Find the number of terms of the A.P. −12, −9, −6, ..., 21.

First term, $a_{1}=-12$

Common difference, $d=a_{2}-a_{1}=-9-(-12)=3$

$a_{n}=21$

$\Rightarrow a+(n-1) d=21$

$\Rightarrow-12+(n-1) \times 3=21$

$\Rightarrow 3 n=36$

$\Rightarrow n=12$

Therefore, number of terms in the given A.P. is 12.

Now, when 1 is added to each of the 12 terms, the sum will increase by 12.

So, the sum of all terms of the A.P. thus obtained

$=S_{12}+12$

$=\frac{12}{2}[2(-12)+11(3)]+12$

$=6 \times(9)+12$

$=66$