Geometric Sequence

**gilagila** · December 18th 2011, 10:23 PM

In a geometric sequence, the sum of first four terms is 16 times the sum of the following four term, find the common ratio?

Sn =( a(1-r^n)) /1-r

anyone can tell me how to start or resolve this ??

Thank you

**Quacky** · December 18th 2011, 10:34 PM

Looks to me to imply:

$S_4=16(S_8-S_4)$ . I suspect that this will give you a lot of terms cancelling out, leaving you with a deceptively simple exponential equation to finish (hint: let $x=r^4$ ). I don't know if there's a better approach.

**e^(i*pi)** · December 19th 2011, 03:52 AM

Originally Posted by gilagila

In a geometric sequence, the sum of first four terms is 16 times the sum of the following four term, find the common ratio?

Sn =( a(1-r^n)) /1-r

anyone can tell me how to start or resolve this ??

Thank you

I'd use the difference of two squares (repeatedly if necessary) to clear the denominator since there is bound to be a common factor somewhere.

If the sum of the first four terms is larger than the sum of the next four then $|r| < 1$ . This will come in useful for checking the answer.

$S_4 = 16(S_8-S_4)$

$S_4 = \dfrac{a(1-r^4)}{1-r} = \dfrac{a(1-r^2)(1+r^2)}{1-r}= a(1+r)(1+r^2)$

$S_8 = \dfrac{a(1-r^8)}{1-r} = \dfrac{a(1-r^4)(1+r^4)}{1-r} = a(1+r^4)(1+r^2)(1+r)$

$S_8 - S_4 = a(1+r^4)(1+r^2)(1+r) - a(1+r)(1+r^2)$

$S_8 - S_4 = \underbrace{a(1+r)(1+r^2)}_{\text{ This is a common factor}}(1+r^4-1) = a(1+r)(1+r^2)r^4$

Now to put this expression back into the original expression:

$\underbrace{a(1+r)(1+r^2)}_{\text{ This is }S_4} = 16\underbrace{a(1+r)(1+r^2)r^4}_{\text{This is }S_8-S_4}$

Do some cancelling and your equation will become clear. Since you have an even power of r there are two possible answers

**Quacky** · December 19th 2011, 04:19 AM

Originally Posted by e^(i*pi)

I'd use the difference of two squares (repeatedly if necessary) to clear the denominator since there is bound to be a common factor somewhere.

Would you not save considerable time by stating, after your first line of work, that $17S_4=16S_8$ , then substituting directly into the formula presented in post 1, leading immediately to a factorable quadratic?

**Archie Meade** · December 19th 2011, 04:27 AM

Originally Posted by gilagila

In a geometric sequence, the sum of first four terms is 16 times the sum of the following four term, find the common ratio?

Sn =( a(1-r^n)) /1-r

anyone can tell me how to start or resolve this ??

Thank you

The geometric sequence is

$a,\;\;ar,\;\;ar^2,\;\;ar^3,\;\;ar^4,\;\;ar^5,\;\;a r^6,\;\;ar^7$

$T_5+T_6+T_7+T_8=r^4S_4$

$S_4=16r^4S_4\Rightarrow\ r^4=\frac{1}{16}$

gives the 2 non-trivial solutions.

**Soroban** · December 19th 2011, 05:37 AM

Hello, gilagila!

In a geometric sequence, the sum of first four terms
is 16 times the sum of the following four term.
Find the common ratio.

We are given that:. $a_1 + a_2 + a_3 + a_4 \;=\;16(a_5+a_6+a_7+a_8)$

Hence: . $a + ar + ar^2 + ar^3 \;=\;16(ar^4 + ar^5 + ar^6 + ar^7)$

Divide by $a\!:\;\;1 + r + r^2 + r^3 \;=\;16(r^4 + r^5 + r^6 + r^7)$

Factor: . $1 + r + r^2 + r^3 \;=\;16r^4(1+r+r^2+r^3)$

. . $16r^4(1+r+r^2+r^3) - (1+r+r^2+r^3) \;=\;0$

Factor: . $(1+r+r^2+r^3)(16r^4-1) \;=\;0$

We have two equations to solve:

$r^3 + r^2 + r + 1 \:=\:0 \quad\Rightarrow\quad r^2(r+1) + (r+1) \:=\:0$

. . $(r+1)(r^2+1) \:=\:0 \quad\Rightarrow\quad \boxed{r \:=\:-1}$

$16r^4-1 \:=\:0 \quad\Rightarrow\quad r^4 \:=\:\tfrac{1}{16} \quad\Rightarrow\quad \boxed{r \:=\:\pm\tfrac{1}{2}}$

**Quacky** · December 19th 2011, 06:41 AM

Originally Posted by Soroban

Hello, gilagila!

We are given that:. $a_1 + a_2 + a_3 + a_4 \;=\;16(a_5+a_6+a_7+a_8)$

Hence: . $a + ar + ar^2 + ar^3 \;=\;16(ar^4 + ar^5 + ar^6 + ar^7)$

Divide by $a\!:\;\;1 + r + r^2 + r^3 \;=\;16(r^4 + r^5 + r^6 + r^7)$

Factor: . $1 + r + r^2 + r^3 \;=\;16r^4(1+r+r^2+r^3)$

. . $16r^4(1+r+r^2+r^3) - (1+r+r^2+r^3) \;=\;0$

Factor: . $(1+r+r^2+r^3)(16r^4-1) \;=\;0$

We have two equations to solve:

$r^3 + r^2 + r + 1 \:=\:0 \quad\Rightarrow\quad r^2(r+1) + (r+1) \:=\:0$

. . $(r+1)(r^2+1) \:=\:0 \quad\Rightarrow\quad \boxed{r \:=\:-1}$

$16r^4-1 \:=\:0 \quad\Rightarrow\quad r^4 \:=\:\tfrac{1}{16} \quad\Rightarrow\quad \boxed{r \:=\:\pm\tfrac{1}{2}}$

With all due respect, within a reasonable time period before your reply, there were already 3 valid solutions presented which guided gilagila towards a solution, whilst allowing him to finish the work for himself. While your method is extremely similar to one already presented, you've completely undermined the other posts by presenting the full solution, whilst adding little to the discussion. I'm not ranting other than to say that I find this slightly irritating.

**Wilmer** · December 19th 2011, 08:01 AM

Originally Posted by Quacky

With all due respect, within a reasonable time period before your reply, there were already 3 valid solutions presented which guided gilagila towards a solution, whilst allowing him to finish the work for himself. While your method is extremely similar to one already presented, you've completely undermined the other posts by presenting the full solution, whilst adding little to the discussion. I'm not ranting other than to say that I find this slightly irritating.

Soroban is well-known for doing this at other sites...looks like he needs hugs

**gilagila** · December 20th 2011, 06:24 AM

thanks for all the people helping me, a simple question but I saw various of solutions, open a new world to me.

thank you very much

**gilagila** · December 20th 2011, 06:34 AM

Originally Posted by Archie Meade

The geometric sequence is

$a,\;\;ar,\;\;ar^2,\;\;ar^3,\;\;ar^4,\;\;ar^5,\;\;a r^6,\;\;ar^7$

$T_5+T_6+T_7+T_8=r^4S_4$

$S_4=16r^4S_4\Rightarrow\ r^4=\frac{1}{16}$

gives the 2 non-trivial solutions.

sorry to ask again

$T_5+T_6+T_7+T_8=r^4S_4$ <--- how u get this ??

**Siron** · December 20th 2011, 06:51 AM

$T_5+T_6+T_7+T_8=ar^4+ar^5+ar^6+ar^7=r^4(a+ar+ar^2+ ar^3)=r^4S_4$

**gilagila** · December 20th 2011, 05:18 PM

Originally Posted by Siron

$T_5+T_6+T_7+T_8=ar^4+ar^5+ar^6+ar^7=r^4(a+ar+ar^2+ ar^3)=r^4S_4$

oops....never noticed that if factories that can be like that, thank you very much

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