Question 1100175
<br>If the sum of the first and last numbers is an integer and the product of the two middle terms is an integer, then almost certainly all the numbers will be integers.<br>
Then if the product of the middle two numbers is 63, the two middle numbers are 7 and 9.<br>
So the common difference is 2; that makes the 6 numbers<br>
3, 5, 7, 9, 11, 13<br><br>
You could, of course, solve the problem using algebra; however, it takes far more work than the logical analysis solution shown above.<br>
Let the first term be a and the common difference be d; then the 6 numbers are
a, a+d, a+2d, a+3d, a+4d, a+5d<br>
Then
(1) {{{2a+5d = 16}}}  the sum of the first and last numbers is 16
(2) {{{(a+2d)(a+3d) = 63}}}  the product of the two middle numbers is 63<br>
Solve equation (1) for a: {{{a = (16-5d)/2}}}<br>
Substitute into equation (2):
{{{((16-d)/2)((16+d)/2) = 63}}}
{{{(256-d^2)/4 = 63}}}
{{{256-d^2 = 252}}}
{{{d^2 = 4}}}
{{{d = 2}}}<br>
Substitute back into equation (1) to find a:
{{{2a+10 = 16}}}
{{{2a = 6}}}
{{{a = 3}}}<br><br>
I think I like the logical analysis solution better....