Question
Three consecutive terms of a Geometry progression have product 343 and sum 49/2, find the numbers.
Answers
Answered by
GPT-5 mini
Let the three consecutive terms be a, ar, ar^2. Then
- Product: a·ar·ar^2 = a^3 r^3 = 343 = 7^3, so ar = 7.
- Sum: a(1 + r + r^2) = 49/2. With a = 7/r this gives
7(1 + r + r^2)/r = 49/2 → 2(1 + r + r^2) = 7r → 2r^2 - 5r + 2 = 0.
Solve: r = (5 ± 3)/4 → r = 2 or r = 1/2. Then a = 7/r:
- r = 2 → a = 7/2, terms = 7/2, 7, 14.
- r = 1/2 → a = 14, terms = 14, 7, 7/2.
(These are the same three numbers in opposite order.)
- Product: a·ar·ar^2 = a^3 r^3 = 343 = 7^3, so ar = 7.
- Sum: a(1 + r + r^2) = 49/2. With a = 7/r this gives
7(1 + r + r^2)/r = 49/2 → 2(1 + r + r^2) = 7r → 2r^2 - 5r + 2 = 0.
Solve: r = (5 ± 3)/4 → r = 2 or r = 1/2. Then a = 7/r:
- r = 2 → a = 7/2, terms = 7/2, 7, 14.
- r = 1/2 → a = 14, terms = 14, 7, 7/2.
(These are the same three numbers in opposite order.)