OCR C2 — Question 8 11 marks

Exam BoardOCR
ModuleC2 (Core Mathematics 2)
Marks11
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicArithmetic Sequences and Series
TypeFind term or common difference
DifficultyModerate -0.3 This is a straightforward two-part question testing standard arithmetic series formulas and basic algebraic manipulation. Part (a) requires direct application of S_n = n/2[2a + (n-1)d] and simple inequality solving. Part (b) involves solving a quadratic equation to find k and then substituting—all routine C2 techniques with no novel problem-solving required. Slightly easier than average due to its mechanical nature.
Spec1.04e Sequences: nth term and recurrence relations1.04h Arithmetic sequences: nth term and sum formulae

  1. An arithmetic series has a common difference of 7. Given that the sum of the first 20 terms of the series is 530, find
    1. the first term of the series, [3]
    2. the smallest positive term of the series. [2]
  2. The terms of a sequence are given by $$u_n = (n + k)^2, \quad n \geq 1,$$ where \(k\) is a positive constant. Given that \(u_2 = 2u_1\),
    1. find the value of \(k\), [4]
    2. show that \(u_3 = 11 + 6\sqrt{2}\). [2]

(a)
(i)
AnswerMarks
\(\frac{20}{7}[2a + (19 \times 7)] = 530\)M1
\(2a + 133 = 53, a = -40\)M1 A1
(ii)
AnswerMarks
\(= -40 + 7k = -40 + 42 = 2\)M1 A1
(b)
(i)
AnswerMarks
\(u_1 = (1 + k)^2, u_2 = (2 + k)^2\)B1
\((2 + k)^2 = 2(1 + k)^2\)
\(4 + 4k + k^2 = 2 + 4k + 2k^2\)
AnswerMarks
\(k^2 = 2\)M1
\(k > 0 \therefore k = \sqrt{2}\)M1 A1
(ii)
AnswerMarks Guidance
\(u_3 = (3 + \sqrt{2})^2 = 9 + 6\sqrt{2} + 2 = 11 + 6\sqrt{2}\)M1 A1 (11)
## (a)

### (i)
$\frac{20}{7}[2a + (19 \times 7)] = 530$ | M1 |

$2a + 133 = 53, a = -40$ | M1 A1 |

### (ii)
$= -40 + 7k = -40 + 42 = 2$ | M1 A1 |

## (b)

### (i)
$u_1 = (1 + k)^2, u_2 = (2 + k)^2$ | B1 |

$(2 + k)^2 = 2(1 + k)^2$
$4 + 4k + k^2 = 2 + 4k + 2k^2$
$k^2 = 2$ | M1 |

$k > 0 \therefore k = \sqrt{2}$ | M1 A1 |

### (ii)
$u_3 = (3 + \sqrt{2})^2 = 9 + 6\sqrt{2} + 2 = 11 + 6\sqrt{2}$ | M1 A1 | (11)

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\begin{enumerate}[label=(\alph*)]
\item An arithmetic series has a common difference of 7.

Given that the sum of the first 20 terms of the series is 530, find

\begin{enumerate}[label=(\roman*)]
\item the first term of the series, [3]
\item the smallest positive term of the series. [2]
\end{enumerate}

\item The terms of a sequence are given by
$$u_n = (n + k)^2, \quad n \geq 1,$$

where $k$ is a positive constant.

Given that $u_2 = 2u_1$,

\begin{enumerate}[label=(\roman*)]
\item find the value of $k$, [4]
\item show that $u_3 = 11 + 6\sqrt{2}$. [2]
\end{enumerate}
\end{enumerate}

\hfill \mbox{\textit{OCR C2  Q8 [11]}}