Help finding the following arithmetic sequence ...plz =)?

Question

Find the arithmetic sequence with first term 1 and common difference not equal to 0, whose second, tenth & thirty-fourth terms are the first 3 terms in a geometric sequence.

SHOW ALL STEPS ... THANK YOU!! .. =D

Answer 1

OK, let's see.

It's an arithmetic sequence, so the difference between consecutive terms is a constant. The first term is 1, so we can express each term as:

x(i) = 1 + k*(i-1), i = 1, 2, 3, ....

Next the problem says that the 2nd, 10th, and 34th terms are the first three terms in a geometric sequence. That means that the ratio of the 10th term to the 2nd term must be the same as the ratio of the 34th term to the 10th, because when you divide any two terms of a geometric sequence, you get the same constant.

i(2) = 1 + 1*k
i(10) = 1 + 9*k
i(34) = 1 + 33*k

So we said i(10)/i(2) must equal i(34)/i(10):

1+9k 1+33k
-------- = --------
1+1k 1+9k

Cross multiplying: (1+9k)^2 = (1+1k)(1+33k).

Then 1 + 18k + 81k^2 = 1 + 34k + 33k^2.

Putting it all on the left side, 48k^2 - 16k = 0, or 16k(3k - 1) = 0.

So either k = 0 (not interesting) or k = 1/3.

Therefore the arithmetic sequence must be:

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

The second term is 1 1/3. The 10th term is 4 (1 + 3 * 1/3). The 34th term is 12 (1 + 33 * 1/3). So the geometric progression would be:

4/3, 4, 12, 48, 192, .....

Answer 2

Let's see:

for the geometric
term, value
1, 1
2, r
3, r^2

arith:
term, value
1, 1
2, d
3, 2d
4, 3d
...
k, (k-1)d
...
10, 9d
34, 33d

so,
arith, geo
2, 1
10, r
34, r^2

is this possible? My indices are probably screwy.
I don't have much confidence in this, but I think trying to tabulate what the problem statement means is a good approach.

Answer 3

An=1+d(n-1); Gn=br^(n-1)
A2=1+d=G1=b
A10=1+d9=G2=br
A32=1+31d=G3=br^2
so...
1+d=b
1+9d=br
1+31d=br^2
three equations three unknowns, solve it.