Question

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

Answer 1

Hint: Here we will first assume the number of means inserted to be 2n and then we will use the concept of arithmetic mean of two numbers and form the equations to solve for n.
The arithmetic mean of two numbers x and y is given by:-
\[AM = \dfrac{{x + y}}{2}\]
Also, the \[n^{th}\] term of an AP is given by:-
\[{T_n} = a + \left( {n - 1} \right)d\]
where a is the first term, n is the total number of terms and d is the common difference.

Complete step-by-step answer:
It is given that the sum of two numbers is \[2\dfrac{1}{6}\].
Let the numbers be x and y then,
\[x + y = 2\dfrac{1}{6}\]
Simplifying it we get:-
\[x + y = \dfrac{{13}}{6}\]……………………………. (1)
Now it is given that even number of arithmetic means are being inserted between them
So, let 2n number of means are inserted, then according to the given condition in the question their sum is 2n+1.
Now we know that,
\[x, x + d, x + 2d........, x + 2n, y\] are in Arithmetic progression.
Now we know that the \[n^{th}\] term of an AP is given by:-
\[{T_n} = a + \left( {n - 1} \right)d\]
Here,\[a = x\], \[d = d\], \[n = 2n + 2\] and \[{T_n} = y\]
Hence putting the respective values in the above formula we get:-
\[y = x + \left( {2n + 2 - 1} \right)d\]
Solving it further we get:-
\[y = x + \left( {2n + 1} \right)d\]……………………….. (2)
Now from equation 1 we get:-
\[y = \dfrac{{13}}{6} - x\]
Putting this value in equation 2 we get:-
\[\dfrac{{13}}{6} - x = x + \left( {2n + 1} \right)d\]
Simplifying it further we get:-
\[2x + \left( {2n + 1} \right)d = \dfrac{{13}}{6}\]………………………. (3)
Now we know that the sum of arithmetic means is given by:-
\[sum = x + d + x + 2d + ...... + x + 2nd\]
Now since it is given that the sum of means is 1 more than the number of means hence,
\[2n + 1 = x + d + x + 2d + ...... + x + 2nd\]
Now we know that the sum of n terms in A.P is given by:-
\[sum = \dfrac{n}{2}\left[ {a + l} \right]\] where a is the first term, l is the last term and n is the total number of terms
Applying this formula in RHS of above equation we get:-
\[2n + 1 = \dfrac{{2n}}{2}\left[ {x + d + x + 2nd} \right]\]
Simplifying it we get:-
\[2n + 1 = n\left[ {2x + \left( {2n + 1} \right)d} \right]\]
Now putting value from equation 3 we get:-
\[2n + 1 = n\left[ {\dfrac{{13}}{6}} \right]\]
Now evaluating the value of n we get:-
\[
  6\left( {2n + 1} \right) = 13n \\
   \Rightarrow 12n + 6 = 13n \\
 \]
Simplifying it further we get:-
\[n = 6\]
Now since 2n means were inserted,
Hence putting the value of n we get:-
Number of means inserted =12

Note: Students should note that the arithmetic mean of n terms is given by:-
\[AM = \dfrac{{{x_1} + {x_2} + {x_3} + ..........{x_n}}}{n}\]
Students should note that we have to take an even number of arithmetic means so we have to multiply the variable taken by 2.
Also, don’t forget to substitute the value of n at the last.