求解！高中數學數列問題?

1)
由1/a，1/b，1/c成等差數列得 1/a + 1/c = 2/b ⇒ b(a+c) = 2ac ... ☆
故 (b+a)/(b-a) + (b+c)/(b-c)
= ((b+a)(b-c) + (b+c)(b-a)) / ((b-a)(b-c))
= (b²+ab-bc-ac + b²+bc-ab-ac) / ((b-a)(b-c))
= (2b² - 2ac) / ((b-a)(b-c))
= (2b² - 2ac) / (b²-ab-bc+ac)
= (2b² - 2ac) / (b²-b(a+c)+ac)
= (2b² - 2ac) / (b²-2ac+ac) ... (由 ☆)
= 2

2)
a + ar + ar² = 19/9
{
1/a + 1/(ar) + 1/(ar²) = 19/4
⇒
a(r² + r + 1) = 19/9
{
r² + r + 1 = 19ar²/4
⇒
a(19ar²/4) = 19/9
(ar)² = 4/9
正數 ar = 2/3 , 代回得:

a + 2/3 + 2r/3 = 19/9
9a + 6 + 6r = 19
9a + 6r = 13
9ar + 6r² = 13r
9(2/3) + 6r² = 13r
6r² - 13r + 6 = 0
(3r - 2)(2r - 3) = 0
r = 2/3 or 3/2
故 ar² = ar r = (2/3)(2/3) = 4/9 或 (2/3)(3/2) = 1 , 則此三數中最小的為 4/9.

第一題(另解,較為簡潔) 請參考