1. **State the problem:** We are given three non-zero numbers $x$, $y$, and $z$ which are consecutive terms of a geometric progression (G.P.) and satisfy $x + y + z = 70$. Additionally, $4x$, $5y$, and $4z$ are consecutive terms of an arithmetic progression (A.P.). We need to find the values of $x$, $y$, and $z$. 2. **Recall formulas and rules:** - For three consecutive terms of a G.P., the middle term squared equals the product of the other two: $$y^2 = xz$$ - For three consecutive terms of an A.P., the middle term is the average of the other two: $$2 imes 5y = 4x + 4z$$ which simplifies to $$10y = 4x + 4z$$ 3. **Use the given sum:** $$x + y + z = 70$$ 4. **Express $z$ in terms of $x$ and $y$ using the G.P. relation:** From $$y^2 = xz$$ we get $$z = \frac{y^2}{x}$$ 5. **Substitute $z$ into the sum equation:** $$x + y + \frac{y^2}{x} = 70$$ Multiply both sides by $x$ to clear the denominator: $$x^2 + xy + y^2 = 70x$$ 6. **Use the A.P. condition:** $$10y = 4x + 4z = 4x + 4 \times \frac{y^2}{x} = 4x + \frac{4y^2}{x}$$ Multiply both sides by $x$: $$10yx = 4x^2 + 4y^2$$ 7. **Rewrite the two equations:** - $$x^2 + xy + y^2 = 70x$$ - $$4x^2 + 4y^2 = 10xy$$ 8. **Simplify the second equation by dividing by 2:** $$2x^2 + 2y^2 = 5xy$$ 9. **Express $y$ in terms of $x$ by setting $t = \frac{y}{x}$ (since $x \neq 0$):** - From the first equation: $$x^2 + x(x t) + (x t)^2 = 70 x$$ $$x^2 + x^2 t + x^2 t^2 = 70 x$$ Divide both sides by $x$: $$x + x t + x t^2 = 70$$ Divide by $x$: $$1 + t + t^2 = \frac{70}{x}$$ - From the second equation: $$2x^2 + 2(x t)^2 = 5 x (x t)$$ $$2x^2 + 2x^2 t^2 = 5 x^2 t$$ Divide both sides by $x^2$: $$2 + 2 t^2 = 5 t$$ 10. **Solve the quadratic in $t$:** $$2 + 2 t^2 = 5 t \implies 2 t^2 - 5 t + 2 = 0$$ Use quadratic formula: $$t = \frac{5 \pm \sqrt{25 - 16}}{4} = \frac{5 \pm 3}{4}$$ So, $$t_1 = 2, \quad t_2 = \frac{1}{2}$$ 11. **Find $x$ for each $t$ using** $$1 + t + t^2 = \frac{70}{x} \implies x = \frac{70}{1 + t + t^2}$$ - For $t=2$: $$x = \frac{70}{1 + 2 + 4} = \frac{70}{7} = 10$$ Then, $$y = t x = 2 \times 10 = 20$$ $$z = \frac{y^2}{x} = \frac{400}{10} = 40$$ - For $t=\frac{1}{2}$: $$x = \frac{70}{1 + \frac{1}{2} + \frac{1}{4}} = \frac{70}{1.75} = 40$$ Then, $$y = t x = \frac{1}{2} \times 40 = 20$$ $$z = \frac{y^2}{x} = \frac{400}{40} = 10$$ 12. **Check both solutions:** Both $(x,y,z) = (10,20,40)$ and $(40,20,10)$ satisfy the conditions. **Final answer:** $$\boxed{(x,y,z) = (10,20,40) \text{ or } (40,20,10)}$$