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If 1/a, 1/b, 1/c are in A.P., prove that: (i) (b+c)/a, (c+a)/b, (a+b)/c are in A.P. (ii) a(b + c), b(c + a), c(a + b) are in A.P.
If 1/a, 1/b, 1/c are in A.P., prove that: (i) (b+c)/a, (c+a)/b, (a+b)/c are in A.P. (ii) a(b + c), b(c + a), c(a + b) are in A.P.
Arithmetic Progressions | CBSE | Class 11 | Exercise 19.5 | Maths | RD Sharma
If 1/a, 1/b, 1/c are in A.P., prove that: (i) (b+c)/a, (c+a)/b, (a+b)/c are in A.P. (ii) a(b + c), b(c + a), c(a + b) are in A.P.

Answer:

(i) 

If a, b, c are in AP

b – a = c – b

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

1/b – 1/a = 1/c – 1/b

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

(c+a)/b – (b+c)/a = (a+b)/c – (c+a)/b

Let us take LCM,

RD Sharma Solutions for Class 11 Maths Chapter 19 – Arithmetic Progressions image - 1

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

1/b – 1/a = 1/c – 1/b

C (b – a) = a (b-c)

Hence, the given terms are in AP.

(ii)

b(c + a) – a(b+c) = c(a+b) – b(c+a)

Consider LHS,

b(c + a) – a(b+c)

Upon simplification,

b(c + a) – a(b+c) = bc + ba – ab – ac

= c (b-a)

 

c(a+b) – b(c+a) = ca + cb – bc – ba

= a (c-b)

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

1/a – 1/b = 1/b – 1/c

Or

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

Hence, the given terms are in AP.

i

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