Question

已知等差数列首项为19 公差为-4 求通项公式AN及前N项和SN

Answer 1

等差数列通项公式an=a1+(n-1)*d
等差数列前n项和sn=na1 + n*(n-1)*d/2=n(a1+an)/2

a1=19 d=-4

可知 an=-4n+23
sn=-2(n^2)+21n

Answer 2

a7/a5=a5/a1
(a1+6d)/(a1+4d)=(a1+4d)/a1
a3=a1+2d=8
(8+4d)/(8+2d)=(8+2d)/(8-2d)
(8+4d)(8-2d)=(8+2d)(8+2d)
64+16d-8d2=64+32d+4d2
12d2+16d=0
3d2+4d=0
d=-4/3
a1=8-2d=32/3
a2=32/3-4/3=28/3
a3=8=24/3
a4=20/3
a5=16/3
a7=8/3
an=(36-4n)/3=
sn=n(34-2n)/3