Question

How many terms of A.P-6,-11/2,-5,... are needed to obtain a sum -25

Answer 1

a=-6
d=a3-a2=a2-a1
d=-5-(-11/2)=(-11/2)-(-6)
d=(-10+11)/2=(-11+12)/2
d=1/2=1/2
sn=n/2(2a+(n-1)d=-25
=n/2(2×-6+(n-1)1/2=-25
=n/2(-12+(n-1)1/2=-25
=n(-12+n/2-1/2=-25×2
=n[(-24+n-1)/2]=-50
=n(-25+n)=-50×2
=-25n+n^2=-100
=n^2-25n+100=0
n^2-20n-5n+100=0
n(n-20)-5(n-20)
(n-5)(n-20)
n-5=0
n=5
n-20=0
n=20
s5=5/2[2×-6+(4×1/2)]
=5/2(-12+2)
=5/2×-10
=5×-5=-25
than
n=5 it is answer