A descending geometric series has first term 1 and the common
ratio r is
positive. The sum of the first 5 terms is twice the sum of terms
from the 6th to
15th inclusive. Prove that r^5 = 1/2 (sqrt 3 – 1).
sqrt= square root
First term = 1
Second term = r
Third term = r^2 .... so on and so for
Thus, nth term = r^(n - 1)
Sum of first 5 terms
= summation of r^(n - 1) from r = 1 to r = 5
= r(r^5 - 1) / (r - 1)
= (r^6 - r) / (r - 1)
Sum of 6th to 15th term
= summation of r^(n - 1) from r = 6 to r = 15
= [summation of r^(n - 1) from r = 1 to r = 15] - [summation of
r^(n - 1) from r = 1 to r = 5]
= [summation of r^(n - 1) from r = 1 to r = 15] - (r^6 -
r) / (r - 1)
= [r(r^15 - 1) / (r - 1)] - [(r^6 - r) / (r - 1)]
= (r^16 - r - r^6 + r) / (r - 1)
= (r^16 - r^6) / (r - 1)
Since the sum of the first 5 terms is twice the sum of the 6th
to 15th term, we have:
(r^6 - r) / (r - 1) = 2(r^16 - r^6) / (r - 1)
r^6 - r = 2(r^16 - r^6)
r^6 - r = 2r^16 - 2r^6
2r^16 - 3r^6 + r = 0
r(2r^15 - 3r^5 + 1) = 0
r = 0 (rejected, since r is positive) or 2r^15 - 3r^5 + 1 =
0
Thus, 2r^15 - 3r^5 + 1 = 0.
Let u = r^5. Note that we can do this, as we don't consider complex
solutions.
We will have:
2u^3 - 3u + 1 = 0
(u - 1)(2u^2 + 2u - 1) = 0
u = 1 (rejected, since r^5 < 1 due to the descending nature of
the GP) or
2u^2 + 2u - 1 = 0
Taking 2u^2 + 2u - 1 = 0, we have:
2(u^2 + u) - 1 = 0
2(u + 1/2)^2 - 1/2 - 1 = 0
2(u + 1/2)^2 - 3/2 = 0
2(u + 1/2)^2 = 3/2
(u + 1/2)^2 = 3/4
u + 1/2 = sqrt(3) / 2 or = -sqrt(3) / 2
u = [sqrt(3) / 2] - 1/2 or = -[sqrt(3) / 2] - 1/2 (rejected, since
r^5 > 0)
u = (1/2)(sqrt(3) - 1)
r^5 = (1/2)(sqrt(3) - 1)