## [高中]等差數列

### [高中]等差數列

n[2a + (n-1)d]/2 = m ...... (1)
m[2a + (m-1)d]/2 = n ...... (2)

(1) ==> 2a = 2m/n - (n-1)d ...... (*)
(2) ==> 2a = 2n/m - (m-1)d

2m/n - (n-1)d = 2n/m - (m-1)d
(m-1)d - (n-1)d = 2n/m - 2m/n
(m-n)d = 2(n² - m²)/(mn)
(m-n)d = 2(n-m)(n+m)/(mn) ...... ∵ m≠n，∴ 此式可同除 m-n
d = -2(n+m)/(mn) ...... (3)

= (m+n)[2a + (m+n-1)d]/2
= m[2a + (m+n-1)d]/2 + n[2a + (m+n-1)d]/2
= m[2a + (m-1)d]/2 + m[nd]/2 + n[2a + (n-1)d]/2 + n[md]/2
= n + mnd/2 + m + mnd/2 ...... By (1) & (2)
= m + n + mnd
= m + n + mn[-2(n+m)/(mn)] ...... By (3)
= m + n - 2(n+m)
= -(m+n) ■

2a = 2m/n - (n-1)[-2(n+m)/(mn)]
a = m/n + (n-1)(n+m)/(mn) ...... (4)

benice

benice 寫到:解：

n[2a + (n-1)d]/2 = m ...... (1)
m[2a + (m-1)d]/2 = n ...... (2)

(1) ==> 2a = 2m/n - (n-1)d ...... (*)
(2) ==> 2a = 2n/m - (m-1)d

2m/n - (n-1)d = 2n/m - (m-1)d
(m-1)d - (n-1)d = 2n/m - 2m/n
(m-n)d = 2(n² - m²)/(mn)
(m-n)d = 2(n-m)(n+m)/(mn) ...... ∵ m≠n，∴ 此式可同除 m-n
d = -2(n+m)/(mn) ...... (3)

= (m+n)[2a + (m+n-1)d]/2
= m[2a + (m+n-1)d]/2 + n[2a + (m+n-1)d]/2
= m[2a + (m-1)d]/2 + m[nd]/2 + n[2a + (n-1)d]/2 + n[md]/2
= n + mnd/2 + m + mnd/2 ...... By (1) & (2)
= m + n + mnd
= m + n + mn[-2(n+m)/(mn)] ...... By (3)
= m + n - 2(n+m)
= -(m+n) ■

2a = 2m/n - (n-1)[-2(n+m)/(mn)]
a = m/n + (n-1)(n+m)/(mn) ...... (4)

n = m[2a+(m-1)d]/2 = ma + m(m-1)d/2
m = n[2a +(n-1)d]/2 = na + m(n-1)d/2

lskuo

lskuo 寫到:

benice