Question
An arithmetic sequence has $\mathrm{u}_1=30, \mathrm{u}_2=35, \mathrm{u}_3=40$.
(a) Find the common difference, $d$.
(b) Find $u_8$.
(c) Find $\mathrm{S}_8$.
Answer/Explanation
Ans:
(a) Let the common difference of the arithmetic sequence be $d$. Then we have:
\begin{align*}
\mathrm{u}_2 – \mathrm{u}_1 &= d = 35 – 30 = 5 \\
\mathrm{u}_3 – \mathrm{u}_2 &= d = 40 – 35 = 5
\end{align*}
Since both expressions are equal to $d$, we can conclude that the common difference of the sequence is $d = 5$.
(b) To find $\mathrm{u}_8$, we can use the formula for the $n$th term of an arithmetic sequence:
$$\mathrm{u}_n = \mathrm{u}_1 + (n-1)d$$
Setting $n=8$, $\mathrm{u}_1=30$, and $d=5$, we get:
$$\mathrm{u}_8 = 30 + (8-1)5 = 65$$
Therefore, $\mathrm{u}_8=65$.
(c) To find $\mathrm{S}_8$, the sum of the first 8 terms of the sequence, we can use the formula for the sum of an arithmetic series:
$$\mathrm{S}_n = \frac{n}{2}(\mathrm{u}_1 + \mathrm{u}_n)$$
Setting $n=8$, $\mathrm{u}_1=30$, $\mathrm{u}_8=65$, and $d=5$, we get:
\begin{align*}
\mathrm{S}_8 &= \frac{8}{2}(30 + 65) \\
&= 4 \times 95 \\
&= 380
\end{align*}
Therefore, $\mathrm{S}_8 = 380$.
Question
Consider the expansion of \({\left( {2x + \frac{k}{x}} \right)^9}\), where k > 0 . The coefficient of the term in x3 is equal to the coefficient of the term in x5. Find k.
▶️Answer/Explanation
Sol:
valid approach to find one of the required terms (must have correct substitution for parameters but accept ${ }^” r^{\prime \prime}$ or an incorrect value for $r$ )
eg $\left(\begin{array}{l}9 \\ r\end{array}\right)(2 x)^{9-r}\left(\frac{k}{x}\right)^r,\left(\begin{array}{l}9 \\ 6\end{array}\right)(2 x)^6\left(\frac{k}{x}\right)^3,\left(\begin{array}{l}9 \\ 0\end{array}\right)(2 x)^0\left(\frac{k}{x}\right)^9+\left(\begin{array}{l}9 \\ 1\end{array}\right)(2 x)^1\left(\frac{k}{x}\right)^8+\ldots$, Pascal’s triangle to
9th row
identifying correct terms (must be clearly indicated if only seen in expansion)
eg for $x^3$ term: $r=3, r=6,7$ th term, $\left(\begin{array}{l}9 \\ 6\end{array}\right),\left(\begin{array}{l}9 \\ 3\end{array}\right),(2 x)^6\left(\frac{k}{x}\right)^3, 5376 k^3$
for $x^5$ term: $r=2, r=7,8$ th term, $\left(\begin{array}{l}9 \\ 7\end{array}\right),\left(\begin{array}{l}9 \\ 2\end{array}\right),(2 x)^7\left(\frac{k}{x}\right)^2, 4608 k^2$
correct equation (may include powers of $x$ )
A1
eg $\left(\begin{array}{l}9 \\ 3\end{array}\right)(2 x)^6\left(\frac{k}{x}\right)^3=\left(\begin{array}{l}9 \\ 2\end{array}\right)(2 x)^7\left(\frac{k}{x}\right)^2$
valid attempt to solve their equation in terms of $k$ only
(M1)
eg sketch, $84 \times 64 k^3-36 \times 128 k^2=0,5376 k-4608=0,\left(\begin{array}{l}9 \\ 3\end{array}\right) 2^6 k^3=\left(\begin{array}{l}9 \\ 2\end{array}\right) 2^7 k^2$
0.857142
$k=\frac{4608}{5376}\left(=\frac{6}{7}\right)($ exact) 0.857