Question 254764: determine the arithmetic sequence with first term 1 and common difference not equal to zero, whose second, tenth, and thirty fourth terms are the first 3 terms in geometric sequence.
Answer by palanisamy(496)   (Show Source):
You can put this solution on YOUR website! Let the arithmetic sequence be 1,1+d,1+2d,..... where d is the common difference.
Second term = 1+d
Tenth term = 1+(10-1)d = 1+9d
Thirty fourth term = 1+(34-1)d = 1+33d
Given, these three terms, 1+d,1+9d,1+33d are the first 3 terms in geometric sequence.
So we get, (1+9d)^2 = (1+d)(1+33d)
1+81d^2+18d = 1+33d+d+33d^2
81d^2+18d+1-33d^2-33d-d-1 = 0
48d^2-16d = 0
16d(3d-1) = 0
d = 0 or d = 1/3
Since d is nonzero, we get d = 1/3
Therefore the arithmetic sequence is 1,1+1/3,1+2/3,..
1,4/3,5/3,...
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