, 2024

The 5th term is obtained by adding $6$ to the 4th term. Thus, the 5th term is
$21+6=27$.
Solution 1:

The 6th term is obtained by adding $6$ to the 5th term. Thus, the 6th term is
$27+6=33$, and so the average of the
4th, 5th and 6th terms is $\frac{21+27+33}{3}=\frac{81}{3}=27$.

Solution 2:

The 4th term is $6$ less than the
5th term, and the 6th term is $6$
more than the 5th term, and so the average of the 4th, 5th and 6th terms
is the 5th term, which is $27$.
The $n$th term ($n\geq2$) is obtained by adding $n-1$ 6s to the first term, $3$. For example, the 2nd term is $3+1\times6$, the 3rd term is $3+2\times6$, the 4th term is $3+3\times6$, and so on.

In general, the $n$th term is given
by $3+(n-1)\times6$.

Therefore, the 20th term is $3+19\times6=3+114=117$.
Solution 1:

Since each new term is obtained by adding $6$ to the previous term and $\frac{1000}{6}\approx166.7$, then it makes
sense to begin by determining the 166th term. The 166th term is $3+165\times6=993$, the next term is $993+6=999$ (still less than $1000$), and so the smallest term that is
greater than 1000 is $999+6=1005$.

(We note that $1005$ is the 168th
term and is equal to $3+167\times6$.)

Solution 2:

From part (c), the $n$th term is
given by the expression $3+(n-1)\times6$.

We want the smallest term that is greater than $1000$. To begin, we find the smallest
possible value of $n$ for which $3+(n-1)\times6>1000$.

Solving this inequality, we get $$\begin{align*} 3+(n-1)\times6&>1000\\ 3+6n-6&>1000\\ 6n-3&>1000\\ 6n&>1003\\ n&>\tfrac{1003}{6}\approx167.2\end{align*}$$ Since $n$ must be an integer, the first term
number to exceed $1000$ is the 168th
term, and its value is $3+167\times6=1005$.