Question

等差数列｛an｝中，已知a1+a2+a3+…+a10

等差数列{an}中,已知a1+a2+a3+…+a10=p,an-9+an-8+…+an=q,则Sn?

Answer 1

a1+a2+...+a10=10a1+45d=p (1)
a(n-9)+a(n-8)+...+an=10a(n-9)+45d=q (2)
(2)-(1)
10[a(n-9)-a1]=q-p
10[a1+(n-10)d-a1]=q-p
d=(q-p)/[10(n-10)]
代入(1)
a1=(p-45d)/10=[p-45(q-p)/(10n-100)]/10=(10np-55p-45q)/(100n-1000)
Sn=na1+n(n-1)d/2
=n(10np-55p-45q)/(100n-1000) +(q-p)n(n-1)/(20n-200)
=[n/(100n-1000)][10np-55p-45q+5(q-p)(n-1)]
=[n/(100n-1000)](5np-50p-50q+5nq)
=n[n(p+q)-10(p+q)]/(20n-200)
=n(n-10)(p+q)/[20(n-10)]
=n(p+q)/20

Answer 2

a1+a2+...+a10=p (1)
a(n-9)+a(n-8)+...+an=q (2)
1+2得
a1+an+a(n-1)+a2+a3+a(n-2)+...+a10+a(n-9)
=10(an+a1)=p+q
Sn=n(a1+an)/2

Answer 3

解：∵数列{an}为等差数列
∴a1+an=a2+an-1=a3+an-2=....=a10+an-9
∴10（a1+an）=p+q，即a1+an=(p+q)/10
∴Sn=(a1+an)n/2=(p+q)n/20

Answer 4

sn=n/10(p+q）×1／2＝n(p+q)/20
可以把前一个式子看成首相，后一个式子看成尾项，那么用公式sn=n{a1+an)/2来做