & \Rightarrow m-1=12k;n-1=17k \\
 & \Rightarrow m=12k+1;n=17k+1 \\
\end{align}$
For $k=1,2,3,...\infty $, we get different values of $m$ and $n$. So, it shows that there is an infinite number of A.P.s will be possible.

Thus, Option (C) is correct.

Note: Here we need to remember that, the given terms are not consecutive terms of the series. So, we can assume any one of them is the first term. But it depends on the increasing or decreasing order since it is in arithmetic progression. Here we choose the increasing order, but we can also go for the decreasing order. In which, the first term becomes $a=13$ since it is the bigger one in the given terms.