The sum of third and seventh term of an ascending arithmetic progression is 12 and their product is 32.

Then find the ninth term?

c'mon, guy - use what you know abot APs:

(a+2d)+(a+6d) = 12
(a+2d)(a+6d) = 32

Now just solve for a and d, then you want a+8d.

Hint: 4*8 = 32

To find the ninth term of the arithmetic progression, we'll need to use the given information about the third and seventh terms.

1. Let's assume that the first term of the arithmetic progression is 'a'.
2. Since it's an ascending arithmetic progression, we can assume the common difference as 'd'.
3. The third term of the arithmetic progression will then be 'a + 2d' (since the common difference is added twice to reach the third term).
4. The seventh term of the arithmetic progression will be 'a + 6d' (since the common difference is added six times to reach the seventh term).

Now, we have two equations based on the given information:

Equation 1: (a + 2d) + (a + 6d) = 12
Equation 2: (a + 2d) * (a + 6d) = 32

Let's solve these equations to find the values of 'a' and 'd'.

From Equation 1:
2a + 8d = 12
a + 4d = 6

From Equation 2:
(a^2) + 8ad + 12d^2 = 32
a^2 + 8ad + 12d^2 - 32 = 0
By simplifying, we get:
a^2 + 4ad + 4ad + 12d^2 - 32 = 0
a(a + 4d) + 4d(a + 4d) - 32 = 0
(a + 4d)(a + 4d) - 32 = 0
(a + 4d)^2 - 32 = 0

Now, we substitute the value of 'a + 4d' from the first equation:
(6)^2 - 32 = 0
36 - 32 = 0
4 = 0

Since the equation is not satisfied, we conclude that there is no solution for 'a' and 'd' that satisfies both equations.

Hence, there is no definite value for the ninth term of the arithmetic progression with the given information.