Question 811879
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.
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The 10 terms are

3, 3r, 3r<sup>2</sup>, 3r<sup>3</sup>, 3r<sup>4</sup>, 3r<sup>5</sup>, 3r<sup>6</sup>, 3r<sup>7</sup>, 3r<sup>8</sup>, 3r<sup>9</sup>

3, 3r<sup>2</sup>, and 3r<sup>4</sup> form an arithmetic progression.

3+d = 3r<sup>2</sup>,
3r<sup>2</sup>+d = 3r<sup>4</sup>

Solve the first for d

d = 3r<sup>2</sup>-3

Substitute in the second equation

3r<sup>2</sup>+3r<sup>2</sup>-3 = 3r<sup>4</sup>

6r<sup>2</sup>-3 = 3r<sup>4</sup>

0 = 3r<sup>4</sup>-6r<sup>2</sup>+3

0 = r<sup>4</sup>-2r<sup>2</sup>+1

0 = (r<sup>2</sup>-1)(r<sup>2</sup>+1)

0 = r<sup>2</sup>-1 = 0;  r<sup>2</sup>+1 = 0
      r<sup>2</sup> = 1;    r<sup>2</sup> = -1
       r = ±1;    r = ±i (imaginary so discard)

d = 3r<sup>2</sup>-3 = 3(±1)<sup>2</sup>-3 = 3(1)-3 = 3-3 = 0

So for r=+1, the sequence is

3, 3r, 3r<sup>2</sup>, 3r<sup>3</sup>, 3r<sup>4</sup>, 3r<sup>5</sup>, 3r<sup>6</sup>, 3r<sup>7</sup>, 3r<sup>8</sup>, 3r<sup>9</sup> is

3, 3(1), 3(1)<sup>2</sup>, 3(1)<sup>3</sup>, 3(1)<sup>4</sup>, 3(1)<sup>5</sup>, 3(1)<sup>6</sup>, 3(1)<sup>7</sup>, 3(1)<sup>8</sup>, 3(1)<sup>9</sup>

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, 3r<sup>2</sup>, 3r<sup>3</sup>, 3r<sup>4</sup>, 3r<sup>5</sup>, 3r<sup>6</sup>, 3r<sup>7</sup>, 3r<sup>8</sup>, 3r<sup>9</sup>

3, 3(-1), 3(-1)<sup>2</sup>, 3(-1)<sup>3</sup>, 3(-1)<sup>4</sup>, 3(-1)<sup>5</sup>, 3(-1)<sup>6</sup>, 3(-1)<sup>7</sup>, 3(-1)<sup>8</sup>, 3(-1)<sup>9</sup>

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</pre>