Question 1106491
<br>
The sum of n terms of an arithmetic series is<br>
(number of terms) times (average of first and last terms)<br>
If the first term is a and the common difference is d, then the sum of the first and last terms is
{{{a + a+(n-1)d = 2a+(n-1)d}}}
and the sum of the first n terms is then
{{{n((2a+(n-1)d)/2)}}}<br>
We are told that the sum of the first n terms of the sequence is 5n^2+4n.  So<br>
{{{n((2a+(n-1)d)/2) = 5n^2+4n}}}
{{{n(2a+dn-d) = 10n^2+8n}}}
{{{dn + (2a-d) = 10n+8}}}<br>
This tells us
{{{dn = 10n}}} which means d is 10, and
{{{2a-d = 8}}}
{{{2a-10 = 8}}}
{{{2a = 18}}}
{{{a = 9}}}<br>
So the arithmetic series has first term 9 and common difference 10.<br>
Then the first three terms are 9, 19, and 29.