1

3n

Then |a2n-a2n- 1|=

1

32n

，|a2n+2-a2n+ 1|=

1

32n+2

∵ sequence {a2n- 1} is a decreasing sequence, and {a2n} is an increasing sequence.

∴ a2n+ 1-a2n- 1 < 0, and a2n+2-a2n > 0,

Then -(a2n+2-a2n) < 0, and the two inequalities are added.

A2n+1-a2n-1-(a2n+2-a2n) < 0, that is, a2n-a2n-1< a2n+2-a2n+1,

∵|a2n-a2n- 1|=

1

32n

> |a2n+2-a2n+ 1|=

1

32n+2

∴ a2n-a2n- 1 < 0, that is, a2n-a2n- 1=-

1

32n

Similarly, A2n+3-A2n+2 < A2n+ 1-A2n,

| a2n+3-a2n+2 |

Then a2n+ 1-a2n=

1

32n+ 1

When the number of items in the sequence {an} is even, let n=2m(m∈N*),

a2-a 1=-

1

32

，a3-a2=

1

33

，…，a2m- 1-a2m-2=

1

32m- 1

，a2m-a2m- 1=-

1

32 meters

When these 2m- 1 equations are added together, a2m-a 1=- (

1

32

+

1

34

+…+

1

32 meters

)+(

1

33

+

1

35

+…+

1

32m- 1

),

∴a2m=a 1-

1

nine

( 1-

1

9m

)

1-

1

nine

+

1

27

( 1-

1

9m- 1

)

1-

1

nine

=

22

24

-

1

four

1

32 meters

.

∴ 12a 10= 12(

22

24

-

1

four

1

3 10

)= 1 1-

1

39

.

Therefore, choose: d.