Expert: Steve Holleran - 5/29/2008

Question
Hi, I need help with this question, Find an arithmetic sequence with first term 1, common difference not equal to zero, and second, tenth and thirty-fourth terms are the first three terms of a geometric sequence.

Answer
Hi Patrick,

Okay, here's what I see:

You have an arithmetic sequence that looks like :


              1 , 1+d, 1+2d, ...

The nth term is 1 + (n-1)d, so term 10 is 1 + 9d

and term 34 is 1 + 33d.

Now, if these are the first  three terms in a geometric sequence,
the geom seq looks like :

          1+d , 1+9d, 1+33d, ...

Geometric sequences have to have a common ratio, which is found by dividing any term by the previous term.  So, for the common ratio to be the same, we have to have

                    (1+9d)/ (1+d) = (1+33d)/(1+9d)

Cross-multiplying gives 1 + 18d + 81d^2 = 1 + 34d + 33d^2

The 1's drop out, and bringing all terms to the left, we have

                     48d^2 - 16d = 0

so                    16d(3d - 1) = 0 so d=0 or d = 1/3.

Since d does not = 0, then d = 1/3.

So the arithmetic sequence is

                   1, 1 1/3, 1 2/3 , 2, ...


Hope this is what you needed to see.

Steve