Question

1.已知等差数列｛an｝满足a2+a7=4，a4+a9=10，求｛an}的前十项和S10

Answer 1

a4+a9=10
a2+2d+a7+2d=10
4+4d=10
4d=6
2d=3

a2+a7=a1+d+a6+d
a2+a7=a1+a6+2d
4=a1+a6+3
a1+a6=1

s10
=a1+a2+a2+a3+....+a10
=a1+a6+a2+a7+........+a5+a10
=a1+a6+a1+a6+2d+a1+a6+4d+a1+a6+6d+..........+a1+a6+8d
=5(a1+a6)+20d
=5*1+10*2d
=5+10*3
=35

Answer 2

a2+a7=4，a4+a9=10 所以可得：
4d=10-4 解得：d=1.5
a2+a7=a1+a8=4
a10=a8+2d=a8+3
所以可得：a1+a10=a1+a8+3=7
S10=(a1+a10)10/2
=7x5
=35

Answer 3

两式相加，得到a2+a9=7即a1+a10=7,所以s10=35

Answer 4

2a1+7d=4,2a1+11d=10,可得d=3/2,a1=-13/4就可以求和了

Answer 5

2a1+7d=4,
2a1+11d=10,
d=3/2,a1=-13/4,a10=41/4
s10=10*(a1+a10)/2
=5*7
=35