已知1/a+1/b+1/c成等差数列,且a+c,a-c,a+c-2b皆为正.求证：lg(a+c),lg(a-c),lg(a+c-2b)也成等差数列.
问题描述：
已知1/a+1/b+1/c成等差数列,且a+c,a-c,a+c-2b皆为正.求证：lg(a+c),lg(a-c),lg(a+c-2b)也成等差数列.
答
1/a,1/b,1/c成等差数列
2/b=(1/a)+(1/c)=(a+c)/ac
b=2ac/(a+c)
要证明lg(a+c),lg(a-c), lg(a+c-2b)成等差数列,
只要证明lg(a+c)+lg(a+c-2b)＝2lg(a-c)
等式左边＝lg(a+c)+lg(a+c-2b)
=lg(a+c)(a+c-2b)
=lg(a+c)[a+c-2(2ac/(a+c)]
=lg[(a+c)^2-4ac]
=lg(a-c)^2
=2lg(a-c)
＝右边