Question

The sum of two numbers is $2\dfrac{1}{6}$ . An even number of arithmetic means are being inserted between them to answer that their sum exceeds their number by $1$. Find the number of means inserted.

Answer 1

To begin with, we will define what arithmetic means is and how they can be inserted between two numbers.
We will assume some variable for two numbers whose sum is given 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 inserted means.
It is given that the sum of the means is one more than the numbers of means.

Complete step-by-step answer:
Assuming any number of arithmetic means can be inserted between two numbers $a$ and $b$, such that the series thus formed is in AP.
It is given that the sum of two numbers is $2\dfrac{1}{6}$
First we have to convert a mixed fraction into improper fraction.
That we just multiply the cross terms and add with the numerator.
$2\dfrac{1}{6} = \dfrac{{13}}{6}$
Let us assume that the numbers given in the question are $A + B = \dfrac{{13}}{6}$
Suppose $x$ arithmetic means are inserted between${\text{A and B}}$, where $x$ is an even number.
Let the means be ${x_1},{x_2},{x_3},........{x_x}.$
We know that the sum of an AP with first and the last term and $l$ be the last term is given by the relation, ${S_n} = \dfrac{n}{2}\left[ {a + l} \right]$, Where $a$ is the first term
$A + {x_1} + {x_2} + {x_3}....... + {x_x} + B = \dfrac{{x + 2}}{2}\left[ {A + B} \right]$
$\Rightarrow {x_1} + {x_2} + {x_3}....... + {x_x} = \dfrac{{x + 2}}{2}\left[ {A + B} \right] - \left[ {A + B} \right]$
$\Rightarrow {x_1} + {x_2} + {x_3}.........{x_x} = \left( {\dfrac{{x + 2}}{2} - 1} \right)\left[ {A + B} \right].............(1)$
Now, it is given that the sum of the means is $1$ more than that number of means.
We also know that
$A + B = \dfrac{{13}}{6}$
Substitute in equation $(1)$, we get
$\Rightarrow x + 1 = \left( {\dfrac{{x + 2}}{2} - 1} \right)\left( {\dfrac{{13}}{6}} \right)$
Taking LCM in RHS, at the first term we get,
$\Rightarrow x + 1 = \left( {\dfrac{{x + 2 - 2}}{2}} \right)\left( {\dfrac{{13}}{6}} \right)$
On subtracting the numerator terms we get,
$\Rightarrow x + 1 = \left( {\dfrac{x}{2}} \right)\left( {\dfrac{{13}}{6}} \right)$
Now multiplying the LHS terms we get,
$\Rightarrow x + 1 = \dfrac{{13x}}{{12}}$
Taking variable as same side and we get,
$\Rightarrow x - \dfrac{{13x}}{{12}} = - 1$
Taking LCM we get,
$\Rightarrow \dfrac{{12x - 13x}}{{12}} = - 1$
On subtracting the numerator terms we get,
$\Rightarrow - \dfrac{x}{{12}} = - 1$
Then we can equate that
$\Rightarrow$ $x = 12$

Therefore, the numbers of means to be inserted are $12.$

Note: We can have many numbers of such series. Depending on the first number, the common difference of the series can be calculated.
Any number of arithmetic means can be inserted between two numbers as long as the series is an AP. If an odd number of means are added. The average of the first and last term is the middle of the series.