Geometric Sequence and Series Word Problems Help

a) How many ancestors, that is, parents, grandparents, great grandparents would a sixth generation have?

b) The sum of the first 6 terms of a geometric progression is nine times the sum of its first three terms. Find the common ratio of the progression.

c) The 3rd, 5th and 8th terms of an arithmetic progression are in geometric progression. Find the common ratio.

2 + 4 + 8 + 16 + 32 =>
2 + 12 + 48 =>
62

At the most, 62 ancestors


ar + ar^2 + ar^3 + ar^4 + ar^5 + ar^6 = 9 * (ar + ar^2 + ar^3)
ar * (1 + r + r^2 + r^3 + r^4 + r^5) = 9 * (ar) * (1 + r + r^2)
1 + r + r^2 + r^3 + r^4 + r^5 = 9 * (1 + r + r^2)
r^3 + r^4 + r^5 = 8 * (1 + r + r^2)
r^3 * (1 + r + r^2) = 8 * (1 + r + r^2)
(r^3 - 8) * (1 + r + r^2) = 0
(r - 2) * (r^2 + 2r + 4) * (r^2 + r + 1) = 0

r = 2

r = (-2 +/- sqrt(4 - 16)) / 2 = (-2 +/- 2i * sqrt(3)) / 2 = -1 +/- i * sqrt(3)

r = (-1 +/- sqrt(1 - 4)) / 2 = (-1 +/- i * sqrt(3)) / 2

Only one real ratio, so r = 2




t[3] = m + 3d
t[5] = m + 5d
t[8] = m + 8d

t[8] / t[5] = t[5] / t[3]
t[8] * t[3] = t[5]^2
(m + 8d) * (m + 3d) = (m + 5d) * (m + 5d)
m^2 + 11md + 24d^2 = m^2 + 10md + 25d^2
md = d^2
m = d

t[5] / t[3] =>
(m + 5d) / (m + 3d) =>
(m + 5m) / (m + 3m) =>
(6m) / (4m) =>
6/4 =>
3/2

The common ratio is 3/2

a)
Geometric Sequence is : a, ar, ar^2, ar^3, ar^4, ar^5, ...,...

S6 = a(r^6 -1) / (r-1) = 9*S3 = 9*[a(r^3-1)/(r-1)]

(r^6-1) = 9(r^3-1), ===> (r^3+1)(r^3-1) = 9(r^3-1)

r^3 = 9 -1 = 8,
SO,
r = 2 >============================< ANSWER

b)
arithmetic progression : a, a+d, a+2d, a+3d, a+4d, a+5d, .., ..,
Now,
a+3d, a+5d, a+7d are in Geometric Sequence, that is,
SO,
r = (a+5d) / (a+3d) = (a+7d) / (a+5d)
(a+5d)^2 = (a+3d)(a+7d)
a^2 + 10d + 25d^2 = a^2 +10d + 21d^2
4d^2 = 0 , ===> d=0,
Hence,

r = a/a = 1 >========================< ANSWER