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Home » If a, b, c are in A.P., then show that: are also in A.P. (ii) b + c – a, c + a – b, a + b – c are in A.P.
If a, b, c are in A.P., then show that: (i){a^2}(b + c),{b^2}(c + a),{c^2}(a + b) are also in A.P. (ii) b + c – a, c + a – b, a + b – c are in A.P.
Arithmetic Progressions | CBSE | Class 11 | Exercise 19.5 | Maths | RD Sharma
If a, b, c are in A.P., then show that: (i){a^2}(b + c),{b^2}(c + a),{c^2}(a + b) are also in A.P. (ii) b + c – a, c + a – b, a + b – c are in A.P.

Answers:

(i) 

If b2(c + a) – a2(b + c) = c2(a + b) – b2(c + a)

b2c + b2a – a2b – a2c = c2a + c2b – b2a – b2c

Given,

b – a = c – b

a, b, c are in AP,

c(b2 – a2 ) + ab(b – a) = a(c2 – b2 ) + bc(c – b)

(b – a) (ab + bc + ca) = (c – b) (ab + bc + ca)

Cancelling ab + bc + ca from both sides,

b – a = c – b

2b = c + a

Hence, the given terms are in AP

(ii)

If (c + a – b) – (b + c – a) = (a + b – c) – (c + a – b)

b + c – a, c + a – b, a + b – c are in A.P.

Consider LHS and RHS,

(c + a – b) – (b + c – a) = (a + b – c) – (c + a – b)

2a – 2b = 2b – 2c

b – a = c – b

a, b, c are in AP,

b – a = c – b

Hence, the given terms are in AP.

i

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