How Can Arithmetic and Geometric Progressions Reveal Values of a and b?

  • Thread starter matadorqk
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In summary, the problem involves finding the values of a and b in an arithmetic and geometric progression where a cannot equal b. The equations for arithmetic and geometric progressions are given, and using basic principles, it is determined that a must equal -2 and b must equal 1. This is verified using the equations and logic to solve the problem. There is also a more systematic approach to solving the problem by eliminating b and solving for a.
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matadorqk
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Arithmetic/Geometric Progression Problem SOLVED

Homework Statement


The Three terms a, 1, b are in arithmetic progression. The three terms 1, a, b are in geometric progression. Find the value of a and of b given that a cannot equal b.

Homework Equations


Arithmetic:
Here, Un is the nth term.U1 is the first term.
Un=U1+(n-1)d
Here, Sn is the Sum of n terms.
Sn=n/2(2U1+(n-1)d)
Sn=n/2(U1+Un)
Geometric
Un=U1 x r^n-1
Below, r can't equal 1.
Sn=U1(r^n-1)/r-1
Sn=U1(1-r^n)/1-r
Below, |r|<1
S=U1/1-r

The Attempt at a Solution


Well, let's establish a couple basic things.
Arithmetic
U1=a
U2=1
U3=b
U2-U1=d
U3-U2=d

U2-U1=U3-U2

So:
1-a=b-1
a+b=2
a=2-b
b=2-a
Then, using the formulas:
Un=U1 + (n-1)d
Sn=n/2(2U1 + (n-1)(d))=n/2(U1+Un)

We can find that:
a1+a2+a3=s3
Which is:
a1+a1+d+a1+2d
Which is:
3a+3d=s3

This is what I have so far, Ill update if I get more.. I am kind of stuck on this and another problem, school starts in 5 hours..(which assumes I won't sleep) so help asap would be appreciated..Thanks!

Now, I went on to the geometric side, where logic, one of my true strongs, hit me with a reasonable answer.
If r=u2/u1 and r=u3/u2
a/1=b/a
a=b/a
The only possible and logical thing that popped in my head was, wel A has to be negative.
So, I plot in -1, but then notice, in arithmetic sequences, b would end up as 1 aswell, which is obviously impossible for an arith.s.
Therefore, -2 seemed fit. I plot it, everything seems fine, my common difference would be 3, so my next term should be 4.
Lets plot that into a couple formulas.
a=b/a
-2=4/-2. YES
a+b=2
-2+4=2 YES

I revised every one, it does work. Sometimes, logic works better for me than anything, I guess that's my way of solving problems.

SOLVED
 
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  • #2
There is, of course, a way to do it without guessing an answer. From the arithmetic series, you obtain the equation b=2-a, and from the geometric, a=b/a, or, b=a^2. Hence, eliminating b yields a^2+a-2=0, which can be solved to give (a+2)(a-1)=0, which gives the result.
 
  • #3
RESPONSE:

Great job on solving the problem! Your use of logic to narrow down the possibilities was a great approach. It's always important to think critically and use all tools at your disposal, not just formulas. Keep up the good work!
 

Related to How Can Arithmetic and Geometric Progressions Reveal Values of a and b?

What is an arithmetic progression?

An arithmetic progression is a sequence of numbers where the difference between any two consecutive terms is constant. For example, 2, 5, 8, 11, 14 is an arithmetic progression with a common difference of 3.

What is a geometric progression?

A geometric progression is a sequence of numbers where the ratio between any two consecutive terms is constant. For example, 2, 6, 18, 54, 162 is a geometric progression with a common ratio of 3.

How do I find the nth term in an arithmetic progression?

The nth term in an arithmetic progression can be found using the formula an = a1 + (n-1)d, where a1 is the first term and d is the common difference. For example, in the sequence 2, 5, 8, 11, 14, the 4th term would be a4 = 2 + (4-1)3 = 11.

How do I find the sum of the first n terms in an arithmetic progression?

The sum of the first n terms in an arithmetic progression can be found using the formula Sn = n/2(2a1 + (n-1)d), where a1 is the first term and d is the common difference. For example, in the sequence 2, 5, 8, 11, 14, the sum of the first 4 terms would be S4 = 4/2(2*2 + (4-1)*3) = 30.

How do I find the sum of an infinite geometric progression?

The sum of an infinite geometric progression can be found using the formula S = a1/(1-r), where a1 is the first term and r is the common ratio. However, this formula only works if the absolute value of r is less than 1. Otherwise, the sum will approach infinity.

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