Question

等差数列中前4项的和为124,后4项的和为156,所有各项的和为210,则数列的项数?

Answer 1

解:前4项和S4=4a1+6d=124,即2a1+3d=62..........(1)
设后4项为a(n-3),a(n-2),a(n-1),an.
其和:S4'=4a(n-3)+6d=156
,用a(n-3)=a1+(n-4)d代入,
得:4[a1+(n-4)d]+6d=4a1+2(2n-5)d=156............(2)
由(1)得2a1=62-3d,代入(2)式,
:2(62-3d)+2(2n-5)d=124+(4n-16)d=156
∴(n-4)d=8..................................(3)
前n项和Sn=na1+n(n-1)d/2=210
用a1=(62-3d)/2代入,得:n(62-3d)/2+n(n-1)d/2=210
故有 (n^2)d-4nd+62n=420......................(4)
由(3)得d=8/(n-4),代入(4)式得:(n^2-4n)[8/(n-4)]+62n=420
化简得n^2-10n+24=0
即(n-6)(n-4)=0
故n=6或n=4(舍去)有6项.

Answer 2

a1+a2+a3+a4+an+a(n-1)+a(n-2)+a(n-3)=124+156,4(a1+an)=280,a1+an=70
sn=(a1+an)*n/2=70n/2=210,n=6