Login/Sign Up
If (b+c)/a, (c+a)/b, (a+b)/c are in AP., prove that: (i) 1/a, 1/b, 1/c are in AP (ii) bc, ca, ab are in AP
If (b+c)/a, (c+a)/b, (a+b)/c are in AP., prove that: (i) 1/a, 1/b, 1/c are in AP (ii) bc, ca, ab are in AP
Arithmetic Progressions | CBSE | Class 11 | Exercise 19.5 | Maths | RD Sharma
If (b+c)/a, (c+a)/b, (a+b)/c are in AP., prove that: (i) 1/a, 1/b, 1/c are in AP (ii) bc, ca, ab are in AP

Answer:

(i) 

If 1/a, 1/b, 1/c are in AP

1/b – 1/a = 1/c – 1/b

Consider LHS,

1/b – 1/a = (a-b)/ab

= c(a-b)/abc

Consider RHS,

1/c – 1/b = (b-c)/bc

= a(b-c)/bc [by multiplying with ‘a’ on both the numerator and denominator]

(b+c)/a, (c+a)/b, (a+b)/c are in AP

RD Sharma Solutions for Class 11 Maths Chapter 19 – Arithmetic Progressions image - 3

(ii) 

If bc, ca, ab are in AP

ca – bc = ab – ca

c (a-b) = a (b-c)

1/a, 1/b, 1/c are in AP

1/b – 1/a = 1/c – 1/b

c (a-b) = a (b-c)

Hence, the given terms are in AP

i

More Material