Question 1070586: Three numbers form an arithmetic sequence. The first term minus the third term is 8. When the 1st, 2nd, and 3rd terms are increased by 3, 5, and 8 respectively, the resulting numbers forms a geometric sequence. Find the common difference of the arithmetic sequence, find the first four terms of the geometric sequence, and the common ratio of the geometric sequence.
Answer by htmentor(1343)   (Show Source):
You can put this solution on YOUR website! The nth term of an arithmetic series can be written a_n = a + (n-1)d where
a is the 1st term and d is the common difference
Given that the 1st term minus the 3rd term is 8, we have
a - (a + 2d) = 8 -> -2d = 8, or d = -4
So a_n = a - 4(n-1)
Adding 3, 5 and 8 to the 1st three terms gives g_1 = a + 3, g_2 = a + 1, and g_3 = a
Since the ratio of successive terms of a geometric sequence is a constant, r, we can write:
r = (a+1)/(a+3) = a/(a+1)
Solve for a:
a^2 + 2a + 1 = a^2 + 3a -> a = 1
The arithmetic sequence is a_n = 1 - 4(n-1)
The common ratio is 1/(1+1) = 1/2
So the 1st 4 terms of the geometric sequence are 4, 2, 1, 1/2
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