Prove that the sum of n arithmetic means between a and b is n times their arithmetic mean.

Let $A_1, A_2, \dots, A_n$ be the $n$ A.Ms between $a$ and $b$.
This forms an Arithmetic Progression (A.P.): $a, A_1, A_2, \dots, A_n, b$.
The first term of this A.P. is $a_1 = a$.
The total number of terms in this A.P. is $N = n+2$.
The last term is $a_{n+2} = b$.

Using the formula for the $k$-th term of an A.P., $a_k = a_1 + (k-1)d$:
For the $(n+2)$-th term:

Solving for $d$:

This matches the derivation in the image.

The sum of an A.P. with $k$ terms is $S_k = \frac{k}{2}(\text{first term} + \text{last term})$.
Here, we are summing the $n$ terms $A_1, \dots, A_n$.
The first term of this sub-sequence is $A_1$, and the last term is $A_n$.
So, the sum is $S_n = \frac{n}{2}(A_1 + A_n)$.

Now, express $A_1$ and $A_n$ in terms of $a$ and $d$:
$A_1$ is the second term of the main A.P., so $A_1 = a_1 + d = a + d$.
$A_n$ is the $(n+1)$-th term of the main A.P., so $A_n = a_1 + nd = a + nd$.

Substitute these into the sum formula:

This matches the image.

Substitute $d = \frac{b-a}{n+1}$ into the expression for $S_n$:

The $(n+1)$ terms cancel out:

This matches the image.

The Arithmetic Mean (A.M.) between $a$ and $b$ is $\frac{a+b}{2}$.
From the previous step, we have:

This completes the proof.

The solution provided in the image is correct.