Question:
Find the number of terms of the A.P. −12, −9, −6, ..., 21. If 1 is added to each term of this A.P., then find the sum of all terms of the A.P. thus obtained.
Solution:
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
$=S_{12}+12$
$=\frac{12}{2}[2(-12)+11(3)]+12$
$=6 \times(9)+12$
$=66$