C1 June 2011 Q9
9.
(a) Calculate the sum of all the even numbers from 2 to 100 inclusive,\[2 + 4 + 6 + \ldots\ldots + 100\] (3)
(b) In the arithmetic series\[k + 2k + 3k + \ldots\ldots + 100\]\(k\) is a positive integer and \(k\) is a factor of 100.
(i) Find, in terms of \(k\), an expression for the number of terms in this series.
(ii) Show that the sum of this series is\[50 + \frac{5000}{k}\]
(4)(c) Find, in terms of \(k\), the 50th term of the arithmetic sequence\[(2k + 1),\ (4k + 4),\ (6k + 7),\ \ldots\ldots,\]giving your answer in its simplest form. (2)
| Scheme | Marks |
|---|---|
| Series has 50 terms | B1 |
| \(S = \dfrac{1}{2}(50)(2 + 100) = 2550\) or \(S = \dfrac{1}{2}(50)(4 + 49\times 2) = 2550\) | M1 A1 |
| (3) |
Notes
(a) B for seeing attempt to use \(n = 50\) or \(n = 50\) stated
M for attempt to use \(\tfrac{1}{2}n(a + l)\) or \(\tfrac{1}{2}n(2a + (n - 1)d)\) with \(a = 2\) and values for other variables (Using \(n = 100\) may earn B0 M1A0)
| Scheme | Marks |
|---|---|
| (i) \(\dfrac{100}{k}\) | B1 |
| (ii) Sum: \(\dfrac{1}{2}\left(\dfrac{100}{k}\right)(k + 100)\) or \(\dfrac{1}{2}\left(\dfrac{100}{k}\right)\left(2k + \left(\dfrac{100}{k} - 1\right)k\right)\) | M1 A1 |
| \(= 50 + \dfrac{5000}{k} \qquad (*)\) | A1 cso |
| (4) |
Notes
(b) M for use of \(a = k\) and \(d = k\) or \(l = 100\) with their value for \(n\), could be numerical or even letter \(n\) in correct formula for sum.
A1: Correct formula with \(n = 100/k\)
A1: NB Answer is printed – so no slips should have appeared in working
| Scheme | Marks |
|---|---|
| 50th term \(= a + (n - 1)d\) \(= (2k + 1) + 49\text{"}(2k + 3)\text{"}\) Or \(2k + 49(2k) + 1 + 49(3)\) | M1 |
| \(= 100k + 148\) | A1 |
| (2) | |
| (9 marks) |
Notes
(c) M for use of formula \(a + 49d\) with \(a = 2k + 1\) and with \(d\) obtained from difference of terms
A1: Requires this simplified answer