Question

find the nth term of the following sequence
4, 7, 12, 19, 28

Answer 1

Answer:

$n^2+3$

Step-by-step explanation:

4, 7, 12, 19, 28

The differences are 7-4=3, 12-7=5, 19-12=7, 28-19=9.

This is an arithmetic sequence since the differences are constant.

There must be $n^2$ in the nth term.

These are the next terms in the sequence:

4, 7, 12, 19, 28, 39, 52, 67, 84, 103, 124, 147, 172, 199, 228, 259, 292, 327, 364, 403, 444, 487, 532, 579, 628, 679, 732, 787, 844, 903, 964, 1027, 1092, 1159, 1228, 1299, 1372, 1447..

$n^2 +c=4$

$1^2+c=4$

$c=4-1$

$c=3$

Therefore, the nth term is $n^2+3$.

Answer 2

Answer:

nth term = n^2 + 3.

Step-by-step explanation:

4 7 12 19 28

The differences  are in an arithmetic sequence so this is a quadratic sequence,

- the differences are 3, 5, 7, 9... which is an A.S.

The  common difference of the A.S. is 2 so there is n^2 in the formula for the nth term.

By inspection , we see that each term is 3 more than  the term number squared.

1^2 + 3 = 4

2^2 + 3 = 7

3^2 + 3 = 12  

4^2 + 3 = 19   and so on.