Find arith. seq. w/ a_1=1, d not zero, so a_2, a_10, a_34 are 1st 3 terms of geo. seq

[nineteen]

Hi, I would appreciate some help on solving this question!

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

 

[tkhunny]

Hi, I would appreciate some help on solving this question!

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

You need some idea how to pick out any specific item. You should have this material at hand.

Arithmetic
d is the common difference.
A0 is what it is.
A1 = A0 + d
A2 = A1 + d = A0 + 2d
A3 = A2 + d = A0 + 3d
Pattern?

Geometric
r is the common ratio.
A0 is what it is.
A1 = A0 * r
A2 = A1 * r = A0 * r^2
A3 = A2 * r = A0 * r^3
Pattern?

You must find the patterns and commit them to memory.

 

[Denis]

Sure not an easy problem!

You'll need to set up 3 equations.

Hint: the ratio (see TKhunny's reference to "r" in geometric series) will equal 3.

Good luck!

 

[Otis]

I would prefer to see tkhunny's post use symbols A and B, instead of using A for both sequences. (For example, A0 in the arithmetic sequence is not the same number as A0 in the geometric sequence.)

… You'll need to set up 3 equations …

Yes, but solving that system wasn't so bad.

First, I used different formulas for the nth-terms:

A_n = A + (n-1)(d)

B_n = (B)(r^n)

We're told that A₁ = 1, so we can immediately solve for A.

Then, my three equations: A₂ = B₁ and A₁₀ = B₂ and A₃₄ = B₃

Using the first equation, I made substitutions in the other two, to obtain a quadratic equation in d.

The solution for d leads to (3)(r)=(r)(r), which I solved by inspection.

The solution for r led to B. :cool:

 

[Denis]

Seems to be exactly what I did:

d = common diff arithmetic series
m = multiplier geometric series
u = 1st term geometric series

1 + d = u [1]
1 + 9d = um [2]
1 + 33d = um^2 [3]