1. The problem asks to find the values of $x, y, z$ such that the numbers $x^2 - yz$, $y^2 - xz$, and $z^2 - xy$ form an arithmetic progression (AP) of 4 terms. 2. Recall that for four terms $a_1, a_2, a_3, a_4$ to be in AP, the difference between consecutive terms is constant: $a_2 - a_1 = a_3 - a_2 = a_4 - a_3$. 3. Since only three expressions are given, we interpret the problem as these three terms being part of a 4-term AP, so the condition is: $$ (y^2 - xz) - (x^2 - yz) = (z^2 - xy) - (y^2 - xz) $$ 4. Simplify the left side: $$ y^2 - xz - x^2 + yz = y^2 - x^2 + yz - xz $$ 5. Simplify the right side: $$ z^2 - xy - y^2 + xz = z^2 - y^2 + xz - xy $$ 6. Set the two sides equal: $$ y^2 - x^2 + yz - xz = z^2 - y^2 + xz - xy $$ 7. Rearrange terms: $$ y^2 - x^2 + yz - xz - z^2 + y^2 - xz + xy = 0 $$ 8. Combine like terms: $$ 2y^2 - x^2 - z^2 + yz - 2xz + xy = 0 $$ 9. This is the condition that $x, y, z$ must satisfy for the given expressions to be in AP. --- 10. The second problem asks to find values of $a, b, c$ such that the numbers $a^2 + ab + b^2$, $a^2 + ac + c^2$, and $b^2 + bc + c^2$ form an arithmetic progression of 4 terms. 11. Using the same AP condition for three terms in a 4-term AP: $$ (a^2 + ac + c^2) - (a^2 + ab + b^2) = (b^2 + bc + c^2) - (a^2 + ac + c^2) $$ 12. Simplify left side: $$ a^2 + ac + c^2 - a^2 - ab - b^2 = ac + c^2 - ab - b^2 $$ 13. Simplify right side: $$ b^2 + bc + c^2 - a^2 - ac - c^2 = b^2 + bc - a^2 - ac $$ 14. Set equal: $$ ac + c^2 - ab - b^2 = b^2 + bc - a^2 - ac $$ 15. Rearrange: $$ ac + c^2 - ab - b^2 - b^2 - bc + a^2 + ac = 0 $$ 16. Combine like terms: $$ 2ac + c^2 - ab - 2b^2 - bc + a^2 = 0 $$ 17. This is the condition that $a, b, c$ must satisfy for the given expressions to be in AP. --- 18. The third problem asks to find values of $a, b, c$ such that the numbers $\frac{1}{a+b}$, $\frac{1}{a+c}$, $\frac{1}{b+c}$ form an arithmetic progression of 4 terms, given that $a^2, b^2, c^2$ form an arithmetic progression of 4 terms. 19. First, the condition for $a^2, b^2, c^2$ in AP of 4 terms is: $$ 2b^2 = a^2 + c^2 $$ 20. For the reciprocals to be in AP of 4 terms: $$ \frac{1}{a+c} - \frac{1}{a+b} = \frac{1}{b+c} - \frac{1}{a+c} $$ 21. Simplify left side: $$ \frac{1}{a+c} - \frac{1}{a+b} = \frac{(a+b) - (a+c)}{(a+c)(a+b)} = \frac{b - c}{(a+c)(a+b)} $$ 22. Simplify right side: $$ \frac{1}{b+c} - \frac{1}{a+c} = \frac{(a+c) - (b+c)}{(b+c)(a+c)} = \frac{a - b}{(b+c)(a+c)} $$ 23. Set equal: $$ \frac{b - c}{(a+c)(a+b)} = \frac{a - b}{(b+c)(a+c)} $$ 24. Multiply both sides by $(a+c)$: $$ \frac{b - c}{a+b} = \frac{a - b}{b+c} $$ 25. Cross multiply: $$ (b - c)(b + c) = (a - b)(a + b) $$ 26. Expand: $$ b^2 - c^2 = a^2 - b^2 $$ 27. Rearrange: $$ a^2 + c^2 = 2b^2 $$ 28. This matches the condition from step 19, confirming the consistency. --- Final answers: - For problem 3 (part ក): $2y^2 - x^2 - z^2 + yz - 2xz + xy = 0$ - For problem 3 (part ខ): $2ac + c^2 - ab - 2b^2 - bc + a^2 = 0$ - For problem 3 (part គ): $a^2 + c^2 = 2b^2$ and the reciprocals condition reduces to the same. - For problem 4, the problem statement is unclear and incomplete, so no solution is provided here.