Question

设{an}是等差数列，a1+a5+a9=39,a2+a6+a10=48,则a7+a11+a15=?

Answer 1

（a2+a6+a10）-（a1+a5+a9）=（a2-a1)+(a6-a5)+(a10-a9）=48-39
3d=48-39=9
d=3
a7+a11+a15=(a2+a6+a10)+5*3d=48+5*9=93

Answer 2

An=A1+(n-1)d

a1+a5+a9=39
a1+a1+4d+a1+8d=39
3a1+12d=39

a2+a6+a10=48
a1+d+a1+5d+a1+9d=48
3a1+15d=48

a1= 1 d= 3

a7+a11+a15= a1+6d+a1+10d+a1+14d = 3a1+30d=3+90=93