The first term of an arithmetic series is $a$ and the common difference is $d$ - Edexcel - A-Level Maths Pure - Question 10 - 2009 - Paper 1

To find $d$, we can subtract equation (1) from equation (2):

$(a + 20d) - (a + 17d) = 32.5 - 25\implies 3d = 7.5\implies d = 2.5.$

Now substituting $d$ back into equation (1):

$a + 17(2.5) = 25\implies a + 42.5 = 25\implies a = 25 - 42.5\implies a = -17.5.$

Thus, we find: $a = -17.5 ext{ and } d = 2.5.$

The sum of the first $n$ terms of an arithmetic series is given by: $S_n = \frac{n}{2}(2a + (n-1)d).$ Given that $S_n = 2750$, we substitute:

$2750 = \frac{n}{2}(2(-17.5) + (n-1)(2.5))$

This simplifies to: $2750 = \frac{n}{2}(-35 + 2.5n - 2.5)\implies 2750 = \frac{n}{2}(2.5n - 37.5)$

Multiplying through by 2: $5500 = n(2.5n - 37.5) \implies 2.5n^2 - 37.5n - 5500 = 0.$

To eliminate the decimal, multiply through by 10: $25n^2 - 375n - 55000 = 0.$

This can be rearranged to give: $n^2 - 15n = 55 \times 40.$

From the equation: $n^2 - 15n - 2200 = 0,$

we can apply the quadratic formula: $n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ where $a = 1, b = -15, c = -2200$.

Calculating the discriminant: $b^2 - 4ac = (-15)^2 - 4 \cdot 1 \cdot (-2200) = 225 + 8800 = 9025.$

Now computing: $n = \frac{15 \pm \sqrt{9025}}{2} \implies n = \frac{15 \pm 95}{2}.$

Calculating the possible values:

1. $n = \frac{110}{2} = 55$
2. $n = \frac{-80}{2} = -40 ext{ (not a valid solution).}$

Thus, the only valid solution is: $n = 55.$