Let Sn = an² + bn + c.

Actually, any Sn in quadratic form will eventually lead to an arithmetic progression later on.

Sn = an² + bn + c
S(n - 1) = a (n - 1)² + b (n - 1) + c

So, the nth term Tn
= Sn - S(n - 1)
= an² + bn + c - [a (n - 1)² + b (n - 1) + c]
= an² + bn + c - [a (n² - 2n + 1) + bn - b + c]
= an² + bn + c - an² + 2an - a - bn + b - c
= an² - an² + bn - bn + c - c + 2an - a + b
= 2an - a + b
= (b - a) + 2an

and this in in the form "a + bn" which is typical of arithmetic progression.

But for this question, let's take it that we DO NOT know that it's going to be an arithmetic progression, so we can only use the difference of sums. Give me a moment to write my answer.

Anyway, for the other question I saw that Nabil has responded to part b already, so I won't interfere unless you need me to do the write up.

Let Sn = an² + bn + c.

Now, without knowing what type of progression it would be, we cannot assume it.

What we can only do here is to find specific terms using the differences of two sums. This holds true for arithmetic, geometric and other progressions.

The first term is 50.
The second term is 45.
The sum of the next 10 terms is 175.

How do we, then, convert this into "sum" statements?

We need to add these.

1. The sum of the first 1 term is 50, so S(1) = 50.
2. The sum of the first 2 terms is 95, since 50 + 45 = 95, so S(2) = 95.
3. The sum of the first 12 terms is 270 since 50 + 45 + 175 = 270, so S(12) = 270.

Let's form the three simuls.

a(1)² + b(1) + c = 50
a(2)² + b(2) + c = 95
a(12)² + b(12) + c = 270

After simplification, we get

a + b + c = 50
4a + 2b + c = 95
144a + 12b + c = 270

We then simplify and solve these simuls.

I obtained a = -2.5, b = 52.5 and c is nil.

Sn = -2.5n² + 52.5n
Sn = -2.5 (n² - 21n)

To prove that Sn is arithmetic, we take Sn - S(n - 1).

S(n - 1) = -2.5 ](n - 1)² - 21 (n - 1)]
S(n - 1) = -2.5 [n² - 2n + 1 - 21n + 21]
S(n - 1) = -2.5 [n² - 23n + 22]

Taking Sn - S(n - 1) gives

Tn = -2.5 [n² - 21n - (n² - 23n + 22)]
Tn = -2.5 [2n - 22]
Tn = -5n + 55

proving that our series is arithmetic, with common difference -5 and first term 50.