<PRE><B>Question 816667</B>
let a&lt;sub&gt;n&lt;/sub&gt; be an arithmetic sequence. if a&lt;sub&gt;4&lt;/sub&gt;=27 and a&lt;sub&gt;9&lt;/sub&gt;=67, what is a&lt;sub&gt;1&lt;/sub&gt;?
&lt;pre&gt;
a&lt;sub&gt;n&lt;/sub&gt; = a&lt;sub&gt;1&lt;/sub&gt; + (n-1)d

a&lt;sub&gt;4&lt;/sub&gt; = a&lt;sub&gt;1&lt;/sub&gt; + (4-1)d
27 = a&lt;sub&gt;1&lt;/sub&gt; + 3d

a&lt;sub&gt;9&lt;/sub&gt; = a&lt;sub&gt;1&lt;/sub&gt; + (9-1)d
67 = a&lt;sub&gt;1&lt;/sub&gt; + 8d

So we solve this system of equations:

27 = a&lt;sub&gt;1&lt;/sub&gt; + 3d
67 = a&lt;sub&gt;1&lt;/sub&gt; + 8d

To eliminate d, multiply the first 
equation through by -8, and multiply
the second equation through by 3

-216 = -8a&lt;sub&gt;1&lt;/sub&gt; - 24d
 201 =  3a&lt;sub&gt;1&lt;/sub&gt; + 24d
-------------------
 -15 = -5a&lt;sub&gt;1&lt;/sub&gt;
   3 = a&lt;sub&gt;1&lt;/sub&gt;

--------------------

To check we find d by substituting
3 for a&lt;sub&gt;1&lt;/sub&gt; in

27 = a&lt;sub&gt;1&lt;/sub&gt; + 3d
27 = 3 + 3d
24 = 3d
 8 = d

So the sequence is found by starting with 
3 and adding 8 each time to get the next
term:

3, 11, 19, &lt;font color=&quot;red&quot;&gt;27&lt;/font&gt;, 35, 43, 51, 59, &lt;font color=&quot;red&quot;&gt;67&lt;/font&gt;

Edwin&lt;/pre&gt; </PRE>