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Series. AP GP. Some confusion. Answer checking needed also.

  • Thread starter SolCon
  • Start date
  • #1
33
1
Hi again.

Alright, I'm having a problem in this question which has 3 parts.

The question is: An arithmetic progression (ap) has first term a and common difference d.

Part 1 of it says to write down expressions for 2nd and 6th terms in terms of a and d.
This is no problem as it is is simply: a (1st term) and a+5d (2nd term).

Part 2 says: The first, second, and sixth terms of this ap are also the first three terms of a geometric progression (gp). Prove that d=3a.
I've got this question done also but not before having seen the initial equation which was:

(a+d)/a = (a+5d)/a+d

This would give us d=3a. However, I don't know why it is written like this and so have actually simply memorized the equation format as "[Small term]/a = [Big term]/[Small term]" . E.g, in another similar question, we are required to express the 5th and 15th term in terms of a and d. This will give us a+4d and a+14d.

So, using the same format:

[Small term]/a = [Big term]/[Small term]

(a+4d)/a = (a+14d)/a+4d.

I would just really appreciate an explanation of this, as I haven't been able to find one.

As for the third part, it says: Given that a=2, find the sum of the first 15 terms of each progression.
Done this but don't have the answers to match (and I could have done it wrong :uhh: ).

The Ap sum is coming as 345. We'll use d=3a (getting d=6) and have the following:

S.n= n/2[2(a)+(n-1)d]
or
S.15= 15/2[2(2)+(15-1)6]

For Gp sum, I'm confused. The gp sum uses infinity formula (s.inf=a/1-r). The common ration (r) we are getting is 4. In part 2 it said that the 1st, 2nd and 6th ap terms are equal to first 3 terms of gp. As a=2 and d=6, 2nd term would be 6+2=8. If we multiply 2 by 4, we get 8, so 'r' is 4.

Then:

s.inf=a/1-r
of
s.inf=2/1-4 which comes as -2/3 which doesn't seem right.
In the other GP formula: Sn=a(1-r^n)/1-r.

We get: S.15=2(1-4^15)/1-4
which is 715,827,882.

Is the ap sum correct? And which of the gp are correct (if any at all)?

Thanks for your help. :smile:
 

Answers and Replies

  • #2
VietDao29
Homework Helper
1,423
2
Hi again.

Alright, I'm having a problem in this question which has 3 parts.

The question is: An arithmetic progression (ap) has first term a and common difference d.

Part 1 of it says to write down expressions for 2nd and 6th terms in terms of a and d.
This is no problem as it is is simply: a (1st term) and a+5d (2nd term).
Be careful, the problem asks for the second term, and the sixth term. :)

Part 2 says: The first, second, and sixth terms of this ap are also the first three terms of a geometric progression (gp). Prove that d=3a.
I've got this question done also but not before having seen the initial equation which was:

(a+d)/a = (a+5d)/a+d

This would give us d=3a. However, I don't know why it is written like this and so have actually simply memorized the equation format as "[Small term]/a = [Big term]/[Small term]" . E.g, in another similar question, we are required to express the 5th and 15th term in terms of a and d. This will give us a+4d and a+14d.

So, using the same format:

[Small term]/a = [Big term]/[Small term]

(a+4d)/a = (a+14d)/a+4d.

I would just really appreciate an explanation of this, as I haven't been able to find one.
I'll explain the first one, the second one is pretty much the same. So, in our AP, we have:
  • First term: a.
  • Second term: a + d.
  • Sixth term: a + 5d.

And these three terms must make the first 3 terms of some other GP. A GP means that each later term is founded by multiplying the former (previous) term by a constant r (known as common ratio). Which means if we divide the later term by the former one, we'll get back r.

So, let's say we have a GP, of which terms are denoted by bi.
[tex]b_1; b_2; b_3; b_4 ...[/tex]
So, we'll have:
[tex]\frac{b_2}{b_1} = \frac{b_3}{b_2} = \frac{b_4}{b_3} = ... = r[/tex]

We know that the first three terms of some GP is:
a; a + d; and a + 5d
So:
[tex]\frac{a + d}{a} = \frac{a + 5d}{a + d}[/tex]

As for the third part, it says: Given that a=2, find the sum of the first 15 terms of each progression.
Done this but don't have the answers to match (and I could have done it wrong :uhh: ).

The Ap sum is coming as 345. We'll use d=3a (getting d=6) and have the following:

S.n= n/2[2(a)+(n-1)d]
or
S.15= 15/2[2(2)+(15-1)6]
You're using the correct formula, however, your result is incorrect. You should re-check it.

For Gp sum, I'm confused. The gp sum uses infinity formula (s.inf=a/1-r). The common ration (r) we are getting is 4. In part 2 it said that the 1st, 2nd and 6th ap terms are equal to first 3 terms of gp. As a=2 and d=6, 2nd term would be 6+2=8. If we multiply 2 by 4, we get 8, so 'r' is 4.

Then:

s.inf=a/1-r
of
s.inf=2/1-4 which comes as -2/3 which doesn't seem right.
You've misunderstood the usage of this formula. This formula is used to calculate the value of Infinite Geometric Series. By infinite, we mean that the number of terms that we take into summation is infinite. In this problem, you're asked for the sum of the first 15 terms, which is, of course, finite.

The theorem is that:
If a GP has the common ratio r, such that: -1 < r < 1. Then the series:
[tex]\sum_{i = 1} ^ {\infty} a.r ^ {i - 1}[/tex] converge. And it's value can be calculated by:

[tex]\sum_{i = 1} ^ {\infty} a.r ^ {i - 1} = a \frac{1}{1 - r}[/tex]

In the other GP formula: Sn=a(1-r^n)/1-r.

We get: S.15=2(1-4^15)/1-4
which is 715,827,882.

Is the ap sum correct? And which of the gp are correct (if any at all)?

Thanks for your help. :smile:
Yup, this is correct. Congratulations. :)
 
  • #3
33
1
Ah yes. :smile:

Thanks for that explanation. Didn't know that it was 'r' that caused the structure of that formula.

Also, I made 2 mistakes above. The first you pointed out; yes, I meant to write sixth term again instead of second term, but got mixed up with the question's arrangement. The second mistake, about the AP sum, I don't know where I went wrong but the answer I'm getting now is 660. Is this right?

Other then this, I thank you for both explanations about the common ration in GP equation and the sum to infinity equation. I was having confusion in them but its cleared up now. :smile: .
 
  • #4
VietDao29
Homework Helper
1,423
2
Ah yes. :smile:

Thanks for that explanation. Didn't know that it was 'r' that caused the structure of that formula.

Also, I made 2 mistakes above. The first you pointed out; yes, I meant to write sixth term again instead of second term, but got mixed up with the question's arrangement. The second mistake, about the AP sum, I don't know where I went wrong but the answer I'm getting now is 660. Is this right?

Other then this, I thank you for both explanations about the common ration in GP equation and the sum to infinity equation. I was having confusion in them but its cleared up now. :smile: .
Looks good to me. :)
 

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