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22 January, 06:14

What is the nth term of quadratic sequence 4 7 12 19 28

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  1. 22 January, 06:37
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    We find the first differences between terms:

    7-4=3; 12-7=5; 19-12=7; 28-19=9.

    Since these are different, this is not linear.

    We now find the second differences:

    5-3=2; 7-5=2; 9-7=2. Then:

    Since these are the same, this sequence is quadratic.

    We use (1/2a) n², where a is the second difference:

    (1/2*2) n²=1n².

    We now use the term number of each term for n:

    4 is the 1st term; 1*1²=1.

    7 is the 2nd term; 1*2²=4.

    12 is the 3rd term; 1*3²=9.

    19 is the 4th term; 1*4²=16.

    28 is the 5th term: 1*5²=25.

    Now we find the difference between the actual terms of the sequence and the numbers we just found:

    4-1=3; 7-4=3; 12-9=3; 19-16=3; 28-25=3.

    Since this is constant, the sequence is in the form (1/2a) n²+d;

    in our case, 1n²+d, and since d=3, 1n²+3.

    The correct answer is n²+3
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