# Geometric Sequences..

• Nov 11th 2010, 08:40 PM
tmas
Geometric Sequences..
I don't understand how the book got the answer of 5/9

The question is:

Given the two terms of the geometric sequence, find the individual term.
t4=5 t8=405 find t2

i used the formula tn = ar^n-1 for this

t8/t4 = 405/5 = ar^7 / ar^3
=81 = ar^4
r=3

5=a(3)^4
5=81a
a = 5/81

t2=(5/81)(3)^1
=5/27
• Nov 11th 2010, 08:58 PM
Prove It
You should know that $\displaystyle \displaystyle t_n = a\,r^{n - 1}$.

Therefore $\displaystyle \displaystyle t_4 = a\,r^3$ and $\displaystyle \displaystyle t_8 = a\,r^7$.

From this we have

$\displaystyle \displaystyle a\,r^7 = 405$

$\displaystyle \displaystyle a\,r^3 = 5$.

Dividing the first equation by the second gives

$\displaystyle \displaystyle r^4 = 81$

$\displaystyle \displaystyle r = \pm 3$.

Substituting back into one of the equations:

$\displaystyle \displaystyle 405 = a\,(\pm 3)^7$

$\displaystyle \displaystyle 405 = \pm 2187 a$

$\displaystyle \displaystyle a = \pm\frac{5}{27}$.

So there are two possible sequences, the first with $\displaystyle \displaystyle a = -\frac{5}{27}$ and $\displaystyle \displaystyle r = -3$, the second with $\displaystyle \displaystyle a = \frac{5}{27}$ and $\displaystyle \displaystyle r = 3$.

Either way, you should find that

$\displaystyle \displaystyle t_2 = a\,r = \frac{5}{27}\cdot 3 = \frac{5}{9}$ or $\displaystyle \displaystyle t_2 = -\frac{5}{27}\left(-3\right) = \frac{5}{9}$.