1)the sum to n terms of a particular series is given by S_n=17n-3n²

a)find an expression for the n term of the series
b)show that the series is an arithmetic progression

2)a particular arithmetic progression has a positive common difference and is such that for any three adjacent terms ,three times the sum of their squares exceeds the square of their sum is 375.Find the common difference

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 n-1 terms, plus a_n . That is

S_n = S_n-1 + a_n

Thus

17n - 3n² = 17(n-1) - 3(n-1)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[x² + (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[x² + (x + d)² + (x + 2d)² ] is 375 more than (x + x + d + x + 2d)², or

3[x² + (x + d)² + (x + 2d)² ] = (x + x + d + x + 2d)² + 375

Simplify and see if you can determine d.

Penny