等差数列an中,a1

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等差数列an中,a1

等差数列an中,a1
等差数列an中,a1

等差数列an中,a1
设公差为q,则
s3 = (a1+a3)*3/2 = (a1+a1+2*q)*3/2 = 3(a1+q)
s16 = (a1+a16)*16/2=(a1+a1+15*q)*8=8(2a1+15q)
3(a1+q)=8(2a1+15q)
-13a1 = 117q
a1=-9q
sn=(a1+a1+(n-1)q)*n/2 = (-19q+nq)*n/2=0.5q(n^2-19n)
=0.5q[(n-9.5)^2-9.5^2]
sn取最小值时显然n为9.5