Question

等差数列{an}中 (1)已知a4+a14=1,求前17项的和【求详细解,

Answer 1

1)S17=a1+a2+...+a17=(a1+a17)+(a2+a16)+...+(a8+a10)+a9

=17*(a4+a14)/2

=17*1/2

=8.5

2)S21=a1+a2+...+a21=21*a11=21*20=420

3)S11=11*a6

a6=S11/11=66/11=6

4)d=(S8-S4-S4)÷16=(6-2-2)/16=1/8

S16=S8+S8+8*8d

=6=6+64x1/8

=20