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
Looks to me to imply:
$\displaystyle S_4=16(S_8-S_4)$. 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 $\displaystyle x=r^4$). I don't know if there's a better approach.
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 $\displaystyle |r| < 1$. This will come in useful for checking the answer.
$\displaystyle S_4 = 16(S_8-S_4)$
$\displaystyle S_4 = \dfrac{a(1-r^4)}{1-r} = \dfrac{a(1-r^2)(1+r^2)}{1-r}= a(1+r)(1+r^2)$
$\displaystyle S_8 = \dfrac{a(1-r^8)}{1-r} = \dfrac{a(1-r^4)(1+r^4)}{1-r} = a(1+r^4)(1+r^2)(1+r)$
$\displaystyle S_8 - S_4 = a(1+r^4)(1+r^2)(1+r) - a(1+r)(1+r^2) $
$\displaystyle S_8 - S_4 = \underbrace{a(1+r)(1+r^2)}_{\text{ This is a common factor}}(1+r^4-1) = a(1+r)(1+r^2)r^4$
Now to put this expression back into the original expression:
$\displaystyle \underbrace{a(1+r)(1+r^2)}_{\text{ This is }S_4} = 16\underbrace{a(1+r)(1+r^2)r^4}_{\text{This is }S_8-S_4}$
Do some cancelling and your equation will become clear. Since you have an even power of r there are two possible answers
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:.$\displaystyle a_1 + a_2 + a_3 + a_4 \;=\;16(a_5+a_6+a_7+a_8)$
Hence: .$\displaystyle a + ar + ar^2 + ar^3 \;=\;16(ar^4 + ar^5 + ar^6 + ar^7)$
Divide by $\displaystyle a\!:\;\;1 + r + r^2 + r^3 \;=\;16(r^4 + r^5 + r^6 + r^7)$
Factor: .$\displaystyle 1 + r + r^2 + r^3 \;=\;16r^4(1+r+r^2+r^3)$
. . $\displaystyle 16r^4(1+r+r^2+r^3) - (1+r+r^2+r^3) \;=\;0$
Factor: .$\displaystyle (1+r+r^2+r^3)(16r^4-1) \;=\;0$
We have two equations to solve:
$\displaystyle r^3 + r^2 + r + 1 \:=\:0 \quad\Rightarrow\quad r^2(r+1) + (r+1) \:=\:0 $
. . $\displaystyle (r+1)(r^2+1) \:=\:0 \quad\Rightarrow\quad \boxed{r \:=\:-1}$
$\displaystyle 16r^4-1 \:=\:0 \quad\Rightarrow\quad r^4 \:=\:\tfrac{1}{16} \quad\Rightarrow\quad \boxed{r \:=\:\pm\tfrac{1}{2}}$
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.