Question

已知数列[an]是等差数列，且3(a3+a5)+2(a7+a10+a13)=24,那么数列[an]的前13项和为

Answer 1

3(a3+a5)+2(a7+a10+a13)=24 (1)
设方差为d
由a3=a1+2d a5=a1+4d a7=a1+6d ...
可以知道(1) => 12a1+72d =24 =>a1+6d =2

a1+a2+...+a13=(a1+a13)*13/2= (a1*2+12d)*13/2=13*(a1+6d)=13*2=26
望采纳，谢谢了~

Answer 2

设公差为D
3(A3+A5)+2(A7+A10+A13)=3(2A1+6D)+2(3A1+27D)=6(A1+3D)+6(A1+9D)
=6(A4+A10)=6(A1+A13)=24
所以A1+A13=4
S13=(A1+A13)*13/2=4*13/2=26

Answer 3

3(a3+a5)+2(a7+a10+a13)=243(a4+a4)+2(a10+a10+a10)=246a4+6a10=24a4+a10=4a1+a13=4S13=13(a1+a13)/2=26

Answer 4

26.