Question 1201408: The first, second and third term of a GP are the first, third and seventh term of an AP respectively. If the seventh term of the AP is 80, find the first term of the AP and the GP.
Answer by greenestamps(13200)   (Show Source):
You can put this solution on YOUR website!
The problem has two clear possible solution strategies:
(1) Show that the first, second, and third terms of the GP form an AP; or
(2) Show that the first, third, and seventh terms of the AP form a GP.
Strategy 1....
The first three terms of the GP are a, ar, and ar^2.
If those three terms are the first, third, and seventh terms of an AP, then the difference between ar^2 and ar is 2 times the difference between ar and a:
We have three possible solutions: a = 0; or r = 1 or r = 2.
a = 0 is not possible, because every term of the GP would be 0, but the third term of the GP is 80.
r = 1 gives us a solution that is not very interesting. A GP with a common ratio of 1 is a constant sequence; since the third term of the GP is 80, every term of the GP (and the AP) is 80.
r = 2 gives us the solution the problem probably wanted us to get. With a common ratio of 2 and a third term of 80, the first three terms of the GP are 20, 40, and 80.
Since those three terms are the first, third, and seventh terms of the AP, the common difference of the AP is 10, and the first seven terms of the AP are 20, 30, 40, 50, 60, 70, and 80.
ANSWER: The first term of the AP and GP is 20
Strategy 2...
Let d be the common difference of the AP. Then, since the 7th term is 80,
1st term of AP = 80-6d
3rd term of AP = 80-4d
The 1st, 3rd, and 7th terms of the AP form a GP, so
 or 
d = 0 actually gives us a solution, but it is not very interesting; the AP and the GP are both constant: 80, 80, 80, ....
d = 10 gives us the answer the problem almost certainly is looking for. The first, third, and seventh terms are 80-6(10) = 20, 80-4d = 40, and 80; and the numbers 20, 40, and 80 form a GP.
ANSWER: The first term of the AP and GP is 20
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