Given,
First term, a = -12
Common difference, d = a2 – a1 = – 9 – (- 12)
d = – 9 + 12 = 3
And, we know that nth term = an = a + (n – 1)d
⟹ 21 = -12 + (n – 1)3
⟹ 21 = -12 + 3n – 3
⟹ 21 = 3n – 15
⟹ 36 = 3n
⟹ n = 12
Thus, the number of terms is 12.
Now, if 1 is added to each of the 12 terms, the sum will increase by 12.
Hence, the sum of all the terms of the A.P. so obtained is
⟹ S12+ 12 = 12/2[a + l] + 12
= 6[-12 + 21] + 12
= 6 × 9 + 12
= 66
Therefore, the sum after adding 1 to each of the terms in the A.P is 66.
