+1833 The sum of the first three terms of a geometric sequence of positive integers is equal to seven times the first term, and the sum of the first four terms is 45. What is the first term of the sequence?
+33616 a + a*r + a*r^2 = 7a (1) (a is the first term)
a + a*r + a*r^2 + a*r^3 = 45 (2)
Divide each term in (1) by a:
1 + r + r^2 = 7 or r^2 +r - 6 = 0 or (r -2)(r + 3) = 0 r = 2 or r = -3
Factor a from (2) and rearrange as:
a = 45/(1 + r + r^2 + r^3)
a = 45/(7 + r^3)
If r = 2 then a = 45/15 or a = 3
If r = -3 then a = 45/(7 - 27) or a = -45/20 or a = -9/4
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+128521 Let a be the first term and the next three terms are ar, ar^2, ar^3.......and we have that
7a = a + ar + ar^2 (1) and
45 = a + ar + ar^2 + ar^3 (2)
Rearranging (1), we have 0 = -6a + ar + ar^2 divide through by a
0 = -6 + r + r^2 factor
0 = (r + 3) ( r -2) so........r = 2 or r = -3
Using (2) and letting r = 2, we have
45 = a + 2a + 4a + 8a
45 = 15a
a = 3
Using (2) and letting r = -3, we have
45 = a - 3a + 9a -27a
45 = -20a
a = -45/20 = -9/4 but, since a has to be a positive integer, reject this
So....the series is
3, 6, 12, 24
