Question

5.
If a, b, care in H.P., then value of c/a+ b-c/b-a +ab+bc/ac

(C) 3
(D)4
(A) 1
(B) 2​

Answer 1

Answer:

this answer is(B) 2

Answer 2

Answer:

Option (B) 2 is correct

Explanation :

Given equation :

c/a + (b-c)/(b-a) + ab + bc/ac

=> c/a + (b-c)/(b-a) + ab + b/a -----(1)

Since a, b and c are in HP, (1/a), (1/b) and (1/c) are in AP.

1/b is the arithmetic mean of (1/a) and (1/c)

-> 2/b = (1/a) + (1/c)

-> 2/b = (a + c)/ac

-> b = 2ac/(a + c)

Use that in the expression [c/a + (b-c) /(b-a) + ab + bc/ac]

We'll do it part by part :

(b - c) = [2ac/(a + c)] - c = (2ac - ac - c^2)/(a + c) = (ac - c^2)/(a + c)

(b - a) = [2ac/(a + c)] - a = (2ac - a^2 - ac) /(a + c) = (ac - a^2)/(a + c)

So, (b - c)/(b - a) = (ac - c^2)/(ac - a^2) = c(a - c)/a(c - a) ------(2)

Now,

From eqn (1) & (2) , we get

eqn(1)=>

c/a + c(a - c)/a(c - a) + ab + b/a

=> c/a - c/a + ab + b/a

=> 2