Moderate -0.8 Part (a) is a standard bookwork proof that appears in every A-level textbook, requiring only writing out terms and pairing them—a routine exercise with no problem-solving. Part (b) is a straightforward application requiring identification of the sequence (7, 14, ..., 497), finding n, then substituting into the formula. This is easier than average as it combines memorized proof with direct formula application.
5. (a) Prove that the sum of the first \(n\) terms of an arithmetic series is given by the formula
$$S _ { n } = \frac { n } { 2 } [ 2 a + ( n - 1 ) d ]$$
where \(a\) is the first term of the series and \(d\) is the common difference between the terms.
(b) Find the sum of the integers which are divisible by 7 and lie between 1 and 500
LHS should be \(2S\). RHS must follow from at least two terms correctly matching; should include at least two terms each correctly \(\{2a+(n-1)d\}\) or \((a+L)\)
\(S_n = \frac{n}{2}[2a+(n-1)d]\)
A1*
Need indication of at least three terms being added and achieve final answer with no errors
Part (b):
Answer
Marks
Guidance
Answer/Working
Marks
Guidance
Uses either \(\frac{n}{2}(2 \times a + (n-1)7)\) or \(\frac{n}{2}(a+497)\) or \(7 \times \sum_{i=1}^{71} i\)
M1
Uses correct formula with their \(a\) and \(n\), with \(d=7\) or with last term correct
\(\frac{71}{2}(2\times7+70\times7)\) or \(\frac{72}{2}(2\times0+71\times7)\) or \(\frac{71}{2}(7+497)\) or \(7\times\frac{71}{2}(72)\)
A1
Uses consistent and correct \(a\) and \(n\)
\(= 17892\)
A1
Correct answer
# Question 5:
## Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $S_n = a + (a+d) + (a+2d) + \ldots + (a+(n-1)d)$ | M1 | List terms including at least first two and a last term which may be $a+nd$ or $a+(n-1)d$ or $L$ |
| $S_n = (a+(n-1)d) + (a+(n-2)d) + \ldots + (a+d) + a$ | M1 | List terms in reverse including at least their last term (or correct last term) and finally their first term |
| $2S_n = (2a+(n-1)d) + (2a+(n-1)d) + \ldots + (2a+(n-1)d)$ | M1 | LHS should be $2S$. RHS must follow from at least two terms correctly matching; should include at least two terms each correctly $\{2a+(n-1)d\}$ or $(a+L)$ |
| $S_n = \frac{n}{2}[2a+(n-1)d]$ | A1* | Need indication of at least three terms being added and achieve final answer with no errors |
## Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Uses either $\frac{n}{2}(2 \times a + (n-1)7)$ or $\frac{n}{2}(a+497)$ or $7 \times \sum_{i=1}^{71} i$ | M1 | Uses correct formula with their $a$ and $n$, with $d=7$ or with last term correct |
| $\frac{71}{2}(2\times7+70\times7)$ or $\frac{72}{2}(2\times0+71\times7)$ or $\frac{71}{2}(7+497)$ or $7\times\frac{71}{2}(72)$ | A1 | Uses consistent and correct $a$ and $n$ |
| $= 17892$ | A1 | Correct answer |
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5. (a) Prove that the sum of the first $n$ terms of an arithmetic series is given by the formula
$$S _ { n } = \frac { n } { 2 } [ 2 a + ( n - 1 ) d ]$$
where $a$ is the first term of the series and $d$ is the common difference between the terms.\\
(b) Find the sum of the integers which are divisible by 7 and lie between 1 and 500\\
\hfill \mbox{\textit{Edexcel C12 2015 Q5 [7]}}