An=4 d =2 sn =-14 find n and a
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An=a+(n-1)d=4
a+(n-1)2=4 (substitute d=2)
a+2n-2=4
a+2n=6
a=6-2n
Sn=n/2(2a+(n-1)d)
-14=n/2(2(6-2n)+(n-1)2) (substitute a=6-2n,d=2)
-14=n/2(12-4n+2n-2)
-14=n/2(10-2n)
-14=n(5-n)
5n-n^2=-14
n^2-5n-14=0
n^2-7n+2n-14=0
6-2n(n-7)+2(n-7)=0
(n-7)(n+2)=0
n=7,-2
n can't be negative
therefore n=7
a=6-2n (substitute n=7)
a=6-2(7)=6-14=-8
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