Hi,
1) Suppose that the terms of the sequence are a_{n} then the sum of the first n terms is the sum of the first n1 terms, plus a_{n} . That is
S_{n} = S_{n1} + a_{n}
Thus
17n  3n^{2 } = 17(n1)  3(n1)^{2 } + a_{n}
Solve for a_{n}
2) If d is the common difference and some term is x then the next two terms are x + d and x + 2d. Hence three times the sum of the squares of these terms, that is
3[x^{2} + (x + d)^{2} + (x + 2d)^{2} ]
exceeds the square of their sum by 375, in other words exceeds
(x + x + d + x + 2d)^{2}
by 375. Thus 3[x^{2} + (x + d)^{2} + (x + 2d)^{2} ] is 375 more than (x + x + d + x + 2d)^{2}, or
3[x^{2} + (x + d)^{2} + (x + 2d)^{2} ] = (x + x + d + x + 2d)^{2} + 375
Simplify and see if you can determine d.
Penny
