Question

if the first term of an ap is -5 and the common difference is 2 find the sum of first 6 terms

Answer 1

Sum of AP=(n/2)[2a+(n-1)d]

where, a=first term of the series, d=common difference, n=number of terms

Here, a=-5, d=2, n=6

Sum= (6/2)[(2*(-5))+(6-1)(2)]

=3*[(-10)+(5*2)]

=3*[-10+10]

=3*0

=0

Sum of the given AP is 0.

Verification:

a1=-5

a2=-5+d=-5+2=-3

a3=-5+2d=-5+(2*2)=-5+4=-1

a4=-5+3d=-5+(3*2)=-5+6=1

a5=-5+4d=-5+(4*2)=-5+8=3

a6=-5+5d=-5+(5*2)=-5+10=5

Sum=a1+a2+a3+a4+a5+a6

=(-5)+(-3)+(-1)+1+3+5

=0