The first term of an arithmetic series is $a$ and the common difference is $d$ - Edexcel - A-Level Maths Pure - Question 11 - 2009 - Paper 1

To solve for $a$ and $d$, we can subtract the first equation from the second:

(a + 20d) - (a + 17d) &= 32.5 - 25 \\ 3d &= 7.5 \\ d &= 2.5. \end{align*}$$ Now substituting $d$ back into the first equation: $$a + 17(2.5) = 25 \\ a + 42.5 = 25 \\ a = 25 - 42.5 \\ a = -17.5.$$

The sum of the first $n$ terms of an arithmetic series can be expressed as:

$S_n = \frac{n}{2} (2a + (n-1)d).$

Given that $S_n = 2750$, we substitute the known values of $a$ and $d$:

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

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

or: $2750 = \frac{n}{2} (2.5n - 37.5)$

Multiplying through by 2 gives: $5500 = n(2.5n - 37.5).$

Thus, rearranging gives: $2.5n^2 - 37.5n - 5500 = 0$ or simplifying: $n^2 - 15n = 55 imes 40.$

We will solve the equation:

$n^2 - 15n - 2200 = 0.$

Using the quadratic formula:

$n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

where $a = 1$, $b = -15$, and $c = -2200$:

$n = \frac{15 \pm \sqrt{(-15)^2 - 4(1)(-2200)}}{2(1)}$ $= \frac{15 \pm \sqrt{225 + 8800}}{2}$ $= \frac{15 \pm \sqrt{9025}}{2}$ $= \frac{15 \pm 95}{2}.$

Calculating the two potential solutions:

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

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