the sum to infinity of a gp is four times the first term. find the common ratio
To find the common ratio (r) of a geometric progression (GP) when the sum to infinity (S∞) is four times the first term (a), we can use the formula for the sum of an infinite geometric series:
S∞ = a / (1 - r)
Given that S∞ = 4a, we can substitute this into the formula:
4a = a / (1 - r)
To solve for the common ratio (r), we can cross-multiply:
4a(1 - r) = a
Now we can simplify and solve for r:
4a - 4ar = a
Multiplying through by -1 to eliminate the negative term:
-4ar + 4a = -a
Factoring out a from both terms:
a(-4r + 4) = -a
Dividing both sides by -a:
-4r + 4 = -1
Subtracting 4 from both sides:
-4r = -5
Dividing both sides by -4:
r = 5/4
Therefore, the common ratio of the geometric progression is 5/4.
To find the common ratio (r) of a geometric progression (GP) in this case, where the sum to infinity is four times the first term, we can use the formula for the sum to infinity of a GP:
S ∞ = a / (1 - r), where S ∞ is the sum to infinity, a is the first term, and r is the common ratio.
In this case, we are given that S ∞ = 4a. Substituting this into the formula, we have:
4a = a / (1 - r).
To solve for r, we can cross-multiply:
4a × (1 - r) = a.
Simplifying the equation:
4a - 4ar = a.
Rearranging the equation:
4ar + a = 4a.
Factoring out 'a' on the left-hand side:
a(4r + 1) = 4a.
Now, we can divide both sides of the equation by 'a':
4r + 1 = 4.
Subtracting 1 from both sides:
4r = 3.
Finally, dividing both sides by 4:
r = 3/4.
The common ratio (r) of the geometric progression is 3/4.
Welcome
sum to infinity = a/(1-r)
but a/(1-r) = 4a
divide both sides by a
1/(1-r) = 4
continue