SOLUTION: The first,third and fifth terms of a geometric sequence form arithmetic sequence.If the first term of the sequence is 3,find the tenth term of the geometric sequence.

The 10 terms are

3, 3r, 3r2, 3r3, 3r4, 3r5, 3r6, 3r7, 3r8, 3r9

3, 3r2, and 3r4 form an arithmetic progression.

3+d = 3r2,
3r2+d = 3r4

Solve the first for d

d = 3r2-3

Substitute in the second equation

3r2+3r2-3 = 3r4

6r2-3 = 3r4

0 = 3r4-6r2+3

0 = r4-2r2+1

0 = (r2-1)(r2+1)

0 = r2-1 = 0;  r2+1 = 0
      r2 = 1;    r2 = -1
       r = ±1;    r = ±i (imaginary so discard)

d = 3r2-3 = 3(±1)2-3 = 3(1)-3 = 3-3 = 0

So for r=+1, the sequence is

3, 3r, 3r2, 3r3, 3r4, 3r5, 3r6, 3r7, 3r8, 3r9 is

3, 3(1), 3(1)2, 3(1)3, 3(1)4, 3(1)5, 3(1)6, 3(1)7, 3(1)8, 3(1)9

3, 3, 3, 3, 3, 3, 3, 3, 3, 3

So the 10th term is 3

And for r=-1, the squence is

3, 3r, 3r2, 3r3, 3r4, 3r5, 3r6, 3r7, 3r8, 3r9

3, 3(-1), 3(-1)2, 3(-1)3, 3(-1)4, 3(-1)5, 3(-1)6, 3(-1)7, 3(-1)8, 3(-1)9

3, -3, 3, -3, -3, 3, 3, -3, 3, -3

The 10th term is -3

Two solutions, the 10th term is either 3 or -3.

Edwin