There are two ways to solve this problem, conventional method and special method. Because it's a fill-in-the-blank question, I'd better talk about the special method first.

I need to add some knowledge before introducing special methods. Usually, when we calculate arithmetic progression's summation formula, we are used to adding *n/2 from beginning to end. But in fact, if the number of items in a series is odd, we can take twice the middle number instead of the added value from beginning to end, in fact, the essence is the same.

For example, S3=3a2 and S5=5a3.

Special methods: a1/b1= s1/t1= 2/4, A2/B2 = S3/T3 = 6/ 10, a3/B3 = S5/t5 =/.

It is observed that if a 1=2, then b 1=4, d 1=4 and d2=6.

So A5 = a1+4d1= 2+4 * 4 =18, B7 = b1+6d1= 4+6 * 6 = 40.

So a7/b9=9/20.

Traditional methods:

A1/b1= s1/t1= 2/4, if a 1=2a. Then b 1=4a.

s 1 = n *(2a 1+(n- 1)d 1)/2，t 1 = n *(2b 1+(n- 1)D2)/2。 If you add cn to both the numerator and denominator, you can know that there is.

sn = n *(2a 1+(n- 1)d 1)/2 = cn *(2n)，( 1)

TN = n *(2b 1+(n- 1)D2)/2 = cn *(3n+ 1)，(2)

If the formula (1) holds, then both sides of the coefficient equation of n are the same, and both sides of the coefficient equation of n 2 must also be the same, that is:

a 1-d 1/2 = 0(3)d 1/2 = 2c(4)

2a 1=d 1 (5), so d 1=4a.

The same principle can also be obtained from (2)

b 1-d2/2=c (6)d2/2=3c (7)

(6)/(7) gives 3b 1=2d2 (8), so d2=6a.

So a5 = a1+6d1= 2a+4 * 4a =18a.

b7=b 1+6d2=4a+6*6a=40a

So a9/b 10=9/20.