In the following sequence, $(1\times2)-2=0$ $(2\times3)-4=2$ $(3\times4)-6=6$ $(4\times5)-8=12$ ... $(a\times12)-b=c$ ... $(d\times e)-f=g$

1. Identify the row pattern and find $a$.

The terms follow the pattern
$(1\times2)-2,\ (2\times3)-4,\ (3\times4)-6,\ldots$
The second factor is always one more than the first factor.

Since the term is $(a\times12)-b=c$, we have
$a+1=12$
$a=11.$

2. Find $b$ and $c$ using the sequence pattern.

The number subtracted is twice the first factor:
$2,4,6,8,\ldots$
So when $a=11$,
$b=2\times 11=22.$
Then
$c=(11\times12)-22$
$c=132-22=110.$

1. Express the other entries in the row in terms of $d$.

From the sequence,

When the first number is $d$, the next number is $d+1$ and the number subtracted is $2d$.

So in $(d\times e)-f=g,$
$e=d+1,\quad f=2d.$

2. Form and simplify the expression for $g$.

$g=d(d+1)-2d$
$=d^2+d-2d$
$=d^2-d$

1. Use the general expression for a result in the sequence and compare with 279.

From part (b),
$g=d^2-d=d(d-1).$
So any result in the sequence must be a product of two consecutive integers.

Near $279$,
$16\times17=272,$
$17\times18=306.$
Since $279$ lies between $272$ and $306$, it is not equal to any product of two consecutive integers. Therefore, $279$ cannot be the result of an equation in this sequence.