Suppose that T= ar^n-1 is a GP with a limiting sum.
a) Find the common ratio r if the limiting sum equals 5 times the first term. (Answer: r=4/5)
b) Find the ratio if the sum of all terms except the first equals 5 times the first term. (Answer: 5/6)
Thanks so much to anyone who helps
Why did the geometric progression go to therapy?
Because it couldn't find its common ratio and it was feeling a bit unbalanced!
Now, let's tackle your questions:
a) To find the common ratio, we know that the limiting sum is 5 times the first term, which can be expressed as 5a. The limiting sum of a geometric progression formula is: S = a / (1 - r), where S represents the limiting sum, a is the first term and r is the common ratio.
So, we have:
5a = a / (1 - r)
To solve for r, let's isolate it:
5(1 - r) = 1
5 - 5r = 1
-5r = 1 - 5
-5r = -4
r = -4 / -5
r = 4/5
Therefore, the common ratio is 4/5.
b) In this case, the sum of all terms except the first is 5 times the first term. Let's denote the sum of all terms except the first as S'.
The sum of all terms in a geometric progression formula (excluding the first term) is: S' = a * (1 - r) / (1 - r)
In this case, we know that S' = 5a. So we have:
5a = a * (1 - r) / (1 - r)
To find r, let's cancel out the a:
5 = (1 - r) / (1 - r)
Now we can cross-multiply:
5(1 - r) = 1 - r
5 - 5r = 1 - r
-5r + r = 1 - 5
-4r = -4
r = -4 / -4
r = 1
Thus, the common ratio is 1.
I hope this helps! Let me know if you have any more questions or if you need another silly joke.
a) To find the common ratio, we need to use the formula for the sum of an infinite geometric series:
S = a / (1 - r)
where S is the limiting sum, a is the first term, and r is the common ratio.
Given that the limiting sum is 5 times the first term, we can write this as:
5a = a / (1 - r)
We can simplify this equation by multiplying both sides by 1 - r:
5a(1 - r) = a
Expanding and simplifying:
5a - 5ar = a
Combining like terms:
4a = 5ar
Dividing both sides by 5a:
4/5 = r
So the common ratio, r, is equal to 4/5.
b) To find the common ratio in this case, we'll again use the formula for the sum of an infinite geometric series:
S = a / (1 - r)
However, this time we need to consider the sum of all terms except the first term, which we can write as:
S' = (a - a * r) / (1 - r)
Given that the sum of all terms except the first is 5 times the first term, we can write this as:
5a = (a - a * r) / (1 - r)
Multiplying both sides by (1 - r) to remove the fraction:
5a(1 - r) = a - a * r
Expanding and simplifying:
5a - 5ar = a - a * r
Combining like terms:
4a = 4ar
Dividing both sides by 4a:
1 = r
So the common ratio, r, is equal to 1.
I apologize for the confusion in my previous response. The correct answer for part (b) is that the common ratio, r, is equal to 1.
To solve these problems, we need to use the formulas for the sum of a geometric progression (GP).
The formula for the sum of an infinite GP with a first term (a) and a common ratio (r), where -1 < r < 1, is given by:
S = a / (1 - r)
a) In this case, the limiting sum is 5 times the first term (T = 5a). So, we have:
T = a / (1 - r)
Substituting T = 5a, we get:
5a = a / (1 - r)
To eliminate the denominator, cross-multiply:
5a(1 - r) = a
Simplifying the equation:
5 - 5r = 1
Rearranging the equation:
5r = 4
Dividing both sides by 5:
r = 4/5
So, the common ratio for the GP is 4/5.
b) In this case, the sum of all terms except the first is 5 times the first term (T = 5a). We need to find the common ratio (r).
To find the sum of all terms except the first, we use the formula for the sum of an infinite GP, subtracting the value of the first term:
S = a / (1 - r) - a
Since S = 5a, we have:
5a = a / (1 - r) - a
To eliminate the denominator, cross-multiply:
5a(1 - r) = a - a(1 - r)
Simplifying the equation:
5 - 5r = 1 - a + ar
Rearranging the equation and simplifying:
5r - ar = 4 - a
Factoring out 'r':
r(5 - a) = 4 - a
Dividing both sides by (5 - a):
r = (4 - a) / (5 - a)
Substituting T = 5a, we get:
r = (4 - T/5) / (5 - T/5)
Simplifying the equation:
r = (4 - T/5) / (5 - T/5)
Multiplying both the numerator and denominator by 5 to get rid of the fraction:
r = (20 - T) / (25 - T)
Since T = 5a, we have:
r = (20 - 5a) / (25 - 5a)
So, the common ratio for the GP is (20 - 5a) / (25 - 5a).
If S = 5 times the 1st, then
a/(1-r) = 5a
solve for r.
If the 1st term is excluded, then
a/(1-r) - a = 5a