Question 949553
WHAT IS THE 30TH TERM OF GEOMETRIC SEQUENCE IF THE 10TH IS 3/512 AND 15TH IS 3/16384
<pre>
Use 

{{{a[n]=a[1]r^(n-1)}}}

Substitute {{{n=10}}}, {{{a[10]=3/512=3/2^9}}}

{{{a[10]=a[1]r^(10-1)}}}
{{{3/2^9=a[1]r^9}}}
Divide both sides by {{{r^9}}}
{{{3/(2^9r^9)=a[1]}}}
{{{3/(2r)^9=a[1]}}}

Substitute {{{n=15}}}, {{{a[15]=3/16384= 3/2^14}}}

{{{a[15]=a[1]r^(15-1)}}}
{{{3/2^14=a[1]r^14}}}
Divide both sides by {{{r^14}}}
{{{3/(2^14r^14)=a[1]}}}
{{{3/(2r)^14=a[1]}}}


Equate the two expressions for {{{a[1]}}}

{{{3/(2r)^9=3/(2r)^14}}}

Divide both sides by 3

{{{1/(2r)^9=1/(2r)^14}}}

Cross-multiply

{{{(2r)^14 = (2r)^9}}}

Divide both sides by {{{(2r)^9}}}

{{{(2r)^5 = 1}}}

Take 5th roots of both sides

{{{2r=1}}}

{{{r=1/2}}}

Substitute 2r=1 in

{{{3/(2r)^9=a[1]}}}

{{{3/(1)^9=a[1]}}}

{{{3/1=a[1]}}}

{{{3=a[1]}}}

Now you finish by substituting n=30, a<sub>1</sub>=3, and {{{r=1/2}}}

into

{{{a[n]=a[1]r^(n-1)}}}

To find a<sub>30</sub>.

Edwin</pre>