Question

Between two numbers whose sum is $\dfrac{13}{6}$ , an even number of A.M.s is inserted, the sum of these means exceeds their number by unity. Find the number of means.

Answer 1

Hint: To begin with, we will define what arithmetic means and how they can be inserted between two numbers. We will assume some variable for the two numbers whose sum is $\dfrac{13}{6}$ and also assume some variables for those means. We will then define the property of a series of arithmetic means and use that property to find the sum of the inserted means. It is given that the sum of the means is one more than the number of means.

Complete step-by-step answer:
Any number of arithmetic means can be inserted between two numbers a and b, such that the series thus formed is in AP.
Therefore, if 3 means are to be inserted between 3 and 11, then those numbers will be such that they form an arithmetic progression with 3 and 11 as the first and the last term.
The means will be 5, 7 and 9 and the AP will be 3, 5, 7, 9, 11.
 Now, let us assume that the numbers given in the question are A and B.
Therefore, A + B = $\dfrac{13}{6}$.
Suppose x arithmetic means are inserted between A and B, where x is an even number. Let the means be ${{x}_{1}},{{x}_{2}},{{x}_{3}}....{{x}_{x}}$.
We know that the series of the means with A and B as the first and the last term are in AP.
Therefore, the series form is A, ${{x}_{1}},{{x}_{2}},{{x}_{3}}....{{x}_{x}}$, B, where A is the first term, B is the last term and number of terms is x + 2.
We know that the sum of an AP with first and the last term is given by the relation ${{S}_{n}}=\dfrac{n}{2}\left[ a+l \right]$, where a is the first term and l is the last term.
$\begin{align}
  & \Rightarrow A+{{x}_{1}}+{{x}_{2}}+{{x}_{3}}+...+{{x}_{x}}+B=\dfrac{x+2}{2}\left[ A+B \right] \\
 & \Rightarrow {{x}_{1}}+{{x}_{2}}+{{x}_{3}}+...+{{x}_{x}}=\dfrac{x+2}{2}\left[ A+B \right]-\left[ A+B \right] \\
 & \Rightarrow {{x}_{1}}+{{x}_{2}}+{{x}_{3}}+...+{{x}_{x}}=\left( \dfrac{x+2}{2}-1 \right)\left[ A+B \right] \\
\end{align}$
Now, it is given that the sum of the means is 1 more than the number of means. We also know that A + B = $\dfrac{13}{6}$.
$\begin{align}
  & \Rightarrow x+1=\left( \dfrac{x+2}{2}-1 \right)\left( \dfrac{13}{6} \right) \\
 & \Rightarrow x+1=\left( \dfrac{x+2-2}{2} \right)\left( \dfrac{13}{6} \right) \\
 & \Rightarrow x+1=\dfrac{x}{2}\left( \dfrac{13}{6} \right) \\
 & \Rightarrow x-\dfrac{13x}{12}=-1 \\
 & \Rightarrow -\dfrac{x}{12}=-1 \\
 & \Rightarrow x=12 \\
\end{align}$
Therefore, the number of means to be inserted are 12.

Note: Any number of arithmetic means can be inserted between two numbers as long as the series is in AP. If an odd number of means are added. The average of the first and the last term is the middle term of the series.