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The derivation process of finding the first n items of arithmetic progression in senior high school mathematics. For example, the derivation process of the ratio of the sum of odd terms to the sum of
The derivation process of finding the first n items of arithmetic progression in senior high school mathematics. For example, the derivation process of the ratio of the sum of odd terms to the sum of even terms and so on. When the number of items is even:

1.S even -S odd =nd

S odd /S even =S odd /(S odd +nd)=(S odd +nd-nd)/(S odd +nd)= 1-nd/(S odd +nd)

S odd number =na 1+n(n- 1)*2d/2 (odd column tolerance is 2)=na 1+n2d-nd.

S odd number +nd=na 1+n2d

Nd/(S odd number +nd)=d/(a 1+nd)

2.S odd /S even = 1-nd/(S odd+nd) = (a1+(n-1) d)/(a1+nd) = an/an+/kloc-0.

When the number of terms is 2n+ 1, the middle term is n+ 1.

3.S odd -S even =a 1+nd=an+ 1=a medium.

S odd /S even =(S even+one middle) /S even = 1+ one middle /S even.

A in =(a2+a2n)/2

S even =n(a2+a2n)/2(2n+ 1 column has n even numbers, n+ 1 odd numbers, and the sum of even numbers is n/2 a2+a2n).

4. So s odd /S even =(S even +a middle) /S even = 1+a middle /S even =1+1/n = (n+1)/n.

When the number of terms is 2n- 1, the middle term is n.

5.S odd -S even =a 1+(n- 1)d=an=a medium.

6. Similarly, it is proved that S odd /S even =(S even +a middle) /S even = 1+a middle /S even =1(n-1) = n/(n-1).