Answers

2016-03-18T12:45:55+05:30
Last term = a+(n-1)d
   12=-12+(n-1)2
   12= -12+2n-2
   26 = 2n
   n =13
∴ number of terms = 13
if 1 is added on both sides
   series = -11,-8,.........13
sum of terms = n/2[2a+(n-1)d]
                        13/2[2(-11) + (13-1)2]
                        13/2[-22+24]
                        13/2[2]
                         13  
2 3 2
2016-03-18T13:44:43+05:30
Let,
A.P => -12, -9, -6,........,12.
first term = a = -12
2nd term = b = -9
last term = T = 12
no. of terms = n
so, common difference = d
d = b - a = (-9) - (-12)
or, d = -9 + 12 = 3

Now,
T = a + (n-1)d
or, 12 = -12 + (n - 1)3
or, (n - 1)3 = 12 + 12 = 24
or, n - 1 = 24/3 = 8
or, n = 8 + 1
n = 9
Total no. of terms = 9
and,sum of n terms(s) = n(a + T)/2
or, s = 9(-12 + 12)3 = 0

a/q,
Sum of n terms if 1 is added to every terms(S) = s + n
S = 0 + 9 = 9
Therefore sum of n terms if 1 is added to every terms is 9.
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