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Between two numbers whose sum is $2\dfrac{1}{6}$ an even number of arithmetic means are inserted; the sum of these means exceeds their number by unity: how many means are there?

Answer Verified Verified
Hint: Use the formula of sum of ‘n’ terms in A.P i.e. ${{S}_{n}}=\dfrac{n}{2}[a+l]$ and subtract the sum of two numbers which is given and then equate it with the given condition to get the final answer.

Complete step-by-step answer:
To solve the above problem we will first assume the two numbers as ‘a’ and ‘b’ respectively,
Therefore from given condition we can write,
$a+b=2\dfrac{1}{6}$
$\therefore a+b=\dfrac{13}{6}$ ……………………………………… (1)
Assume the even number of arithmetic means inserted between ‘a’ and ‘b’ be ‘2N’.
As given in the problem that the sum of means exceeds their number by unity therefore we can write,
Sum of Arithmetic Means = 2N+1 …………………… (2)
Now to satisfy above condition we should find the sum of arithmetic means and for that we should know the formula given below,
Formula:
Sum of ‘n’ numbers for an Arithmetic Progression is given by,
${{S}_{n}}=\dfrac{n}{2}\left[ a+l \right]$
Where, a- First term of A.P.
     l - Last term of A.P.
As the arithmetic means are inserted in between the two numbers therefore we can say that ‘a’ ‘2N arithmetic means’ and ‘b’ all are in Arithmetic Progression with, ‘a’ as its first term and ‘b’ as its last term.
Therefore by using the formula given above we can write the sum of A.P as shown below,
${{S}_{b}}=\dfrac{b}{2}\left[ a+b \right]$
Now we can express the number of ${{b}^{th}}$ term as [2N+2] therefore we can replace it with b in above equation,
${{S}_{[2N+2]}}=\dfrac{[2N+2]}{2}\left[ a+b \right]$
If we put the value of equation (1) in above equation we will get,
$\therefore {{S}_{[2N+2]}}=\dfrac{[2N+2]}{2}\left[ \dfrac{13}{6} \right]$
$\therefore {{S}_{[2N+2]}}=\dfrac{2[N+1]}{2}\left[ \dfrac{13}{6} \right]$
$\therefore {{S}_{[2N+2]}}=[N+1]\left[ \dfrac{13}{6} \right]$………………………………………. (3)
Now we can easily find out the sum of arithmetic means by simply just subtracting the sum of ‘a’ and ‘b’ from ${{S}_{[2N+2]}}$ therefore we will get,
Sum of Arithmetic Means $={{S}_{[2N+2]}}-(a+b)$
By substituting the values of equation (1) and equation (3) in above equation we will get,
Sum of Arithmetic Means $=[N+1]\left[ \dfrac{13}{6} \right]-\left[ \dfrac{13}{6} \right]$
By taking $\dfrac{13}{6}$ common we will get,
Sum of Arithmetic Means $=\dfrac{13}{6}\left[ \left( N+1 \right)-1 \right]$
Sum of Arithmetic Means $=\dfrac{13}{6}N$……………………………… (4)
Now to get the final answer we will put the value of equation (4) in equation (2) therefore we will get,
$\therefore \dfrac{13}{6}N=2N+1$
$\therefore \dfrac{13}{6}N-2N=1$
$\therefore N\left( \dfrac{13}{6}-2 \right)=1$
$\therefore N\left( \dfrac{13-6\times 2}{6} \right)=1$
\[\therefore N\left( \dfrac{13-12}{6} \right)=1\]
\[\therefore N\left( \dfrac{1}{6} \right)=1\]
\[\therefore N=6\]
As we have assumed the number of means inserted to be 2N therefore we can write,
Total numbers of means inserted = 2N \[=2\times 6\] \[=12\]
Therefore the total number of means inserted between the two numbers is 16.


Note: Do remember to subtract the sum of two numbers from ${{S}_{[2N+2]}}$ otherwise the answer will become wrong. If you won’t get the method then you can solve this problem by writing an A.P. of all the means and the two numbers, like, a, A,A+d,A+2d,………….,A+(2N-1)d, b. Where ‘A’ is the first mean and ‘A+(2N-1)d’ be the ‘2N th’ mean and likewise you can apply the conditions.

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