Question 1152187
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You have received two responses to your question showing algebraic methods for obtaining the quadratic function that produces the given 2nd, 3rd, and 5th terms; and from that quadratic function you can determine the second difference and the first term.<br>
Note that the question does not require you to find the quadratic function; it only asks you to find the second difference and the first term.<br>
Here is how you can do that, using the method of finite differences -- without having to determine the formula for the quadratic sequence.<br>
We know the sequence is quadratic and therefore there is a common second difference.  Let that common difference be d; and let the missing 1st and 4th terms be a and b.  Using the method of finite differences with the information we have at the beginning gives us the following diagram showing the original sequence and the first and second differences:<br><pre>
    a    -1    -6     b    -14
       x    -5     y     z 
          d     d     d</pre>
We can use the method of finite differences to get expressions for x, y, and z in terms of d:<br>
-5-x = d  -->  x = -d-5<br>
y-(-5) = d  -->  y+5 = d  -->  y = d-5<br>
z-y = d  -->  z-(d-5) = d  -->  z = 2d-5<br>
Our finite differences diagram now looks like this:<br><pre>
    a       -1    -6       b       -14
       -d-5    -5     d-5     2d-5 
             d     d       d</pre>
We can now get an expression for a in terms of d:<br>
-1-a = -d-5  -->  a = d+4<br>
We can also get TWO DIFFERENT equations relating b and d; that will allow us to solve for d.<br>
-14-b = 2d-5<br>
b+6 = d-5<br>
Add the two equations: -8 = 3d-10  -->  d = 2/3<br>
We have the answer to one of the two questions: the common second difference is d = 2/3.<br>
And we have an expression for the first term a in terms of d: a = d+4 = 2/3+4 = 14/3.<br>
ANSWERS:
the second difference for the quadratic sequence is 2/3
the first term of the sequence is 14/3<br>