<PRE><B>Question 846949</B>
The third term of an AP is ten more than the first term while the fifth term is 15 more than the second term. find the sum of the 8th and 15th term of the Ap if the 7th term is 7 times the first term.
&lt;pre&gt;
We use the formula for the nth term:

a&lt;sub&gt;n&lt;/sub&gt; = a&lt;sub&gt;1&lt;/sub&gt; + (n-1)d
&lt;/pre&gt;
&gt;&gt;The third term of an AP is ten more than the first term...&lt;&lt;
&lt;pre&gt;
a&lt;sub&gt;1&lt;/sub&gt; + 2d = a&lt;sub&gt;1&lt;/sub&gt; + 10
     2d = 10
      d = 5     
&lt;/pre&gt;
&gt;&gt;...the 7th term is...&lt;&lt;
&lt;pre&gt;
a&lt;sub&gt;7&lt;/sub&gt; = a&lt;sub&gt;1&lt;/sub&gt; + (7-1)(5) = a&lt;sub&gt;1&lt;/sub&gt; + (6)(5) =  a&lt;sub&gt;1&lt;/sub&gt; + 30
&lt;/pre&gt;
...the 7th term is 7 times the first term
&lt;pre&gt;
a&lt;sub&gt;7&lt;/sub&gt; = 7a&lt;sub&gt;1&lt;/sub&gt; 

So we have the system of equations:

a&lt;sub&gt;7&lt;/sub&gt; = 7a&lt;sub&gt;1&lt;/sub&gt;
a&lt;sub&gt;7&lt;/sub&gt; = a&lt;sub&gt;1&lt;/sub&gt; + 30

7a&lt;sub&gt;1&lt;/sub&gt; = a&lt;sub&gt;1&lt;/sub&gt; + 30
6a&lt;sub&gt;1&lt;/sub&gt; = 30
a&lt;sub&gt;1&lt;/sub&gt; = 5

That makes things very easy.  We don&#39;t need the formula anymore,
for now we can find any term easily.  That&#39;s because the 
arithmetic sequence is just counting by fives:

5,10,15,20,25,30,...

So to find the nth term you just multiply n by 5.
&lt;/pre&gt;
&gt;&gt;...find the sum of the 8th and 15th term...&lt;&lt;
&lt;pre&gt;
8(5) + 15(5) = 40 + 75 = 115

Edwin&lt;/pre&gt;</PRE>