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.

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.

∴ l = a_n = 21   ...[a_n = a + (n – 1)d]

⇒ –12 + (n – 1) × 3 = 21   

⇒ 3n – 15 = 21

⇒ 3n = 21 + 15 = 36

⇒ 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

= `12/2 (-11 + 22)`   ...`[S_n = n/2 (a + l)]`

= 6 × 11

= 66 

Hence, the required sum is 66.