An arithmetic series is given by $$\sum_{r=5}^{20} (4r + 1)$$ 10 (a) (i) Write down the first term of the series - AQA - A-Level Maths Mechanics - Question 10 - 2020 - Paper 1

To find the common difference, we need to examine the general term of the series.
The given expression is $4r + 1$.
The common difference, which is the difference between successive terms, can be calculated as follows:

$d = (4(r + 1) + 1) - (4r + 1) = 4$

The first term is 5 and the last term is 20, inclusive.
We can find the number of terms ($n$) in the series using the formula for the $n$-th term of an arithmetic series given by:

$n = \frac{\text{last term - first term}}{\text{common difference}} + 1$

Thus,
$n = \frac{20 - 5}{4} + 1 = \frac{15}{4} + 1 = 4.75 + 1 = 5$
Hence, the number of terms is 16.

Given the sum of the series, we can express the sum of the terms as:

$S_n = \frac{n}{2} (2a + (n - 1)d)$
In the context of this question, we have $n = 91$, $a = 10b + c$, and $d = 1$.
Therefore, substituting all in:
$S_{100} = 7735$
Then we arrive at the equation $55b + 6c = 85$.

From the information provided regarding the 40th and 2nd terms:
The 40th term can be expressed as $40b + c$ and the 2nd term as $2b + c$.
Since the 40th term is 4 times the 2nd:

$40b + c = 4(2b + c)$
This can be solved alongside our previous equation to reveal $b$ and $c$.
Combining gives us the system of equations to find the values accordingly.