Question

公差不为0的等差数列an中，a4=10，且a3，a6，a10成等比数列，求1/anan+1的前n项和Tn

Answer 1

(a6)^2=a3*a10
(a4+2d)^2=(a4-d)*(a4+6d)
(10+2d)^2=(10-d)*(10+6d)
4d^2+40d+100=100+50d-6d^2
4d^2+40d=50d-6d^2
10d^2-10d=0
d(d-1)=0
d=0或d=1
公差不为0
所以d=1
a4=a1+3d
10+a1+3*1
a1=7

an=a1+(n-1)d
=7+n-1
=n+6

s=1/7*8+1/8*9+1/9*10+......+1/(n+6)(n+7)
=1/7-1/8+1/8-1/9+1/9-1/10+.....+1/(n+6)-1/(n+7)
=1/7-1/(n+7)
=(n+7)/[7(n+7)]-7/[7(n+7)]
=(n+7-7)/(7n+49)
=n/(7n+49)

Answer 2

an=a1+(n-1)d
a4=a1+3d=10 a1=10-3d
a3=a1+2d=10-3d+2d=10-d
a6=a1+5d10-3d+5d=10+2d
a10=a1+9d=10-3d+9d=10+6d
a3，a6，a10成等比数列,a6/a3=a10/a6
a6*a6=a3*a10
(10+2d)^2=(10-d)(10+6d)
d=1
a1=10-3=7
bn=1/anan+1=1/(7+n-1)(7+n)=1/(6+n)(7+n)=1/(6+n)-1/(7+n)
sbn=(1/7-1/8)+(1/8-1/9)+……（1/(6+n)-1/(7+n)
=1/7-1/(7+n)

Answer 3

公差不为0的等差数列{an}中,a4=10,且a3,a6,a10成等比数列,求数列{an}前20项的和S20
解：设公差为d，则a3=10-d，a6=10+2d，a10=10+6d
依题：（10+2d）^2=(10-d)(10+6d)
即:10d^2-10d=0
d=1或d=0(舍去)
d=1,
a1=7,
a20=26,
S20=(7+26)*20/2=330

仅供参考

Answer 4

我提示你一下
把首项设成x
等差设为d
用x、d分别把a4、a3、a6、a10表示出来