Answer :

The given AP is-

-12, -9, -6, …., 21

Let the first term and common difference of given AP be a and d respectively.

first term(a) = -12 and,

common difference(d) = (n+1)th term - nth term = -9-(-12)

= -9+12

= 3

last term or nth term(a_{n}) = 21

Let the total no. of terms in above A.P be n.

∴ a_{n} = a+(n-1)×d

⇒ 21 = -12+(n-1)×3

⇒ 33 = 3n-3

⇒ 3n = 36

∴ n = 12

After adding 1 to each of the 12 terms of AP, the sum of all terms of new AP formed will increase by 12.

Sum of all the 12 terms of the given AP is given by-

S_{12} =(12/2)(-12+21)

[∵S_{n} = (n/2)(a+l) =(n/2)[(2a+(n-1)d]

=6×9

=54

Thus, the sum of all the terms of New AP formed = 54+12 = 66.

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