Find the number of terms of the AP −12, −9, −6, ..., 21. If 1 is added to each term of this AP then find the sum of all terms of the

The given AP is −12, −9, −6, ..., 21.

Here, a = −12, d = −9 − (−12) = −9 + 12 = 3 and l = 21

Suppose there are n terms in the AP.

$\therefore l=a_{n}=21$

$\Rightarrow-12+(n-1) \times 3=21 \quad\left[a_{n}=a+(n-1) d\right]$

$\Rightarrow 3 n-15=21$

$\Rightarrow 3 n=21+15=36$

$\Rightarrow n=12$

Thus, there are 12 terms in the AP.

If 1 is added to each term of the AP, then the new AP so obtained is −11, −8, −5, ..., 22.

Here, first term, A = −11; last term, L = 22 and n = 12

∴ Sum of the terms of this AP

$=\frac{12}{2}(-11+22) \quad\left[S_{n}=\frac{n}{2}(a+l)\right]$

$=6 \times 11$

$=66$

Hence, the required sum is 66.