If the sum of the first and the second terms of an infinite Geometric progression is equal to 3/8, and the sum of its term is equal to 1/2. find the common ratio of the first term of the progression
To solve this problem, we need to set up an equation using the given information.
Let's denote the first term as 'a' and the common ratio as 'r'.
From the given information:
The sum of the first and the second terms is equal to 3/8:
a + ar = 3/8
The sum of all the terms is equal to 1/2:
a + ar + ar^2 + ar^3 + ... = 1/2
Now, there's a useful formula to find the sum of an infinite geometric series when the absolute value of the common ratio is less than 1. It's given by:
Sum = a / (1 - r)
Using this formula, we can rewrite the second equation:
a / (1 - r) = 1/2
Now we have a system of equations:
a + ar = 3/8 (equation 1)
a / (1 - r) = 1/2 (equation 2)
To solve this system of equations, we can use substitution or elimination method. Let's use the substitution method.
Rearrange equation 2 to solve for 'a':
a = (1/2)(1 - r)
Substitute this value of 'a' in equation 1:
(1/2)(1 - r) + (1/2)(1 - r)r = 3/8
Simplify the equation:
(1 - r + r - r^2)/2 + (1 - r)r/2 = 3/8
(1 - r - r^2 + r - r^2)/2 = 3/8
(2 - 2r - 2r^2)/2 = 3/8
2 - 2r - 2r^2 = 3/4
8 - 8r - 8r^2 = 3
Rearrange the equation to form a quadratic equation:
8r^2 + 8r - 5 = 0
Now, we can use the quadratic formula to solve for 'r':
r = [-b ± sqrt(b^2 - 4ac)] / 2a
where a = 8, b = 8, and c = -5.
Plugging in the values:
r = [-8 ± sqrt(8^2 - 4 * 8 * -5)] / 2 * 8
r = [-8 ± sqrt(64 + 160)] / 16
r = [-8 ± sqrt(224)] / 16
Simplifying further:
r = (-8 ± 4√14) / 16
r = -1/2 ± √14/4
So, the common ratio of the first term of the progression can be either (-1/2 + √14/4) or (-1/2 - √14/4).