Question

The sum of the squares of three consecutive integers is 194. Find the integers.

• A. 7, 8, 9
• B. 6, 7, 8
• C. 5, 6, 7
• D. 8, 9, 10

Hint:

Use a variable for the middle integer and solve the quadratic formed by the sum of squares.

Answer 1

The Correct Option is A

- Step 1: Set up equation. Let integers be $n-1$, $n$, $n+1$. Sum of squares:
\[ (n-1)^2 + n^2 + (n+1)^2 = 194. \] - Step 2: Expand. $(n-1)^2 = n^2 - 2n + 1$, $(n+1)^2 = n^2 + 2n + 1$. So:
\[ (n^2 - 2n + 1) + n^2 + (n^2 + 2n + 1) = 3n^2 + 2 = 194. \] - Step 3: Solve. $3n^2 + 2 = 194 \implies 3n^2 = 192 \implies n^2 = 64 \implies n = \pm 8$.
- Step 4: Case $n = 8$. Integers: $7, 8, 9$. Check: $7^2 + 8^2 + 9^2 = 49 + 64 + 81 = 194$.
- Step 5: Case $n = -8$. Integers: $-9, -8, -7$. Check: $81 + 64 + 49 = 194$.
- Step 6: Select positive. Options suggest positive, so $7, 8, 9$.
- Step 7: Conclusion. Option (1) 7, 8, 9 is correct.