Let `S_(k) = (1+2+3+...+k)/(k)`. If `S_(1)^(2) + s_(2)^(2) +...+S_(10)^(2) = (5)/(12)A`, then A is equal to

Options

A. 303
B. 283
C. 156
D. 301

Answer

To solve the problem, we need to find the value of \( A \) given that

\[ S_1^2 + S_2^2 + \ldots + S_{10}^2 = \frac{5}{12} A \]

where

\[ S_k = \frac{1 + 2 + 3 + \ldots + k}{k} \]

### Step-by-Step Solution:

1. **Calculate \( S_k \)**:
The sum of the first \( k \) natural numbers is given by the formula:

\[ 1 + 2 + 3 + \ldots + k = \frac{k(k + 1)}{2} \]

Therefore,

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

2. **Calculate \( S_k^2 \)**:
Now, we need to find \( S_k^2 \):

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

3. **Sum \( S_k^2 \) from \( k = 1 \) to \( k = 10 \)**:
We need to compute:

\[ S_1^2 + S_2^2 + \ldots + S_{10}^2 = \sum_{k=1}^{10} S_k^2 = \sum_{k=1}^{10} \frac{(k + 1)^2}{4} \]

This can be simplified as:

\[ = \frac{1}{4} \sum_{k=1}^{10} (k + 1)^2 = \frac{1}{4} \sum_{k=2}^{11} k^2 \]

4. **Calculate \( \sum_{k=2}^{11} k^2 \)**:
The formula for the sum of the squares of the first \( n \) natural numbers is:

\[ \sum_{k=1}^{n} k^2 = \frac{n(n + 1)(2n + 1)}{6} \]

Thus,

\[ \sum_{k=1}^{11} k^2 = \frac{11 \cdot 12 \cdot 23}{6} = \frac{3036}{6} = 506 \]

Now, we need to subtract \( 1^2 \) from this sum:

\[ \sum_{k=2}^{11} k^2 = 506 - 1 = 505 \]

5. **Final Calculation**:
Now substituting back, we have:

\[ S_1^2 + S_2^2 + \ldots + S_{10}^2 = \frac{1}{4} \cdot 505 = \frac{505}{4} \]

Setting this equal to \( \frac{5}{12} A \):

\[ \frac{505}{4} = \frac{5}{12} A \]

6. **Solve for \( A \)**:
Cross-multiplying gives:

\[ 505 \cdot 12 = 5 \cdot 4A \]

Simplifying:

\[ 6060 = 20A \implies A = \frac{6060}{20} = 303 \]

Thus, the value of \( A \) is

\[ \boxed{303} \]