Question

在等差数列{an}中，已知a4+a7+a10=17,a4+a5+a6+……+a14=77。若ak=13，k=?

Answer 1

解：等差数列 所以a4+a7+a10=17

a4 + a10 = a1+ a14 = 2a7 =17 a7 = 17/3

a4+a5+a6+……+a14=77 a4+a14 = a1+a18 = 2a9

2a9*(14-4+1)/2=77 a9 = 7

公差为 a9 - a7 /2 =2/3

(ak - a9)/(2/3) = 9

那么 ak是 a9 + 9d 即 K为18

Answer 2

等差数列 所以a4+a7+a10=17

a4 + a10 = a1+ a14 = 2a7 =17 a7 = 17/3

a4+a5+a6+……+a14=77 a4+a14 = a1+a18 = 2a9

2a9*(14-4+1)/2=77 a9 = 7

公差为 a9 - a7 /2 =2/3

所以a1+6d=17/3 a1+8d=7
所以a1=5/3
5/3+(k-1)*2/3=13
k=21

Answer 3

a4+a7+a10==3a7=17
a7=17/3

a4+a5+a6+……+a14=11a9=77
a9=7

2d=a9-a7=4/3
d=2/3

a1=a7-6d=5/3
ak=a1+(k-1)d=5/3+(k-1)*2/3=13
k=18

Answer 4

an=a+(n-1)d

17=a4+a7+a10=3a+18d
77=a4+a5+a6+……+a14=11a+88d
7=a+8d

a=5/3
d=2/3

an=5/3+(2/3)(n-1)=(2/3)n-1
ak=(2/3)k-1=13
k=14*3/2=21