Unit 6

6 marks · Hard

The sequence is defined by , for .

(a)

Write an explicit expression for the sum $S_m$ of the first $m$ terms of the sequence.

[2 marks]

(b)

For which value of $m$ is $S_m = 0$?

[2 marks]

(c)

Let $A = \{n\,|\, a_n > 0,\ n\in \mathbb{N}\}$ and $B = \{x \in \mathbb{Z} \mid x = 2y+1,\ y \in \mathbb{N}, x < 7\}$. List all the elements of $A \cap B$.

[2 marks]

Work this out on paper, then compare with the answer and worked solution below. With a free account you can submit written answers for marking.

Show answer

(a) S_m = m(7 - m) · must show correct formula for sum in terms of m

(b) m = 7

Common mistakes: incorrect sum formula, not using $a_n$ correctly; wrong set intersection

Show mark scheme and worked solution

Marking points

M1 (a) · 1 mark · Identifies correct method to sum arithmetic sequence of form $a_n$.

A1 (a) · 1 mark · Obtains explicit formula $S_m = m(7 - m)$.

M2 (b) · 1 mark · Sets $S_m = 0$ and solves for $m$.

A2 (b) · 1 mark · Finds $m = 7$.

M3 (c) · 1 mark · Identifies both sets and finds elements meeting both descriptions.

A3 (c) · 1 mark · Lists $1, 3, 5$ as $A\cap B$.

Worked solution

a) $a_n = 6 - 2n$\ This is an arithmetic sequence with first term $a_1 = 4$, common difference $d = -2$. Sum of $m$ terms: $S_m = \frac{m}{2}[2a_1 + (m-1)d]$\ $= \frac{m}{2}[2\times 4 + (m-1)\times (-2)]$\ $= \frac{m}{2}[8 - 2m + 2]$\ $= \frac{m}{2}(10 - 2m)$\ $= m(5 - m)$\ However, starting with $a_1 = 4$ would make $a_n = 6 - 2n$, so $a_1 = 6 - 2 \times 1 = 4$. Alternatively, calculate: $a_1 = 4$, $a_2 = 2$, $a_3 = 0$, $a_4 = -2$ etc.

b) Set $S_m = 0$: $m(7 - m) = 0$\ So $m = 0$ or $7 - m = 0 \implies m = 7$\ Discard $m = 0$ as $m$ is a positive integer, so $m = 7$.

c) $A = \{ n\mid 6 - 2n > 0, n \in \mathbb{N}\}$\ So $6 > 2n\implies n < 3$\ But check explicitly: $n=1, a_1 = 4$\ $n=2, a_2 = 2$\ $n=3, a_3 = 0$ (so not included). So $A = \{1, 2\}$. But $a_n > 0$ for $n = 1, 2$ only. For $B$, $x = 2y + 1, y \in \mathbb{N}, x < 7$\ So $y = 1 \rightarrow 3$\ $y = 2 \rightarrow 5$\ $y = 3 \rightarrow 7$ (not < 7); so $B = \{3, 5\}$\ So $A \cap B = \{1, 2\} \cap \{3, 5\} = \emptyset$. Correction: Actually, need to reconsider: $A$ was $\{1,2,3\}$ because $a_3 = 0$ not $>0$, so only $1$ and $2$. If $B$ is $n \in \mathbb{N}$, $n$ odd and $n < 7$: $n=1,3,5$. $A = \{1,2\}$, so $A \cap B = \{1\}$.

But $a_1 = 4$, $a_2 = 2$, $a_3 = 0$, so only $1,2$ in $A$, intersection with $B$ is $1$ only. So $A \cap B = \{1\}$.