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.
The n-th term of a geometric sequence with initial value a and common ratio r is given by:
an = a * r ^ ( n - 1 )
a is the first term
r is common ratio
In this case:
a1 = a * r ^ ( 1 - 1 ) = a * r ^ 0 = a * 1 = a
a2 = a * r ^ ( 2 - 1 ) = a * r ^ 1 = a * r
a3 = a * r ^ ( 3 - 1 ) = a * r ^ 2
a4 = a * r ^ ( 4 - 1 ) = a * r ^ 3
The third term is greater than the first by 12 mean:
a3 = a1 + 12
The fourth is greater than the second by 36 mean:
a4 = a2 + 36
So:
a3 = a1 + 12
a * r ^ 2 = a + 12
a4 = a2 + 36
a * r ^ 3 = a * r + 36
Now you must solve system of 2 equations with 2 unknow:
a * r ^ 2 = a + 12
a * r ^ 3 = a * r + 36
a * r ^ 2 = a + 12 Subtract a to both sides
a * r ^ 2 - a = a + 12 - a
a * r ^ 2 - a = 12
a * ( r ^ 2 - 1 ) = 12
a * r ^ 3 = a * r + 36 Subtract a * r to both sides
a * r ^ 3 - a * r = a * r + 36 - a * r
a * r ^ 3 - a * r = 36
a * r * ( r ^ 2 - 1 ) = 36
Your system become:
a * ( r ^ 2 - 1 ) = 12
a * r * ( r ^ 2 - 1 ) = 36
a * ( r ^ 2 - 1 ) = 12 Divide both sides by a
r ^ 2 - 1 = 12 / a
a * r * ( r ^ 2 - 1 ) = 36 Divide both sides by a * r
r ^ 2 - 1 = 36 / a * r
r ^ 2 - 1 = r ^ 2 - 1
12 / a = 36 / a * r Multiply both sides by a
12 * a / a = 36 * a / a * r
12 = 36 / r Multiply both sides by r
12 * r = 36 * r / r
12 r = 36 Divide both sides by 12
r = 36 / 12
r = 3
Replace this value in equation:
a * ( r ^ 2 - 1 ) = 12
a * ( 3 ^ 2 - 1 ) = 12
a * ( 9 - 1 ) = 12
a * 8 = 12
8 a = 12 Divide both sides by 8
a = 12 / 8
a = 4 * 3 / ( 4 * 2 )
a = 3 / 2
a = 3 / 2 , r = 3 so:
a1 = a
a1 = 3 / 2
a2 = a * r
a2 = ( 3 / 2 ) * 3 = 9 / 2
a3 = a * r ^ 2
a3 = ( 3 / 2 ) * 3 ^ 2 = ( 3 / 2 ) * 9 = 27 / 2
a4 = a * r ^ 3
a4 = ( 3 / 2 ) * 3 ^ 3 = ( 3 / 2 ) * 27 = 81 / 2
Your geometric progression:
3 / 2 , 9 / 2 , 27 / 2 , 81 / 2
Proof:
a3 = a1 + 12
27 / 2 = 3 / 2 + 12
27 / 2 = 3 / 2 + 12 * 2 / 2
27 / 2 = 3 / 2 + 24 / 2
a4 = a2 + 36
81 / 2 = 9 / 2 + 36
81 / 2 = 9 / 2 + 36 * 2 / 2
81 / 2 = 9 / 2 + 72 / 2
ar^2 = a+12
ar^3 = ar+36
Nice work. I'd have changed things a bit here:
Your system become:
a * ( r ^ 2 - 1 ) = 12
a * r * ( r ^ 2 - 1 ) = 36
Divide the bottom by the top to get
r = 36/12 = 3
a*8 = 12
a = 3/2
To find the four numbers that form a geometric progression, we'll use the information given in the problem. Let's denote the first term as 'a' and the common ratio as 'r'.
According to the problem, the third term is greater than the first term by 12. We can express this as:
a * r^2 = a + 12 ----------- Equation 1
Similarly, the fourth term is greater than the second term by 36:
a * r^3 = a * r + 36 ----------- Equation 2
Now, we have two equations with two unknowns (a and r). Let's solve them to find the values.
First, we can rearrange Equation 1 to isolate 'a':
a * r^2 - a = 12
a * (r^2 - 1) = 12
a = 12 / (r^2 - 1)
Substitute this value of 'a' in Equation 2:
(12 / (r^2 - 1)) * r^3 = (12 / (r^2 - 1)) * r + 36
12 * r = (12 * r^3) / (r^2 - 1) + 36
We can then simplify this equation by cross-multiplying:
12 * r * (r^2 - 1) = 12 * r^3 + 36 * (r^2 - 1)
Now, we have a quadratic equation:
12r^3 - 12r + 36r^2 - 36 = 12r^3 + 36r^2 - 36
We can cancel out the common terms:
12r + 36r^2 = 0
Dividing both sides by 12r gives us:
r + 3r^2 = 0
Factoring out an 'r' from the left side:
r * (1 + 3r) = 0
This equation has two solutions:
r = 0 or r = -1/3
As the common ratio of a geometric sequence cannot be zero, we can discard r=0 as a solution. Therefore, r = -1/3.
Now, substitute this value of 'r' into Equation 1 to find 'a':
a = 12 / ((-1/3)^2 - 1)
a = 12 / (1/9 - 1)
a = 12 / (-8/9)
a = -108/8
a = -13.5
Hence, the four numbers that form a geometric progression, satisfying the given conditions, are:
-13.5, -4.5, -1.5, -0.5