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geometrical progressions?
The third and fifth terms of a geometric series S are 45 and 405 respectively. Find
a) two possible values of the common ratio of S,
b) the first term of s
given that the sum of the third, fourth, and fifth terms is 315
c)find the first twelve terms of S
another series T is formed with all its terms positive. the first term of T is the same as the first term of S. The common ratio of T is the reciprocal of one of the answers to part (a).
d) find the sum to infinity of T
1 Answer
- BrianLv 79 years agoFavorite Answer
a) In general for a geometric sequence the nth term is s(n) = s(1)*r^(n-1), where
r is the common ratio. So we are given that s(3) = s(1)*r^(3-1) = s(1)*r^2 = 45 and
s(5) = s(1)*r^(5-1) = s(1)*r^4 = 405. Now
s(5) / s(3) = s(1)*r^4 / (s(1)*r^2) = r^2 = 405/45 = 9, so r = +/- 3.
b) Now with s(3) = s(1)*r^2 = s(1)*9 = 45, we have s(1) = 5.
c) If the sum of the 3rd, 4th and 5th terms is 315 < (45 + 450), the 4th term must
be negative and so r = -3, and a(4) = -3 * 45 = -135. So the first 12 terms are
5, -15, 45, -135, 405, -1215, 3645, -10935, 32805, -98415, 295245, -885735.
d) T(1) = s(1) = 5. If all the terms of T(n) are positive, then the ratio must be
R = 1/3. Thus the
sum(n=1 to infinity)(T(n)) = T(1)/(1 - (R)) = 5/(1 - (1/3)) = 5/(2/3) = 15/2.