Question

已知等差数列{an}中，a1-a4-a8-a12+a15=2,则a3+a13=

Answer 1

根据公式an=a1+(n-1)d得
a4=a1+3d
a8=a1+7d
a12=a1+11d
a15=a1+14
a1-a4-a8-a12+a15=a1-(a1+3d)-(a1+7d)-(a1+11d)+(a1+14d)=-a1-7d=2
a3+a13=a1+2d+a1+12d=2a1+14d
-2(-a1-7d)=2a1+14d=-4

Answer 2

解：a1-a4-a8-a12+a15=a1+a15-(a4+a12+a8）=2a8-（2a8+a8）=-a8=2
a8=-2
a3+a13=2*a8=2*（-2）=-4

Answer 3

a1-a4-a8-a12 a15=2
m n=p q则am an=ap aq得
-a8-(a4 a12) (a15 a1)=2
a3 a13=a4 a12=a15 a1=4a4
原式=-a8=2所以a16=-4

Answer 4

解：∵a1-a4-a8-a12+a15=2
又∵a1+a15=a4+a12=2a8
∴a8=-2
∴a3+a13=2a8=-4
故答案为：-4

Answer 5

a1-a4-a8-a12+a15=a1-(a1+3d)-(a1+7d)-(a1+11d)+(a1+14d)=-a1-7d=2
a1+7d=-2
a3+a13=a1+2d+a1+12d=2a1+14d=2(a1+7d)=2*(-2)=-4

Answer 6

a1-a4-a8-a12+a15=a1-(a1+3d)-(a1+7d)-(a1+11d)+(a1+14d)=-a1-7d=2
a1+7d=-2
a3+a13=a1+2d+a1+12d=2a1+14d=2(a1+7d)=2*(-2)=-4