Arithmetic and Geometric Progressions

bummielove · **Joined:** Thu, 7 Jul 2011 18:07:39 UTC **Posts:** 11

I need some help here! We've been given the formula for Arithmetic and Geometric Progressions but the problems below, I cannot apply the formula. Its kinda tricky. Would you help me? These are sample questions for our long test. I need these so I can reviewT_T

1 Find an arithmetic progression if the sum of its first four terms is equal to 26 and the product of the same terms equals 880.

2. An arithmetic progression consists 20 terms. The sum of the terms occupying even places equals 259 while that of the terms occupying odd places equals 220. Find the two medium terms of the progression.

3 Find the four numbers forming a geometric progression in which the sum of the extremes is 27 and the product of the means is 72.

THANK YOU VERY MUCH.

Shadow · **Posted:** Fri, 26 Aug 2011 01:53:32 UTC

bummielove wrote:

I need some help here! We've been given the formula for Arithmetic and Geometric Progressions but the problems below, I cannot apply the formula. Its kinda tricky. Would you help me? These are sample questions for our long test. I need these so I can reviewT_T

1 Find an arithmetic progression if the sum of its first four terms is equal to 26 and the product of the same terms equals 880.

2. An arithmetic progression consists 20 terms. The sum of the terms occupying even places equals 259 while that of the terms occupying odd places equals 220. Find the two medium terms of the progression.

3 Find the four numbers forming a geometric progression in which the sum of the extremes is 27 and the product of the means is 72.

THANK YOU VERY MUCH.

Remember, an arithmetic progression is a_n=a_0+nd

where d is the common difference. So if we sum up all of the first FOUR terms say, that's $a_0+a_0+d+a_0+2d+a_0+3d=4a_0+(1+2+3)d=4d+{3(3+1)\over 2}d$ This means 4a_0+6d=26

, now divide both sides by 2 and you get 2a_0+3d=13

.

So we make the reasonable guess that a_0=2

, which by (*)

gives us

. Now we check, does this indeed work? Well, we get: 2*(2+3)*(2+6)*(2+9)=2*5*8*11=2^4*5*11=880

, so we're good.

(If this had not worked, we'd next try a_0=5

which would give d=1

, but that one doesn't work. Basically try the nice numbers until you get one that works.

2. This time I'll start the first term calling it a_1

and you'll see why, then the even terms are:

$a_2+a_4+\ldots +a_{20}$
=(a_1+d)+(a_1+3d)+

$\ldots + (a_1+11d)$
=10a_1+(1+3+5+

$\ldots + 11)d$
=10a_1+6^2d

which is impossible because the left side is even and the right side is not. Even if you assume the first term is the zeroth term, you'll end up with an even thing on the left because it will be $10a_0+2{9(9+1)\over 2}d$

3. Again, four numbers, label the common ratio as r, and then we get: a, ar, ar^2, ar^3

The sum of the extremes is a+ar^3

, but I don't understand what you mean by "means" which mean are you talking about? Which numbers are you averaging to get these means?

bummielove · **Joined:** Thu, 7 Jul 2011 18:07:39 UTC **Posts:** 11

Shadow wrote:

THANK YOU SHADOW FOR YOUR HELP!

3. Again, four numbers, label the common ratio as r, and then we get: a, ar, ar^2, ar^3

The sum of the extremes is a+ar^3

, but I don't understand what you mean by "means" which mean are you talking about? Which numbers are you averaging to get these means?

Maybe the means are the middle two terms and the extremes are the outer two terms (?). I have no idea. =))

Shadow · **Posted:** Fri, 26 Aug 2011 03:28:53 UTC

bummielove wrote:

Shadow wrote:

THANK YOU SHADOW FOR YOUR HELP!

3. Again, four numbers, label the common ratio as r, and then we get: a, ar, ar^2, ar^3

The sum of the extremes is a+ar^3

, but I don't understand what you mean by "means" which mean are you talking about? Which numbers are you averaging to get these means?

Maybe the means are the middle two terms and the extremes are the outer two terms (?). I have no idea. =))

The means are definitely the first and last, but I'm fairly confident means signifies what I think it does, which is something techinical.

Shadow · **Posted:** Fri, 26 Aug 2011 03:39:08 UTC

Hmm, okay so possibly the numbers involved are not integers, but then that's a really wacky question to ask for someone at your level. I'll try and think on it more to make sure there's really not much you can do.

bummielove · **Joined:** Thu, 7 Jul 2011 18:07:39 UTC **Posts:** 11

Shadow wrote:

Hmm, okay so possibly the numbers involved are not integers, but then that's a really wacky question to ask for someone at your level. I'll try and think on it more to make sure there's really not much you can do.

Thank You Shadow!

Shadow · **Posted:** Fri, 26 Aug 2011 03:45:00 UTC

Oh, Okay, I got it. I was silly and forgot a bunch of terms, sorry about that.

Here we go.

You're supposed to subtract.

the sum of the ten evens is 10a_1+10^2d=10a_1+100d

and the sum of the first ten evens is:

$10a_1+2{9(9+1)\over 2}=10a_1+90d$

subtract and you get:

10d=259-220=39

so

From this it is easy to get the two middle terms, you can just use 10a_1+100(3.9)=259

to solve for a_1

then get the two middle terms directly.

(note: I get the number of times I add

by using the fact that the sum of the first n odd numbers is n^2

and the sum of the first n even numbers (starting at 0) is 2{n(n-1)/2}

)

Soroban · **Posted:** Fri, 26 Aug 2011 17:29:27 UTC

Hello, bummielove!

Quote:

2. An arithmetic progression consists 20 terms.
The sum of the terms occupying even places equals 259,
while that of the terms occupying odd places equals 220.
Find the two medium ? terms of the progression.
I assume you mean the two middle terms.

The sequence of "even" terms has first term a_2

, common difference

, and

terms.
. . Its sum is: . $\frac{10}{2}[2a_2 + 9(2d)] \:=\:259 \quad\Rightarrow\quad a_2 + 9d \:=\:25.9$ .[1]

The sequence of "odd" terms has first term a_1

, common difference

, and

terms.
. . Its sum is: . $\frac{10}{2}[2a_1 + 9(2d)] \:=\:220 \quad\Rightarrow\quad a_1 + 9d \:=\:22$ .[2]

Subtract [1] - [2]: . $a_2 - a_1 \:=\:3.9 \quad\Rightarrow\quad \text{ Hence: }\:d \:=\:3.9$

Substitute into [2]: . $a_1+9(3.9) \:=\:22 \quad\Rightarrow\quad a_1 \:=\:\text{-}13.1$

Therefore: . $\begin{Bmatrix}a_{10} &=& \text{-}13.1 + 9(3.9) &=& 22 \\ a_{11} &=& \text{-}13.1 + 10(3.9) &=& 25.0 \end{Bmatrix}$

Shadow · **Posted:** Fri, 26 Aug 2011 19:03:46 UTC

Soroban wrote:

Hello, bummielove!

Quote:

2. An arithmetic progression consists 20 terms.
The sum of the terms occupying even places equals 259,
while that of the terms occupying odd places equals 220.
Find the two medium ? terms of the progression.
I assume you mean the two middle terms.

The sequence of "even" terms has first term a_2

, common difference

, and

terms.
. . Its sum is: . $\frac{10}{2}[2a_2 + 9(2d)] \:=\:259 \quad\Rightarrow\quad a_2 + 9d \:=\:25.9$ .[1]

The sequence of "odd" terms has first term a_1

, common difference

, and

terms.
. . Its sum is: . $\frac{10}{2}[2a_1 + 9(2d)] \:=\:220 \quad\Rightarrow\quad a_1 + 9d \:=\:22$ .[2]

Subtract [1] - [2]: . $a_2 - a_1 \:=\:3.9 \quad\Rightarrow\quad \text{ Hence: }\:d \:=\:3.9$

Substitute into [2]: . $a_1+9(3.9) \:=\:22 \quad\Rightarrow\quad a_1 \:=\:\text{-}13.1$

Therefore: . $\begin{Bmatrix}a_{10} &=& \text{-}13.1 + 9(3.9) &=& 22 \\ a_{11} &=& \text{-}13.1 + 10(3.9) &=& 25.0 \end{Bmatrix}$

Um, if those are consecutive terms, shouldn't they vary by 3.9? 25.0-22.0=3.0. . .

Soroban · **Posted:** Sat, 27 Aug 2011 04:29:47 UTC

A silly typo! . . . Struck "0" instead of "9".

It should have been:

. . Therefore: . $\begin{Bmatrix}a_{10} &=& \text{-}13.1 + 9(3.9) &=& 22 \\ a_{11} &=& \text{-}13.1 + 10(3.9) &=& 25.9 \end{Bmatrix}$

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Arithmetic and Geometric Progressions

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