Edexcel C12 2017 October — Question 6 7 marks

Exam BoardEdexcel
ModuleC12 (Core Mathematics 1 & 2)
Year2017
SessionOctober
Marks7
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicArithmetic Sequences and Series
TypeReal-world AP: find term or total
DifficultyModerate -0.8 This is a straightforward arithmetic sequence question requiring only direct application of standard formulas (nth term and sum). Part (a) is a simple linear equation, part (b) uses the sum formula with given values, and part (c) equates two sums. All steps are routine with no problem-solving insight needed, making it easier than average but not trivial due to the multi-part structure.
Spec1.04e Sequences: nth term and recurrence relations1.04h Arithmetic sequences: nth term and sum formulae

  1. Each year Lin pays into a savings scheme. In year 1 she pays in \(\pounds 600\). Her payments then increase by \(\pounds 80\) a year, so that she pays \(\pounds 680\) into the savings scheme in year \(2 , \pounds 760\) in year 3 and so on. In year \(N\), Lin pays \(\pounds 1000\) into the savings scheme.
    1. Find the value of \(N\).
    2. Find the total amount that Lin pays into the savings scheme from year 1 to year 15 inclusive.
    Saima starts paying into a different savings scheme at the same time as Lin starts paying into her savings scheme. In year 1 she pays in \(\pounds A\). Her payments increase by \(\pounds A\) each year so that she pays \(\pounds 2 A\) in year \(2 , \pounds 3 A\) in year 3 and so on. Given that Saima and Lin have each paid, in total, the same amount of money into their savings schemes after 15 years,
  2. find the value of \(A\).

Question 6:
Part (a):
AnswerMarks Guidance
Answer/WorkingMarks Guidance
Uses \(1000 = 600 + 80(N-1) \Rightarrow N = 6\)M1 Attempts to use \(u_n = a + (n-1)d\) to find value of \(n\); evidence would be \(1000 = 600 + 80(N-1)\); alternatively attempts \(\frac{1000-600}{80}+1\) or repeated addition of £80 onto £600 until £1000 is reached
\(N = 6\)A1 Accept "the 6th year" or similar; answer alone scores both marks
Part (b):
AnswerMarks Guidance
Answer/WorkingMarks Guidance
Uses \(\frac{15}{2}(2\times600+(15-1)\times80) = \text{(£)}17400\)M1 Uses correct sum formula \(S = \frac{n}{2}(2a+(n-1)d)\) with \(n=15, a=600, d=80\); alternatively \(S=\frac{n}{2}(a+l)\) with \(l=600+14\times80=1720\); accept sum of 15 terms starting \(600+680+760+840+\ldots\)
\((£)17400\)A1 cao
Part (c):
AnswerMarks Guidance
Answer/WorkingMarks Guidance
Total for Saima \(= \frac{15}{2}(2A+14A) = 120A\)B1 Finds sum for Saima; accept unsimplified forms \(\frac{15}{2}(2A+14A)\) or \(\frac{15}{2}(A+15A)\) or simplified \(120A\); isw following correct answer
Sets \(120A = 17400 \Rightarrow A = 145\)M1 Sets their \(120A\) equal to answer from (b); must attempt sums not terms
\(A = 145\)A1 cao
## Question 6:

### Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Uses $1000 = 600 + 80(N-1) \Rightarrow N = 6$ | M1 | Attempts to use $u_n = a + (n-1)d$ to find value of $n$; evidence would be $1000 = 600 + 80(N-1)$; alternatively attempts $\frac{1000-600}{80}+1$ or repeated addition of £80 onto £600 until £1000 is reached |
| $N = 6$ | A1 | Accept "the 6th year" or similar; answer alone scores both marks |

### Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Uses $\frac{15}{2}(2\times600+(15-1)\times80) = \text{(£)}17400$ | M1 | Uses correct sum formula $S = \frac{n}{2}(2a+(n-1)d)$ with $n=15, a=600, d=80$; alternatively $S=\frac{n}{2}(a+l)$ with $l=600+14\times80=1720$; accept sum of 15 terms starting $600+680+760+840+\ldots$ |
| $(£)17400$ | A1 | cao |

### Part (c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Total for Saima $= \frac{15}{2}(2A+14A) = 120A$ | B1 | Finds sum for Saima; accept unsimplified forms $\frac{15}{2}(2A+14A)$ or $\frac{15}{2}(A+15A)$ or simplified $120A$; isw following correct answer |
| Sets $120A = 17400 \Rightarrow A = 145$ | M1 | Sets their $120A$ equal to answer from (b); must attempt sums not terms |
| $A = 145$ | A1 | cao |

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\begin{enumerate}
  \item Each year Lin pays into a savings scheme. In year 1 she pays in $\pounds 600$. Her payments then increase by $\pounds 80$ a year, so that she pays $\pounds 680$ into the savings scheme in year $2 , \pounds 760$ in year 3 and so on. In year $N$, Lin pays $\pounds 1000$ into the savings scheme.\\
(a) Find the value of $N$.\\
(b) Find the total amount that Lin pays into the savings scheme from year 1 to year 15 inclusive.
\end{enumerate}

Saima starts paying into a different savings scheme at the same time as Lin starts paying into her savings scheme.

In year 1 she pays in $\pounds A$. Her payments increase by $\pounds A$ each year so that she pays $\pounds 2 A$ in year $2 , \pounds 3 A$ in year 3 and so on.

Given that Saima and Lin have each paid, in total, the same amount of money into their savings schemes after 15 years,\\
(c) find the value of $A$.

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\hfill \mbox{\textit{Edexcel C12 2017 Q6 [7]}}