How was this answer achieved in this geometric sequence question? Find Tn of the geometric sequence 1, 1.4, 1/16, 1/64

''Find Tn of the geometric sequence # 1, 1/4, 1/16, 1/64, ... # ''

# Tn = a.r^(n-1) #

=#1.(1/4)^(n-1)#

=#1/4^(n-1)#

#Tn = 4^(-n+1)#

I do not understand why and how #1.(1/4)^(n-1)# became #1/4^(n-1)#

and why or how #1/4^(n-1)# became #Tn = 4^(-n+1)#

1 Answer
Somebody N.
Feb 3, 2018

See below.

Explanation:

If #a ,b ,c# are in geometric sequence, then:

#b/a=c/b# This is known as the common ratio.

From the example:

#1,1/4,1/16,1/64#

#(1/4)/1=(1/16)/(1/4)=1/4#

So the common ratio is #1/4#

The nth term of a geometric sequence is given by:

#ar^(n-1)#

Where, #a# is the 1st term, #r# is the common ratio and #n# is the nth term.

In this case:

#a=1# and #r=1/4#

No term is specified so we leave #n-1# as it is.

So:

#1*(1/4)^(n-1)#

Notice that:

#(1/4)^(n-1)=(1^(n-1))/(4^(n-1)#

#1# raised to any power is always #1#

So:

#(1^(n-1))/(4^(n-1))=1/(4^(n-1)#

And:

#1*1/(4^(n-1))=1/(4^(n-1)#

Why

#1/(4^(n-1))=4^(-1+n)#

#a^-1<=>1/a#

So:

#1/(4^(n-1))=4^-(n-1)#

And:

#-(n-1)=-n+1#

Then:

#4^-(n-1)=4^(-n+1)#

These are just the laws of indices. It would do you well to study these. They are of great importance in all areas of mathematics.

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