The sum of the first three terms of a geometric sequence is 54. The sum of the first six terms is 70. Find the common ratio. (All the terms are real numbers.)
+129852 Let a be the first term
We have that
a + ar + ar^2 = 54
a ( 1 + r + r^2) = 54
a = 54 / (1 +r + r^2) ⇒ (1 + r + r^2) = 54/a (1)
And
54 + ar^3 + ar^4 + ar^5 = 70
ar^3 / (1 + r + r^2) = 16 ⇒ (1 + r + r^2) = 16/ (ar^3) (2)
Equating (1) , (2)
54 / a = 16 / (ar^3) [ a ≠ 0 ]
54 = 16/r^3
r^3 = 16/54 = 8/ 27
r = (8/27)^(1/3) = 2 / 3
CORRECTION ....THX, CATMG !!!
+2407 I think that there may be a small mistake.
The sum of the first six terms is 70, so
a + ar + ar^2 + ar^3 + ar^4 + ar^5 = 70
54 + ar^3 + ar^4 + ar^5 = 70
ar^3 + ar^4 + ar^5 = 16
I think your method is really smart since by keeping the amount of terms in both expressions constant, you can just divide. :))
=^._.^=
