SOLUTION: Find the sum of the first n terms of the given arithmetic series. a2=-6, a5=-18 (n=8)

Question 39509: Find the sum of the first n terms of the given arithmetic series.
a2=-6, a5=-18 (n=8)
Found 2 solutions by venugopalramana, Fermat:
Answer by venugopalramana(3286) (Show Source): You can put this solution on YOUR website!
Find the sum of the first n terms of the given arithmetic series.
a2=-6, a5=-18 (n=8)
A2=A+D=-6......................I
A5=A+4D=-18.................II
3D=-18+6=-12
D=-4
A=-6+4=-2
SN=(N/2){2*-2+(N-1)(-4)}=(N/2){-4-4N+4}= -2N^2
S8=-2*8^2=-128
Answer by Fermat(136) (Show Source): You can put this solution on YOUR website!
a2 = -6, a5 = -18
(a5 - a2) = -18 - (-6) = -18 + 6 = -12
difference between any two consectutive terms in an AP is d.
a5 - a2 = 3d
.: 3d = -12
d = -4
======
since d = -4 and a2 = -6,
then,
a1 = a2 - d = -6 - (-4) = -2
a0 = a1 - d = -2 - (-4) = 2
a0 = 2
======
For an AP, the sum of the 1st n terms is,
Sn = (n/2)[2a0 + (n-1)d]
Using n = 8, d = -4 and a0 = 2,
S8 = (8/2)[2*2 + (8-1)*(-4)]
S8 = 4[4 - 7*4]
S8 = 4[-24]
S8 = -96
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