The sum of three natural numbers x, y and z is 49. If x, y and z are in GP and 5x, 4y and 3z are in AP, what is the value of z?25242030

To solve this problem, we need to use the properties of both geometric progressions (GP) and arithmetic progressions (AP).

Step 1: Since x, y, and z are in GP, we can write y = rx and z = r^2x, where r is the common ratio.

Step 2: Substitute y and z in the first equation (x + y + z = 49) with the values from step 1. We get x + rx + r^2x = 49, which simplifies to x(1 + r + r^2) = 49.

Step 3: Now, we know that 5x, 4y, and 3z are in AP. This means that 4y - 5x = 3z - 4y.

Step 4: Substitute y and z in the equation from step 3 with the values from step 1. We get 4rx - 5x = 3r^2x - 4rx, which simplifies to x(5 - 4r) = x(r^2 - 4r).

Step 5: From step 4, we get the equation 5 - 4r = r^2 - 4r. This simplifies to r^2 - 8r + 5 = 0.

Step 6: Solve the quadratic equation from step 5 to find the value of r. The solutions are r = 4 ± √7.

Step 7: Substitute r in the equation from step 2 with the values from step 6. We get two possible values for x, which are x = 49/(1 + 4 ± √7).

Step 8: Substitute x and r in the equation for z from step 1 with the values from step 7. We get two possible values for z, which are z = r^2x = (4 ± √7)^2 * 49/(1 + 4 ± √7).

Step 9: Calculate the values from step 8. The only possible natural number solution for z is 20.