SOLUTION: 1,3,5 are the first difference of a quadratic sequence the 7th is equal to 35 therefore how do I calculate 5th term and the 6th term please help me

Question 1071688: 1,3,5 are the first difference of a quadratic sequence the 7th is equal to 35 therefore how do I calculate 5th term and the 6th term please help me
Answer by Boreal(15235) (Show Source): You can put this solution on YOUR website!
an^2+bn+c
divide the second difference (2, so it is a quadratic sequence) by 2 to get 1
that is a
n^2+bn+c
Now, when n=7 we have 49+bn+c=35
We also know that the first term is 1+b+c
the second term is 4+2b+c
The difference between those two is 3+b, and it equals 1
3+b=1
b=-2
so 1+-2+c=-1 the first term, and c must be 0.
second term is 4+2b+c=0, difference 1
third term is 9+3b+c=3, difference 3
fourth term is 16+4b+c=8, difference 5
fifth term is 25+5b+c=15, difference 7
sixth term is 36+6b+c=24, difference 9
seventh term is 49-14=35

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