Question 353570
The third and fourth terms of a sequnce are 26 and 40. If the second differences are a constant 4, what are the first five terms of the sequence
<pre><b>
We have this information.  Put blanks for the unknown values:

n  a<sub>n</sub>  1st diffs   2nd diffs
1  __                  
          __
2  __                  4
          __
3  26                  4
          __
4  40                  4 
          __
5  __

Since we have two consecutive terms, the 3rd and 4th,
we can find the first difference between them by subtracting
26 from 40, getting 14, so we fill that in


n  a<sub>n</sub>  1st diffs   2nd diffs
1  __                  
          __
2  __                  4
          __
3  26                  4
          14
4  40                  4 
          __
5  __


We can fill in the 2nd first difference by adding 4 to 14 getting 18:


n  a<sub>n</sub>  1st diffs   2nd diffs
1  __                  
          __
2   8                  4
          18
3  26                  4
          14
4  40                  4 
          __
5  __


We can fill in the 1st first difference by adding 4 to 18 getting 22:

n  a<sub>n</sub>  1st diffs   2nd diffs
1  __                  
          22
2  __                  4
          18
3  26                  4
          14
4  40                  4 
          __
5  __


We can fill in the last 1st difference by subtracting 4 from 14 getting 10:


n  a<sub>n</sub>  1st diffs   2nd diffs
1  __                  
          22
2  __                  4
          18
3  26                  4
          14
4  40                  4 
          10
5  __


We can fill in a<sub>5</sub> by adding 10 to 40 getting 50:

n  a<sub>n</sub>  1st diffs   2nd diffs
1  __                  
          22
2  __                  4
          18
3  26                  4
          14
4  40                  4 
          10
5  50

We can fill in a<sub>2</sub> by subtracting 18 from 26 getting 8:

n  a<sub>n</sub>  1st diffs   2nd diffs
1  __                  
          22
2   8                  4
          18
3  26                  4
          14
4  40                  4 
          10
5  50

Finally we can fill in a<sub>1</sub> by subtracting 22 from 8 getting -14:

n  a<sub>n</sub>  1st diffs   2nd diffs
1 -14                  
          22
2   8                  4
          18
3  26                  4
          14
4  40                  4 
          10
5  50

So the first 5 terms of the sequence are

a<sub>1</sub>=-14, a<sub>2</sub>=8, a<sub>3</sub>=26, a<sub>4</sub>=40, a<sub>5</sub>=50.

Later on you'll have to find the general term, which is

a<sub>n</sub> = -2nē+28n-40

Edwin</pre>