Finding Nth Term

• Feb 11th 2009, 10:14 AM
AlgebraicallyChallenged
Finding Nth Term
Dear Forum , I have to Find the nth term of the geometric sequence for the following,

8,-4/3,2/9...

I was curious if my following work is correct , please let me know what you think:

8,(-4)/(3),(2)/(9)

Remove all extra parentheses from the expression.
8,-(4)/(3),(2)/(9)

This is a geometric sequence since there is a common ratio between each term. In this case, multiplying the previous term in the sequence by -(1)/(6) gives the next term. In other words, an=r*an-1.
Geometric Sequence: r=-(1)/(6)

This is the form of a geometric sequence.
an=a1r^(n-1)

Substitute in the values of a1=8 and r=-(1)/(6).
an=8*(1)/(6^(n-1))

Multiply 8 by (1)/(6^(n-1)) to get (8)/(6^(n-1)).
an=(8)/(6^(n-1))

Substitute in the value of n to find the nth term.
a2=(8)/(6^((2)-1))

Remove the parentheses around the expression 2.
a2=(8)/(6^(2-1))

Subtract 1 from 2 to get 1.
a2=(8)/(6)

Reduce the expression (8)/(6) by removing a factor of 2 from the numerator and denominator.
a2=(4)/(3)
• Feb 11th 2009, 11:34 AM
Plato
That is the correct basic idea. But you forgot the signs.
$8\left(\frac{{\color{red}-}1}{6}\right)^{n-1}$