SOLUTION: A sequence of three real numbers forms an arithmetic progression with a first term of 9. If 2 is added to the second term and 20 is added to the third term, the three resulting nu
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Question 1064171: A sequence of three real numbers forms an arithmetic progression with a first term of 9. If 2 is added to the second term and 20 is added to the third term, the three resulting numbers form a geometric progression. What is the smallest possible value for the third term of the geometric progression? Found 2 solutions by stanbon, MathTherapy:Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website! A sequence of three real numbers forms an arithmetic progression with a first term of 9. If 2 is added to the second term and 20 is added to the third term, the three resulting numbers form a geometric progression. What is the smallest possible value for the third term of the geometric progression?
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arithmetic progression:: a; a+d; a+2d
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a = 9
a+d + 2 = d+11
a+2d + 20 = 2d+ 29
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geometric progression:: a; ar; ar^2
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a = 9
ar = d+11
ar^2 = 2d+29
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r = ar/a = (d+11)/9
r = ar^2/ar = (2d+29)/(d+11)
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Equation:
r = r
(d+11)/9 = (2d+29)/(d+11)
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18d + 261 = d^2 + 22d + 121
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d^2 + 4d -40 = 0
d = 4.63
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Ans: 2d+29 = 2*4.63+29 = 38.26
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Cheers,
Stan H.
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You can put this solution on YOUR website! A sequence of three real numbers forms an arithmetic progression with a first term of 9. If 2 is added to the second term and 20 is added to the third term, the three resulting numbers form a geometric progression. What is the smallest possible value for the third term of the geometric progression?
Adding 2 to the 2nd term of the AP makes the GP’s 2nd term: 9 + d + 2, or 11 + d
Adding 20 to the 3rd term of the AP makes the GP’s 3rd term: 9 + 2d + 20, or 29 + 2d
Since this is now a GP, the common ratio is calculated as: =======> ------ Cross-multiplying
(d - 10)(d + 14) = 0
d, or common difference = 10, or – 14
As the SMALLER 3rd term is being sought, we use d = - 14
Therefore, the SMALLER 3rd term of the GP =
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