If a, b, c are in HP, then what is 1/(b-a) + 1/(b-c) equal to? 1. 2/b 2. 1/a + 1/c 3. (1/2)(1/a + 1/b + 1/c). Select the correct answer using the code given below:

If a, b, c are in HP, then 1/a, 1/b, 1/c are in AP, so 2/b = 1/a + 1/c, which gives b = 2ac/(a+c). Compute b-a = 2ac/(a+c) - a = (2ac - a² - ac)/(a+c) = (ac - a²)/(a+c) = a(c-a)/(a+c). Similarly b-c = c(a-c)/(a+c). So 1/(b-a) + 1/(b-c) = (a+c)/[a(c-a)] + (a+c)/[c(a-c)] = (a+c)/(c-a) · [1/a - 1/c] = (a+c)/(c-a) · (c-a)/(ac) = (a+c)/(ac) = 1/a + 1/c. So statement 2 is correct.