Edexcel C12 2014 June — Question 9 7 marks

Exam BoardEdexcel
ModuleC12 (Core Mathematics 1 & 2)
Year2014
SessionJune
Marks7
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicArithmetic Sequences and Series
TypeSigma notation: arithmetic series evaluation
DifficultyModerate -0.8 Part (i) is a direct application of the arithmetic series formula with clear first term and common difference. Part (ii) is a standard geometric series sum to infinity requiring simple algebraic manipulation. Both are routine textbook exercises with no problem-solving insight required, making this easier than average but not trivial since it tests two distinct series concepts.
Spec1.04h Arithmetic sequences: nth term and sum formulae1.04j Sum to infinity: convergent geometric series |r|<1

9. (i) Find the value of \(\sum _ { r = 1 } ^ { 20 } ( 3 + 5 r )\) (ii) Given that \(\sum _ { r = 0 } ^ { \infty } \frac { a } { 4 ^ { r } } = 16\), find the value of the constant \(a\).

Question 9(i):
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(\sum_{r=1}^{20}(3+5r) = 8+13+18+\ldots+103\)M1 Minimal evidence of arithmetic sum; accept first 3 terms or \(8+13+\ldots+103\)
Use \(S_n = \frac{n}{2}(2a+(n-1)d)\) or \(S_n=\frac{n}{2}(a+l)\) with \(a=3\) or \(8\), \(n=19\) or \(20\), \(d=5\), \(l=103\)M1 Correct formula with correct values
\(S_{20} = \frac{20}{2}(8+103) = 1110\)A1 Accept for all 3 marks if no incorrect working seen
Question 9(ii):
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(r = \frac{1}{4}\)B1 For stating or implying \(r=\frac{1}{4}\); accept series showing \(\times\frac{1}{4}\) or \(\times 0.25\)
Use \(S_\infty = \frac{a}{1-r}\) with \(0<\r\ <1\) and \(S_\infty=16\)
\(16 = \frac{a}{1-r'} \Rightarrow a = \ldots\)dM1 Dependent on previous M; proceed to find \(a\)
\(a = 12\)A1 Correct answer
# Question 9(i):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\sum_{r=1}^{20}(3+5r) = 8+13+18+\ldots+103$ | M1 | Minimal evidence of arithmetic sum; accept first 3 terms or $8+13+\ldots+103$ |
| Use $S_n = \frac{n}{2}(2a+(n-1)d)$ or $S_n=\frac{n}{2}(a+l)$ with $a=3$ or $8$, $n=19$ or $20$, $d=5$, $l=103$ | M1 | Correct formula with correct values |
| $S_{20} = \frac{20}{2}(8+103) = 1110$ | A1 | Accept for all 3 marks if no incorrect working seen |

# Question 9(ii):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $r = \frac{1}{4}$ | B1 | For stating or implying $r=\frac{1}{4}$; accept series showing $\times\frac{1}{4}$ or $\times 0.25$ |
| Use $S_\infty = \frac{a}{1-r}$ with $0<\|r\|<1$ and $S_\infty=16$ | M1 | Correct formula with correct $S_\infty$ |
| $16 = \frac{a}{1-r'} \Rightarrow a = \ldots$ | dM1 | Dependent on previous M; proceed to find $a$ |
| $a = 12$ | A1 | Correct answer |
9. (i) Find the value of $\sum _ { r = 1 } ^ { 20 } ( 3 + 5 r )$\\
(ii) Given that $\sum _ { r = 0 } ^ { \infty } \frac { a } { 4 ^ { r } } = 16$, find the value of the constant $a$.\\

\hfill \mbox{\textit{Edexcel C12 2014 Q9 [7]}}