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Question 494677: Could you help me solve those problems!!!
1. When the same constant is added to the numbers 60, 100, and 150, a three-term geometric sequence arises. What is the constant ratio of the resulting sequence?
2. Find all positive integers a and n such that n is greater or equal to 2 and
a + (a + 1) + (a + 2) + . . . + (a + n - 1) = 100
Thank you for your help!!!
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! 1. When the same constant is added to the numbers 60, 100, and 150, a three-term geometric sequence arises. What is the constant ratio of the resulting sequence?
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Let the constant ratio be "r":
(100+x)/(60+x) = r
(150+x)/(100+x) = r
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Substitute:
(100+x)/(60+x) = (150+x)/(100+x)
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(100+x)^2 = (60+x)(150+x)
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100^2 + 200x + x^2 = 9000 + 210x + x^2
1000 -10x = 0
x = 100
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Solve for "r":
(100+x)/(60+x) = r
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200/160 = r
r = 5/4
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2. Find all positive integers a and n such that n is greater or equal to 2 and
a + (a + 1) + (a + 2) + . . . + (a + n - 1) = 100
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There are n terms
Sum = na + (0+1+2+3+...+n-1) = 100
I'll leave the rest to you
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Cheers,
Stan H.
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