1. Let Sn be the sum of the top n items in arithmetic progression {{An}}. If S3/S6= 1/3, what is S6/S 12?

Let the first term of arithmetic progression be A and the tolerance be D, and you can calculate.

S3=3a+3d

S6=6a+ 15d

S 12= 12a+66d

S3/S6= 1/3

Then (3a+3d)/(6a+15d) =1/3.

The solution is a=2d.

So S6/S 12=3/ 10.

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Arithmetic progression S3, S6-S3, S9-S6 and S 12-S9 also became arithmetic progression.

S3/S6= 1/3，S6=3S3，S6-S3=2S3

S9-S6=3S3，S9=6S3

S 12-S9=4S3，S 12= 10S3

So S6/S 12=3/ 10.

2.Sn is the sum of the first n terms of arithmetic progression {an}. It is known that S9= 18, Sn=240 and a(n-4)=30(n is greater than 9). How to find n?

Let the first term a, tolerance d.

(a+a+8d)*9/2= 18

[a+a+(n- 1)d]*n/2=240

a+(n-5)d=30

Hardware decoding, a =-50/3, d = 14/3, n = 15.

There is a better way:

(a+a+8d)*9/2= 18→a+4d=2

a+(n-5)d=30

Adding the two formulas, we can get 2a+nd-d=32.

Comparison [a+a+(n-1) d] * n/2 = 240 → (2a+nd-d) * n = 480.

Divided by n=480/32= 15.

Note: In fact, S9= 18, which means a5= 18/9=2.