(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 ?
3 answers
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)
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
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
1 answer