The sum of the second and third terms of a geometric progression is six times the fourth term.find the two possible values of the common ratio
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Why did the geometric progression go to therapy? Because it suffered from an identity crisis!
Now, let's get down to business. Let's assume that the first term of the geometric progression is "a" and the common ratio is "r".
According to the given information, the second term would be "ar" and the third term would be "ar^2". The sum of the second and third terms is then:
(ar) + (ar^2) = 6(ar^2)
Expanding this equation, we get:
ar + ar^2 = 6ar^2
Simplifying further:
ar^2 - 5ar - ar = 0
ar^2 - 6ar = 0
Factoring out an "ar" from both terms:
ar(r - 6) = 0
This gives us two possible solutions for "r":
- Either r = 0 (which makes the progression constant)
- Or r = 6 (which keeps the progression going)
So, the two possible values for the common ratio are 0 and 6.
To find the possible values of the common ratio in a geometric progression, let's consider the given information:
We have a geometric progression with terms that we can label as:
a, ar, ar^2, ar^3, ...
According to the problem, the sum of the second and third terms is six times the fourth term. So, we can write this as an equation:
ar + ar^2 = 6(ar^3)
Now, let's simplify this equation step by step to find the possible values of the common ratio (r).
First, divide both sides of the equation by ar to remove the common factor:
1 + r = 6(r^2)
Next, expand the equation:
r^2 - 6r + 1 = 0
Now, we have a quadratic equation. To solve this equation, we can either use the quadratic formula or factor the equation:
(r - 1)(r - 1) = 0
Simplifying further:
(r - 1)^2 = 0
From this equation, we can see that there is only one possible value for the common ratio, which is r = 1.
Therefore, there is only one possible value for the common ratio in this given geometric progression, which is 1.
impatient much?
Use your basic GP formulas ...
ar + ar^2 = 6ar^3
6r^2 - r - 1 = 0
Surely you can finish it off from here ...