Question

已知等差数列{An}中，A2=5,A5=14求数列{An}的通项公式An和　前N项和Sn

Answer 1

A2=5,
A1+d=5 (1)
A5=14

A1+4d=14 (2)
(2)-(1)得
3d=9
d=3， A1=2
则 An=2+3(n-1)
即 An=3n-1
Sn=3n(n+1)/2-n
即 Sn=n(3n+1)/2
【OK？】

Answer 2

A5=A2+3d=5+3d=14
3d=14-5=9
d=3
An=A2+(n-2)d
=5+(n-2)*3
=3n-1

A2=A1+d
=A1+3=5
A1=2
Sn=nA1+n(n-1)d/2
=2n+n(n-1)*3/2
=(6n+3n^2-3n)/2
=3n(n+1)/2

Answer 3

A5=A2+3*d=5+3*d=14,d=3,
A1=A2-d=5-3=2,
An=A1+(n-1)d=2+3(n-1)=3n-1
Sn=n*(A1+An)/2=n*(2+3n-1)/2=n*(3n+1)/2

Answer 4

d=(A5-A2)÷(5-2)=(14-5)÷3=3
A1=A2-3=5-3=2
An=A1+(n-1)d=2+3(n-1)=3n-1
Sn=(2+3n-1)n/2=(3n²+n)/2

Answer 5

A2=5,A5=14
A5-A2=3d=9
d=3
A1=A2-d=5-3=2
An=A1+(n-1)d=3n-1
Sn=nA1+n(n-1)d/2=(3n²+n)/2