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 these expressions as \( k \). Thus, we can write:

\[ a + 1 = k \quad (1) \]
\[ b + 2 = k \quad (2) \]
\[ c + 3 = k \quad (3) \]
\[ d + 4 = k \quad (4) \]
\[ e + 5 = k \quad (5) \]
\[ a + b + c + d + e + 3 = k \quad (6) \]

From equations (1) to (5), we can express \( a, b, c, d, \) and \( e \) in terms of \( k \):

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

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

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

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

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

Now, substituting these expressions into equation (6):

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

Simplifying the left-hand side:

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

Now, we can isolate \( k \):

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

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

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. \( e = k - 5 = 3 - 5 = -2 \)

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

\[ a^2 + b^2 + c^2 + d^2 + e^2 = 2^2 + 1^2 + 0^2 + (-1)^2 + (-2)^2 \]
\[ = 4 + 1 + 0 + 1 + 4 \]
\[ = 10 \]

Thus, the final answer is:

\[ \boxed{10} \]