Question 331346
The third and fourth terms of a sequence are 26 and 40. If the second differences are a constant 4, what are the first five terms of the sequence?
<pre><b>
     1st          2nd
a<sub>n</sub>  diffs        diffs    
---------------------------------------------------------
a<sub>1</sub> |       |
   | a<sub>2</sub>-a<sub>1</sub> |
a<sub>2</sub> |       | (26-a<sub>2</sub>)-(a<sub>2</sub>-a<sub>1</sub>) = 26-a<sub>2</sub>-a<sub>2</sub>+a1 = 26-2a<sub>2</sub>+a<sub>1</sub> = 4, or a<sub>1</sub>-2a<sub>2</sub>=-22
   | 26-a<sub>2</sub> |
26 |       | 14-(26-a<sub>2</sub>) = 14-26+a<sub>2</sub> = -12+a<sub>2</sub> = 4, or a<sub>2</sub>=16
   |  14   |
40 |       | a<sub>5</sub>-40-14 = a<sub>5</sub>-54 = 4, or a<sub>5</sub> = 58
   | a<sub>5</sub>-40 |
a<sub>5</sub> |       |


So we have the system of equations:

{{{system(a[1]-2a[2]=-22, a[2]=16, a[5]=58)}}}

Susbstitute 16 for {{{a[2]}}} in the first equation:

{{{a[1]-2a[2]=-22}}}

{{{a[1]-2(16)=-22}}}

{{{a[1]-32=-22}}}

{{{a[1]=10}}}

So {{{a[1]=10}}}, {{{a[2]=16}}}, {{{a[3]=26}}}, {{{a[4]=40}}}, {{{a[5]=58}}}
<sub>1</sub>
Checking:

   1st 2nd 
   dif dif
10
    6
16      4
    10 
26      4
    14
40      4
    18 
58

Edwin</pre>