Let's tackle each part of the question step by step.
(a) Write down the four shaded numbers and their sum.
The four shaded numbers provided are 2, 4, 12, and 14. To find their sum:
[tex]\[ 2 + 4 + 12 + 14 = 32 \][/tex]
So, the sum of the shaded numbers is 32.
(b) Write down an expression for [tex]\( a \)[/tex] in terms of [tex]\( n \)[/tex].
Consider a [tex]\(3 \times 3\)[/tex] square placed on the number grid. If [tex]\( n \)[/tex] is the top-left number of this [tex]\(3 \times 3\)[/tex] square, the layout of the numbers in the square will be:
[tex]\[
\begin{array}{|c|c|c|}
\hline
n & n+1 & n+2 \\
\hline
n+5 & n+6 & n+7 \\
\hline
n+10 & n+11 & n+12 \\
\hline
\end{array}
\][/tex]
The central element of this [tex]\(3 \times 3\)[/tex] square is [tex]\( a = n + 6 \)[/tex].
So, the expression for [tex]\( a \)[/tex] in terms of [tex]\( n \)[/tex] is:
[tex]\[ a = n + 6 \][/tex]
(c) Write down an expression for the total of all numbers in the [tex]\(3 \times 3\)[/tex] square in terms of [tex]\( n \)[/tex] and simplify your expression.
The sum of the numbers in the [tex]\(3 \times 3\)[/tex] square is the sum of:
[tex]\[
\begin{array}{|c|c|c|}
\hline
n & n+1 & n+2 \\
\hline
n+5 & n+6 & n+7 \\
\hline
n+10 & n+11 & n+12 \\
\hline
\end{array}
\][/tex]
Summing these numbers:
[tex]\[ n + (n+1) + (n+2) + (n+5) + (n+6) + (n+7) + (n+10) + (n+11) + (n+12) \][/tex]
Combining all the [tex]\( n \)[/tex]'s and constants separately:
[tex]\[ 9n + (1 + 2 + 5 + 6 + 7 + 10 + 11 + 12) \][/tex]
The sum of the constants is:
[tex]\[ 1 + 2 + 5 + 6 + 7 + 10 + 11 + 12 = 54 \][/tex]
Therefore, the total can be expressed as:
[tex]\[ 9n + 54 \][/tex]
So, the simplified expression for the total of the numbers in the [tex]\(3 \times 3\)[/tex] square in terms of [tex]\( n \)[/tex] is:
[tex]\[ \text{Total} = 9n + 54 \][/tex]
