Question

find the no. 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

we have an=21, a1=-12, a2=-9
so d= -9-(-12)
d=3
hance an=a1+(n-1)d
21=-12+(n-1)3
21=-12+3n-3
3n=36
n=36/3=12
hence no. of terms is equal to 12.

now if 1 is added to each term a.p. will be like -11,-8............22
now a=-11, d=3, n=12
hence sum of all terms will be Sn= n(a1+an)/2
where an is last term=22
substituting values we get
Sn= 12(-11+22)/2
Sn=12×11/2
sn=66