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Hint: Use the formula of sum of â€˜nâ€™ 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 â€˜aâ€™ and â€˜bâ€™ respectively,

Therefore from given condition we can write,

$a+b=2\dfrac{1}{6}$

$\therefore a+b=\dfrac{13}{6}$ â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦ (1)

Assume the even number of arithmetic means inserted between â€˜aâ€™ and â€˜bâ€™ be â€˜2Nâ€™.

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 â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦ (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 â€˜nâ€™ 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 â€˜aâ€™ â€˜2N arithmetic meansâ€™ and â€˜bâ€™ all are in Arithmetic Progression with, â€˜aâ€™ as its first term and â€˜bâ€™ 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,

$\therefore {{S}_{[2N+2]}}=\dfrac{[2N+2]}{2}\left[ \dfrac{13}{6} \right]$

$\therefore {{S}_{[2N+2]}}=\dfrac{2[N+1]}{2}\left[ \dfrac{13}{6} \right]$

$\therefore {{S}_{[2N+2]}}=[N+1]\left[ \dfrac{13}{6} \right]$â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦. (3)

Now we can easily find out the sum of arithmetic means by simply just subtracting the sum of â€˜aâ€™ and â€˜bâ€™ 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 $=\dfrac{13}{6}N$â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦ (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 wonâ€™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,â€¦â€¦â€¦â€¦.,A+(2N-1)d, b. Where â€˜Aâ€™ is the first mean and â€˜A+(2N-1)dâ€™ be the â€˜2N thâ€™ mean and likewise you can apply the conditions.

Complete step-by-step answer:

To solve the above problem we will first assume the two numbers as â€˜aâ€™ and â€˜bâ€™ respectively,

Therefore from given condition we can write,

$a+b=2\dfrac{1}{6}$

$\therefore a+b=\dfrac{13}{6}$ â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦ (1)

Assume the even number of arithmetic means inserted between â€˜aâ€™ and â€˜bâ€™ be â€˜2Nâ€™.

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 â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦ (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 â€˜nâ€™ 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 â€˜aâ€™ â€˜2N arithmetic meansâ€™ and â€˜bâ€™ all are in Arithmetic Progression with, â€˜aâ€™ as its first term and â€˜bâ€™ 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,

$\therefore {{S}_{[2N+2]}}=\dfrac{[2N+2]}{2}\left[ \dfrac{13}{6} \right]$

$\therefore {{S}_{[2N+2]}}=\dfrac{2[N+1]}{2}\left[ \dfrac{13}{6} \right]$

$\therefore {{S}_{[2N+2]}}=[N+1]\left[ \dfrac{13}{6} \right]$â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦. (3)

Now we can easily find out the sum of arithmetic means by simply just subtracting the sum of â€˜aâ€™ and â€˜bâ€™ 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 $=\dfrac{13}{6}N$â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦ (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 wonâ€™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,â€¦â€¦â€¦â€¦.,A+(2N-1)d, b. Where â€˜Aâ€™ is the first mean and â€˜A+(2N-1)dâ€™ be the â€˜2N thâ€™ mean and likewise you can apply the conditions.

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