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matadorqk
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Arithmetic/Geometric Progression Problem SOLVED
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.
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
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
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
Last edited: