SOLUTION: The sequence 4,6,11.....forms the quadratic sequence : (a)Write down the 4th term of the sequence. (b)which term of the sequence is equal to 81

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Question 1114820: The sequence 4,6,11.....forms the quadratic sequence :
(a)Write down the 4th term of the sequence.
(b)which term of the sequence is equal to 81
Found 2 solutions by rothauserc, greenestamps:
Answer by rothauserc(4718)   (Show Source): You can put this solution on YOUR website!
4, 6, 11
:
we know we have an n^2 term
:
general form of quadratic sequence is
:
An^2 +Bn +C
:
using the sequence elements, we have 3 equations in 3 unknowns
:
A + B + C = 4
:
4A +2B +C = 6
:
9A +3B +C = 11
:
solve using a 3 by 3 linear system method or solver
:
A = 1.5, B = -2.5, C = 5
:
**********************
X(n) = 1.5n^2 -2.5n +5
**********************
:
a) x(4) = 1.5 * 4^2 -2.5 * 4 +5 = 19
:
b) 1.5n^2 -2.5n +5 = 81
:
1.5n^2 -2.5n -76 = 0
:
use quadratic formula to solve for n
:
n = (-(-2.5) +square root((-2.5)^2 -4 * 1.5 * (-76)) / (2*1.5) = 8
:
n = (-(-2.5) -square root((-2.5)^2 -4 * 1.5 * (-76)) / (2*1.5) = -6.3333
:
we reject the negative value for n
:
******
n = 8
******
:

Answer by greenestamps(13203)   (Show Source): You can put this solution on YOUR website!


You can find the formula for the quadratic sequence using the basic equation

and the given first three terms of the sequence:

















The equation for the sequence is


To find the 4th term...


To find which term of the sequence is 81...






Answers:
(a) the 4th term is 19
(b) the 8th term is 81

Since in this problem we are only looking for early terms in the sequence, we can also solve the problem without finding the explicit formula for the sequence by knowing that, in a quadratic sequence, the second differences are constant.

This sequence and the first and second differences are as follows:

   4   6   11
     2   5
       3


Knowing that the second differences are constant, we can write a bunch of 3's on the row of second differences and then work back up the array of numbers to find the next several terms in the sequence:
   4   6   11   19   30   44   61   81  104  130  159 ...
     2   5    8   11   14   17   20   23   26   29 ...
       3    3    3    3    3    3    3    3    3 ...


We can then see that the 4th term is 19, and that 81 is the 8th term.
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