If real numbers a, b, c, d, e satisfy `a+1=b + 2 = c+ 3 = d + 4 = e + 5 = a + b + c + d + e + 3` then find the value of `a^2 + b^2 + c^2 + d^2 + e^2`

Answer

To solve the problem, we start with the given equations:

1. \( a + 1 = b + 2 = c + 3 = d + 4 = e + 5 = a + b + c + d + e + 3 \)

Let's denote the common value of all these expressions as \( k \). Therefore, we can write:

\[ a + 1 = k \\ b + 2 = k \\ c + 3 = k \\ d + 4 = k \\ e + 5 = k \\ a + b + c + d + e + 3 = k \]

From these equations, we can express \( a, b, c, d, e \) in terms of \( k \):

1. From \( a + 1 = k \):
\[ a = k - 1 \]

2. From \( b + 2 = k \):
\[ b = k - 2 \]

3. From \( c + 3 = k \):
\[ c = k - 3 \]

4. From \( d + 4 = k \):
\[ d = k - 4 \]

5. From \( e + 5 = k \):
\[ e = k - 5 \]

Now we substitute these expressions for \( a, b, c, d, e \) into the equation \( a + b + c + d + e + 3 = k \):

\[ (k - 1) + (k - 2) + (k - 3) + (k - 4) + (k - 5) + 3 = k \]

Simplifying the left side:

\[ (k - 1 + k - 2 + k - 3 + k - 4 + k - 5 + 3) = k \\ 5k - 15 + 3 = k \\ 5k - 12 = k \]

Now, we can bring \( k \) to one side:

\[ 5k - k = 12 \\ 4k = 12 \\ k = 3 \]

Now that we have \( k \), we can find the values of \( a, b, c, d, e \):

1. \( a = k - 1 = 3 - 1 = 2 \)
2. \( b = k - 2 = 3 - 2 = 1 \)
3. \( c = k - 3 = 3 - 3 = 0 \)
4. \( d = k - 4 = 3 - 4 = -1 \)
5. \( e = k - 5 = 3 - 5 = -2 \)

Next, we need to find \( a^2 + b^2 + c^2 + d^2 + e^2 \):

\[ a^2 + b^2 + c^2 + d^2 + e^2 = 2^2 + 1^2 + 0^2 + (-1)^2 + (-2)^2 \]

Calculating each term:

\[ = 4 + 1 + 0 + 1 + 4 = 10 \]

Thus, the final answer is:

\[ \boxed{10} \]