Here are the first five terms of a quadratic sequence. 10 21 38 61 90 Find an expression, in terms of n, for the nth term of this sequence.

  an = 3n^2 +2n +5

Step-by-step explanation:

If the n-th term is given by ...

  an = a·n^2 +bn +c

Then we can write 3 equations in a, b, c:

  10 = a·1 +b·1 + c

  21 = a·4 +b·2 +c

  38 = a·9 +b·3 +c

Subtracting each equation from the next gives ...

  11 = 3a +b

  17 = 5a +b

Then subtracting the first of these from the second, we have ...

  6 = 2a

  3 = a

Substituting in to the first of the reduced equations, we find ...

  11 = 3(3) +b   ⇒   b = 2

And substituting these values into the first equation gives ...

  10 = 3 +2 +c   ⇒   c = 5

The expression for the n-th term is ...

  an = 3n^2 +2n +5

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i believe it’s 165 because if u add 137 and 58 u get 195 and then subtract that from 360 and there’s ur answer