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Na 1 mathematics
For the first n terms and formulas of arithmetic progression:

Sn=na 1+n(n- 1)d/2。 ①

s2n = na 1+2n(2n- 1)d/2。 ②

So:

sn/S2n =[na 1+n(n- 1)d]/[2na 1+2n(2n- 1)d]。 ③

Expand it:

sn/S2n =[a 1+(N- 1)d]/[2a 1+(4n-2)d]。 Omit n.

= 1-(3n- 1)d/[2a 1+(4n-2)d]。 ④

So as long as the following fraction is constant.

And let this constant be k (k≠0, because if k=0, there must be d=0).

Sn/S2n= 1-k .⑤

Then:

(3n- 1)d = k[2a 1+(4n-2)d]。 ⑥

Sorted mobile items:

a 1 =[(3n- 1)/2k-2n+ 1]d .⑦

A 1 and d are constants, and the proportional relationship between them must have nothing to do with n.

So there is

3n/2k-2n=0 (making the term related to n in Equation 7 zero)

Do you understand:

k=3/4

Substitute ⑤ ⑦:

a 1=d/3

Sn/S2n= 1-k= 1/4