If a,b,c be in G.P and a + x, b + x, c + x, in H.P ., then the value of x is (a,b,c, are distinct numbers)

(1) c

(2) b

(3) a

(4) None of these

Solution:

Since a, b, c are in GP

b² = ac …(i)

a + x, b + x, c + x, in H.P

So 1/(a+x) , 1/(b+x), 1/(c+x) are in AP.

1/(b+x) – 1/(a+x) = 1/(c+x) – 1/(b+x)

(a-b)/(a+x)(b+x) = (b-c)/(c+x)(b+x)

Cancel (b+x) from both sides

(a-b)/(a+x) = (b-c)/(c+x)

Cross multiply

(a-b)(c+x) = (b-c)(a+x)

ac + ax – bc – bx = ab – ac + bx -cx

2ac – bc – ab + ax – 2bx + cx= 0

2b² -bc -ab + x( a – 2b + c) = 0 (From (i) b² = ac)

b(2b – c – a) + x( a – 2b + c) = 0

b(2b – c – a) = x( -a + 2b – c)

b(2b – c – a) = x(2b – c – a)

So b = x

Hence option (2) is the answer.