| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2014 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Arithmetic Sequences and Series |
| Type | Arithmetic progression with parameters |
| Difficulty | Moderate -0.3 This is a straightforward application of the arithmetic series formula S_n = n/2[2a + (n-1)d]. Students set up S_200 = 4S_100, simplify algebraically to find d = a/100, then use the nth term formula. It requires standard technique with minimal problem-solving insight, making it slightly easier than average but not trivial due to the algebraic manipulation involved. |
| Spec | 1.04h Arithmetic sequences: nth term and sum formulae |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(200/2(2a + 199d) = 4 \times 100/2(2a + 99d)\) | M1A1 [3] | Correct formula used (once) M1, correct eqn A1 |
| \(d = 2a\) cao | A1 | |
| (ii) \(a + 99d = a + 99 \times 2a\) | M1 A1 [2] | Sub. their part(i) into correct formula |
| \(199a\) cao |
(i) $200/2(2a + 199d) = 4 \times 100/2(2a + 99d)$ | M1A1 [3] | Correct formula used (once) M1, correct eqn A1
$d = 2a$ cao | A1 |
(ii) $a + 99d = a + 99 \times 2a$ | M1 A1 [2] | Sub. their part(i) into correct formula
$199a$ cao |5 An arithmetic progression has first term $a$ and common difference $d$. It is given that the sum of the first 200 terms is 4 times the sum of the first 100 terms.\\
(i) Find $d$ in terms of $a$.\\
(ii) Find the 100th term in terms of $a$.
\hfill \mbox{\textit{CAIE P1 2014 Q5 [5]}}