(a) Determine the number of terms, n, if 3+3^2+3^3+..........+3^n= 9840.
(b) Three numbers, a, b and c form a geometric series so that a+b+c = 35 and abc = 1000. What are the values of a, b and c ?
A fractal is created: A circle is drawn with radius 8 cm. Another circle is drawn with half the radius of the previous circle. The new circle is tangent to the previous circle. Suppose this pattern continues through five steps. What is the sum of the areas of the circles? Express your answer as an exact fraction.
9840 = 3(1-3^n)/(1-3)
n = 8
a+ar+ar^2 = 35
a*ar*ar^2 = 1000
a(1+r+r^2) = 35
a^3 r^3 = 1000
by inspection, since 1000 = 2^3 * 5^3, a and r are 2 and 5.
the only factors of 35 are 5 and 7, so
a=5
r=2
(a,b,c) = (5,10,20)
if only one circle of each radius is drawn, the radii are
8,4,2,1,1/2,1/4
pi(8^2 + 4^2 + 2^2 + 1^2 + (1/2)^2 + (1/4)^2) = 1365/16 pi
Note that since
sum(1,infinity) 1/n^2 = pi^2/6, if the sequence were carried out forever, the area would be
pi * 64 * pi^2/6
(a) To determine the number of terms, n, in the given equation 3 + 3^2 + 3^3 + ... + 3^n = 9840, we can use the formula for the sum of a geometric series. The formula is given by:
S = a * (1 - r^n) / (1 - r)
In this equation, S represents the sum of the series, a is the first term, r is the common ratio, and n is the number of terms.
For the given equation, the first term (a) is 3, and the common ratio (r) is also 3. The sum of the series (S) is 9840.
Now, substitute these values into the equation:
9840 = 3 * (1 - 3^n) / (1 - 3)
Simplifying further:
3280 = (1 - 3^n) / -2
To get rid of the negative sign on the right-hand side, we multiply both sides by -2:
-2 * 3280 = 1 - 3^n
-6560 = 1 - 3^n
Rearranging the equation:
3^n = 6561
Now, we need to find the value of n that satisfies this equation. Taking the logarithm (base 3) of both sides will help us do that:
n = log3(6561)
Using a logarithmic calculator, we find that n is equal to 8.
Therefore, the number of terms, n, in the given equation is 8.
(b) To find the values of a, b, and c in the geometric series where a + b + c = 35 and abc = 1000, we need to set up a system of equations.
Let the three numbers be a, ar, and ar^2, where r is the common ratio.
According to the given conditions, we have:
a + ar + ar^2 = 35 --(1)
a * ar * ar^2 = 1000 --(2)
To solve these equations, we can express ar and ar^2 in terms of a:
ar = (35 - a) - ar^2 --(3) (subtracting ar^2 from both sides of equation (1))
ar * ar^2 = (35 - a) * ar^2 - ar^3 --(4) (multiplying both sides of equation (3) by ar^2)
Substituting equation (4) into equation (2):
(35 - a) * ar^2 - ar^3 = 1000
Now, we simplify the equation:
35ar^2 - aar^2 - ar^3 = 1000
35ar^2 - (a * ar^2) - ar^3 = 1000
35ar^2 - a^2r^2 - ar^3 = 1000
We can factor out the common term ar^2:
ar^2 (35 - a - r) = 1000
From this equation, we can see that either ar^2 equals zero (thus a or r is zero) or (35 - a - r) equals zero.
If ar^2 is zero, it means either a or r is zero, which does not satisfy the given conditions since we are looking for non-zero values of a, b, and c.
Therefore, we have:
35 - a - r = 0
Simplifying further:
a + r = 35 --(5)
Using equation (5), we can express r in terms of a:
r = 35 - a
Now, we substitute this expression into equation (3):
ar = (35 - a) - a(35 - a)^2
Simplifying:
ar = (35 - a) - a(1225 - 70a + a^2)
We can expand the equation:
ar = 35 - a - 70a + a^2
ar = a^2 - 71a + 35
Substituting the value of r from equation (5):
a(35 - a) = a^2 - 71a + 35
Expanding:
35a - a^2 = a^2 - 71a + 35
Rearranging the equation:
2a^2 - 35a = 0
Factoring out a:
a(2a - 35) = 0
From this equation, we can see that either a equals zero or (2a - 35) equals zero.
If a is zero, it does not satisfy the given conditions since we are looking for non-zero values of a, b, and c.
Therefore, we have:
2a - 35 = 0
Solving this equation:
2a = 35
a = 17.5
Now, substitute this value of a into equation (5) to find r:
17.5 + r = 35
r = 35 - 17.5
r = 17.5
Therefore, the values of a, b, and c in the geometric series that satisfy the given conditions are:
a = 17.5
b = 17.5
c = 17.5