<PRE><B>Question 830892</B>
the 3rd and 6th term of a geometric progression are 48 and 14 2/9 respectively.
&lt;pre&gt;
a&lt;sub&gt;1&lt;/sub&gt;, a&lt;sub&gt;2&lt;/sub&gt;, a&lt;sub&gt;3&lt;/sub&gt;=48, a&lt;sub&gt;4&lt;/sub&gt;, a&lt;sub&gt;5&lt;/sub&gt;, a&lt;sub&gt;6&lt;/sub&gt;={{{14&amp;2/9}}}

You need this formula for the nth term:

a&lt;sub&gt;n&lt;/sub&gt; = a&lt;sub&gt;1&lt;/sub&gt;·r&lt;sup&gt;(n-1)&lt;/sup&gt;

[Some books and teachers use &quot;t&quot; for &quot;term&quot; instead of &quot;a&quot;.
I&#39;ll use &quot;a&quot;.]

&lt;/pre&gt;
the 3rd and 6th term of a geometric progression are 48 and 14 2/9 respectively.
&lt;pre&gt;
a&lt;sub&gt;3&lt;/sub&gt; = a&lt;sub&gt;1&lt;/sub&gt;·r&lt;sup&gt;(3-1)&lt;/sup&gt;
48 = a&lt;sub&gt;1&lt;/sub&gt;·r&lt;sup&gt;2&lt;/sup&gt;

a&lt;sub&gt;6&lt;/sub&gt; = a&lt;sub&gt;1&lt;/sub&gt;·r&lt;sup&gt;(6-1)&lt;/sup&gt;
{{{14&amp;2/9}}} = a&lt;sub&gt;1&lt;/sub&gt;·r&lt;sup&gt;5&lt;/sup&gt;

Change {{{14&amp;2/9}}} to an improper fraction {{{128/9}}}

{{{128/9}}} = a&lt;sub&gt;1&lt;/sub&gt;·r&lt;sup&gt;5&lt;/sup&gt;

Clear of fractions by multipolying both sides by 9

128 = 9a&lt;sub&gt;1&lt;/sub&gt;·r&lt;sup&gt;5&lt;/sup&gt;

So you have this system of equations:

 48 = a&lt;sub&gt;1&lt;/sub&gt;·r&lt;sup&gt;2&lt;/sup&gt;
128 = 9a&lt;sub&gt;1&lt;/sub&gt;·r&lt;sup&gt;5&lt;/sup&gt;

Solve the first for a&lt;sub&gt;1&lt;/sub&gt;

{{{48/r^2}}}{{{&quot;&quot;=&quot;&quot;}}}{{{(a[1]r^2)/r^2}}}

{{{48/r^2}}}{{{&quot;&quot;=&quot;&quot;}}}{{{(a[1]cross(r^2))/cross(r^2)}}}

{{{48/r^2}}} = a&lt;sub&gt;1&lt;/sub&gt;

Substitute in

128 = 9{{{(48/r^2)}}}·r&lt;sup&gt;5&lt;/sup&gt;

128 = {{{432r^5/r^2}}}

Divide on the right by subtracting exponents:

128 = 432rł

Divide both sides by 432

{{{128/432}}} = {{{432r^3/432}}}

Reduce the fraction on the left and cancel on the right:

{{{8/27}}} = {{{cross(432)r^3/cross(432)}}}

{{{8/27}}} = rł

The cube root of 8 is 2 and the cube root of 27 is 3, so
taking cube roots of both sides:

{{{2/3}}} = r

Substitute for r in

{{{48/r^2}}} = a&lt;sub&gt;1&lt;/sub&gt;

{{{48/(2/3)^2}}} = a&lt;sub&gt;1&lt;/sub&gt;

{{{48/((4/9))}}} = a&lt;sub&gt;1&lt;/sub&gt; 

To divide by a fraction, invert it and multiply

{{{48*(9/4)}}} = a&lt;sub&gt;1&lt;/sub&gt;

4 goes into 48 12 times:

 &lt;sub&gt;12&lt;/sub&gt;
{{{cross(48)*(9/cross(4))}}} = a&lt;sub&gt;1&lt;/sub&gt;
     &lt;sup&gt;1&lt;/sup&gt;
 
108 = a&lt;sub&gt;1&lt;/sub&gt;
&lt;/pre&gt;
Write down the 1st 4 terms of the geometric progression.
&lt;pre&gt;
1st term = {{{108}}},
2nd term = {{{108*expr(2/3)}}}{{{&quot;&quot;=&quot;&quot;}}}{{{72}}}
3rd term = {{{72*expr(2/3)}}}{{{&quot;&quot;=&quot;&quot;}}}{{{48}}}
4th term = {{{48*expr(2/3)}}}{{{&quot;&quot;=&quot;&quot;}}}{{{32}}}

the 3rd term checks, but to check the whole thing,
lets see if the 6th term is {{{14&amp;2/9}}}

5th term = {{{32*expr(2/3)}}}{{{&quot;&quot;=&quot;&quot;}}}{{{64/3}}}{{{&quot;&quot;=&quot;&quot;}}}{{{21&amp;1/3}}}

6th term = {{{expr(64/3)*expr(2/3)}}}{{{&quot;&quot;=&quot;&quot;}}}{{{128/9}}}{{{&quot;&quot;=&quot;&quot;}}}{{{14&amp;2/9}}}

So it checks.

Edwin&lt;/pre&gt; </PRE>