Let $$S_k = \frac{1+2+3+\ldots+k}{k}$$. If $$S_1^2 + S_2^2 + \ldots + S_{10}^2 = \frac{5}{12}A$$, then A is equal to:

First, let's simplify the expression for $$S_k$$. The numerator is the standard sum of the first $$k$$ natural numbers:

$$S_k = \frac{\frac{k(k+1)}{2}}{k} = \frac{k+1}{2}$$

Now, square $$S_k$$ to find the general term of our series:

$$S_k^2 = \left( \frac{k+1}{2} \right)^2 = \frac{(k+1)^2}{4}$$

We need to evaluate the sum from $$k=1$$ to $$10$$:

$$\sum_{k=1}^{10} S_k^2 = \frac{1}{4} \sum_{k=1}^{10} (k+1)^2$$

$$= \frac{1}{4} (2^2 + 3^2 + 4^2 + \dots + 11^2)$$

This is simply the sum of the squares of the first 11 natural numbers, minus the first term ($$1^2$$).

Let's use the standard formula $$\sum n^2 = \frac{n(n+1)(2n+1)}{6}$$ for $$n=11$$:

$$\sum_{n=1}^{11} n^2 = \frac{11(12)(23)}{6} = 11 \times 2 \times 23 = 506$$

So, the sum of our series is:

$$\frac{1}{4} (506 - 1^2) = \frac{505}{4}$$

We are given that this sum equals $$\frac{5}{12} A$$. Equating the two:

$$\frac{505}{4} = \frac{5}{12} A$$

$$A = \frac{505 \times 12}{4 \times 5}$$

$$A = 101 \times 3 = 303$$