Question

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 AP thus obtained. [CBSE 2013]

Answer 1

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.

$$\begin{gathered} \therefore l=a_n=21 \\ \Rightarrow -12+\left(n-1\right)\times 3=21\left[a_n=a+\left(n-1\right)d\right] \\ \Rightarrow 3n-15=21 \\ \Rightarrow 3n=21+15=36 \end{gathered}$$
$\Rightarrow n=12$

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

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

$$\begin{gathered} =6\times 11 \\ =66 \end{gathered}$$

Hence, the required sum is 66.