Question 1164687
50,49,47,44,40,35,...
<pre>
Make the set of first and second differences

50  49  47  44  40  35
  -1  -2  -3  -4  -5
    -1  -1  -1  -1  
  
and since the second differences are constant, we
assume a 2nd degree polynomial in n for the general term:

[If the kth differences are constant, the the general term
will be a polynomial in n of degree k.

{{{a[n]=An^2+Bn+C}}}

Then we substitute the first 3 terms for a<sub>1</sub>, a<sub>2</sub>, a<sub>3</sub> and 1,2,3 for n

 {{{system(50=A(1)^2+B(1)+C,49=A(2)^2+B(2)+C,47=A(3)^2+B(3)+C=47)}}}

or

{{{system(A+B+C=50,4A+2B+C=49,9A+3B+C=47)}}}

Solve that system and get 

A = -1/2, B = 1/2, and C = 50

So the general term is

{{{a[n]=expr(-1/2)n^2+expr(1/2)n+50}}}

You can find the next three terms either by extending the
difference table:

50  49  47  44  40  35  <font color="red"><b>29  22  14</b></font>
  -1  -2  -3  -4  -5<font color="red"><b>  -6  -7  -8</b></font>  
    -1  -1  -1<font color="red"><b>  -1  -1  -1  -1</b></font>

or by substituting 7, 8, and 9 in the general formula:

{{{a[n]=expr(-1/2)n^2+expr(1/2)n+50}}}

{{{a[7]=expr(-1/2)7^2+expr(1/2)7+50=29}}}

{{{a[8]=expr(-1/2)8^2+expr(1/2)8+50=22}}}

{{{a[9]=expr(-1/2)9^2+expr(1/2)9+50=14}}}

Edwin</pre>