Find four numbers that form a geometric progression such that the third term is greater than the first by 12 and the fourth is greater than the second by 36.
ar^2 = a+12
ar^3 = ar+36
Find a,r and then the four terms.
To find the four numbers that form a geometric progression, let's denote the first term as 'a' and the common ratio as 'r.'
We are given two conditions:
1) The third term is greater than the first by 12: a * r^2 = a + 12.
2) The fourth term is greater than the second by 36: a * r^3 = a * r + 36.
We can use these two equations to solve for 'a' and 'r' simultaneously.
Step 1: Solve the first equation for 'r':
a * r^2 = a + 12
Divide both sides by 'a' (assuming 'a' is non-zero):
r^2 = 1 + 12/a
Step 2: Solve the second equation for 'r':
a * r^3 = a * r + 36
Divide both sides by 'a' (assuming 'a' is non-zero):
r^3 = r + 36/a
Step 3: Equate the expressions for 'r' from Step 1 and Step 2:
1 + 12/a = r + 36/a
Combine like terms:
r = 1 + 36/a - 12/a
r = 1 + 24/a
Step 4: Substitute the expression for 'r' back into the first equation:
(a + 24) * (1 + 24/a)^2 = a + 12
Expand and simplify the equation:
(a + 24) * (1 + 48/a + 576/a^2) = a + 12
Multiply through by 'a^2':
(a^3 + 24a^2) + 48a + 576 = a^3 + 12a^2
Combine like terms:
12a^2 - 24a + 576 = 0
Divide through by 12:
a^2 - 2a + 48 = 0
Step 5: Solve the quadratic equation for 'a':
Using the quadratic formula:
a = (-b ± sqrt(b^2 - 4ac))/(2a)
a = (-(-2) ± sqrt((-2)^2 - 4(1)(48)))/(2(1))
a = (2 ± sqrt(4 - 192))/2
a = (2 ± sqrt(-188))/2
Since the discriminant is negative, there are no real solutions for 'a'.
Therefore, there are no four real numbers that form a geometric progression satisfying the given conditions.