To solve this problem, we'll use the properties of arithmetic sequences and the Pythagorean theorem.

Given Information:

• The first element of the arithmetic sequence is $a_1 = 4$.
• $a_{10}$, $a_{31}$, and $a_{34}$ are the sides of a right triangle.
• $a_{34}$ is the hypotenuse.

Step 1: Express $a_{10}$, $a_{31}$, and $a_{34}$ in terms of the first term $a_1$ and the common difference $d$.

For the $n$-th term of an arithmetic sequence, the formula is: $a_n = a_1 + (n - 1) \cdot d$

So, we have: $a_{10} = a_1 + (10 - 1) \cdot d = 4 + 9d$ $a_{31} = a_1 + (31 - 1) \cdot d = 4 + 30d$ $a_{34} = a_1 + (34 - 1) \cdot d = 4 + 33d$

Step 2: Apply the Pythagorean Theorem

Since $a_{10}$, $a_{31}$, and $a_{34}$ are the sides of a right triangle with $a_{34}$ as the hypotenuse: $a_{10}^2 + a_{31}^2 = a_{34}^2$

Substituting the expressions we found: $(4 + 9d)^2 + (4 + 30d)^2 = (4 + 33d)^2$

Step 3: Expand the Squares

Expand each term: $(4 + 9d)^2 = 16 + 72d + 81d^2$ $(4 + 30d)^2 = 16 + 240d + 900d^2$ $(4 + 33d)^2 = 16 + 264d + 1089d^2$

Step 4: Substitute and Simplify the Equation

Substitute these into the Pythagorean equation: $(16 + 72d + 81d^2) + (16 + 240d + 900d^2) = 16 + 264d + 1089d^2$ Combine like terms: $32 + 312d + 981d^2 = 16 + 264d + 1089d^2$ Move all terms to one side of the equation: $981d^2 - 1089d^2 + 312d - 264d + 32 - 16 = 0$ Simplify: $-108d^2 + 48d + 16 = 0$ Divide the equation by $-4$ to simplify: $27d^2 - 12d - 4 = 0$

Step 5: Solve the Quadratic Equation

To solve $27d^2 - 12d - 4 = 0$, we use the quadratic formula: $d = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ where $a = 27$, $b = -12$, and $c = -4$.

Substitute the values: $d = \frac{-(-12) \pm \sqrt{(-12)^2 - 4(27)(-4)}}{2(27)}$ $d = \frac{12 \pm \sqrt{144 + 432}}{54}$ $d = \frac{12 \pm \sqrt{576}}{54}$ $d = \frac{12 \pm 24}{54}$

Step 6: Simplify the Solutions

This gives us two potential solutions for $d$: $d = \frac{12 + 24}{54} = \frac{36}{54} = \frac{2}{3}$ $d = \frac{12 - 24}{54} = \frac{-12}{54} = -\frac{2}{9}$

Conclusion

The common difference $d$ can be $\boxed{\frac{2}{3}}$ or $\boxed{-\frac{2}{9}}$.

Would you like more details on this solution, or do you have any questions?

Related Questions:

1. What is the sum of the first 50 terms of the arithmetic sequence with $a_1 = 4$ and $d = \frac{2}{3}$?
2. How can you determine if three terms in an arithmetic sequence form a right triangle?
3. What would be the sequence if the common difference $d$ were negative?
4. How does the Pythagorean theorem apply to other types of sequences, such as geometric sequences?
5. If $a_{10}$, $a_{31}$, and $a_{34}$ formed an isosceles triangle instead, how would you find the common difference?

Tip:

In arithmetic sequences, quadratic equations often arise when dealing with terms related by non-linear conditions, such as those involving squares or right triangles. Solving the quadratic equation correctly is crucial to finding the right common difference.