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 Pure - Question 10 - 2020 - Paper 1

The common difference in an arithmetic series can be identified by observing the sequence generated by the formula:

The general term is given by $4r + 1$.

Calculating the common difference by substituting $r$ values:

• For $r = 5$: $4(5) + 1 = 21$
• For $r = 6$: $4(6) + 1 = 25$

The common difference:

$d = 25 - 21 = 4$

Thus, the common difference is 4.

To find the number of terms in the series from $r = 5$ to $r = 20$, we use the formula:

$n = ext{Last term} - ext{First term} + 1$

Here, the last term corresponds to $r = 20$, thus:

$n = 20 - 5 + 1 = 16$

Therefore, there are 16 terms in the series.

To find this equation, we analyze the given series. The sum of an arithmetic series from $10$ to $100$:

The number of terms is:

$n = 100 - 10 + 1 = 91$

Now, applying the formula for the sum:

$S_n = \frac{n}{2} (\text{first term} + \text{last term})$

First term when $r = 10$ is:

$10b + c$

Last term when $r = 100$ is:

$100b + c$

Thus:

$7735 = \frac{91}{2} ((10b + c) + (100b + c))$

Simplifying yields:

$7735 = \frac{91}{2} (110b + 2c)$

Multiplying through by 2 gives:

$15470 = 91(110b + 2c)$

Upon simplification, one of the derived equations becomes:

$55b + c = 85$

Using the equations derived, we set up the system:

1. $55b + c = 85$

2. The relation derived from the 40th term being 4 times the 2nd term:

   • 40th term: $40b + c$
   • 2nd term: $2b + c$

   $40b + c = 4(2b + c)$
   Simplifying gives: $40b + c = 8b + 4c$
   Rearranging leads to:
   $32b - 3c = 0$\

   We simplify and rearrange this system to find:

   From $c = 55b - 85$, substituting yields: $32b - 3(55b - 85) = 0$
   Solving this gives: $32b - 165b + 255 = 0$
   Thus, combining yields: $-133b + 255 = 0$
   So, $b = \frac{255}{133} = 1.912 ext{ (approximately)}$
   Supplying back to find $c$: $c = 55(1.912) - 85 = 2.5.$
   Thus, the solutions are: b ≈ 1.912, c = 2.5.