# geometric series help asap!!! would be so appppreciated

• November 22nd 2005, 05:19 AM
emydawn
geometric series help asap!!! would be so appppreciated

a geometric series has first term a and common ratio r where r>1

the sum of the first n terms of the series is denoted by Sn.

given that S4= 10 x S2,

a) find the value of r

given that S3= 26,

b)find the value of a,

c) show that s6 = 728
• November 22nd 2005, 02:46 PM
ThePerfectHacker
The formula for a geometric series is Sn=a[(1-r^n)/(1-r)]
Given that:
S4=10S2
S3=26
r>1 (This is only important to avo1d division by zero)

Solution:
Since, S4=10S2 we have:
a[(1-r^4)/(1-r)]=10[(1-r^2)/(1-r)] multiply by (1-r):
a[(1-r^4)]=10[(1-r^2)] Factor (1-r^4):
a[(1-r^2)(1+r^2)]=10[(1-r^2)] divide by (1-r^2):
a(1+r^2)=10 THIS IS YOUR FIRST EQUATION.
Next:
Since S3=26 we have:
a[(1-r^3)/(1-r)]=26 Long divide 1-r^3 by 1-r to get:
a(1+r+r^2)=26 THIS IS YOUR SECOND EQUATION.
Next:
Divide the first equation by the second:
(1+r^2)/(1+r+r^2)=10/26 or 5/13. Cross Multiply:
13+13r^2=5+5r+5r^2 Collect Terms:
8r^2-5r+8=0
Which has no real solution thus your problem is impossible
• November 23rd 2005, 09:47 AM
CaptainBlack
Quote:

Originally Posted by ThePerfectHacker

Since, S4=10S2 we have:
a[(1-r^4)/(1-r)]=10[(1-r^2)/(1-r)] multiply by (1-r):

$S(4)\ =\ 10.S(2)$ implies:

$a.\frac{1-r^4}{1-r}\ =\ 10.a.\frac{1-r^2}{1-r}$

or, cancelling a and (1-r):

$(1-r^4)\ =\ 10.(1-r^2)$

or:

$\frac{1-r^4}{1-r^2}\ =\ 10$.

Which seems to have solution $r\ =\ 3$.

RonL
• November 23rd 2005, 09:58 AM
CaptainBlack
Next:
Since $S(3)\ =\ 26$ we have:

$a \frac{1-r^3}{1-r}=26$,
but from the first part we know that $r\ =\ 3$, so:

$13.a\ =\ 26$.
so $a\ =\ 2$.

Finally:

$S(6)\ =\ 2.\frac{1-3^6}{1-3}\ =\ 728$.

RonL
• November 23rd 2005, 06:39 PM
ThePerfectHacker
I new I did something wrong I forget the "a" term!