If a, b, c are in AP, show that
(i) (b + c – a), (c + a – b), (a + b – c) are in AP.
(ii) (bc – a2), (ca – b2), (ab – c2) are in AP.
(i) (b + c – a), (c + a – b), (a + b – c) are in AP.
To prove: (b + c – a), (c + a – b), (a + b – c) are in AP.
Given: a, b, c are in A.P.
Proof: Let d be the common difference for the A.P. a,b,c
Since a, b, c are in A.P.
⇒ b – a = c – b = common differnce
⇒ a – b = b – c = d
⇒ 2(a – b) = 2(b – c) = 2d … (i)
Considering series (b + c – a), (c + a – b), (a + b – c)
For numbers to be in A.P. there must be a common difference between them
Taking (b + c – a) and (c + a – b)
Common Difference = (c + a – b) - (b + c – a)
= c + a – b – b – c + a
= 2a – 2b
= 2(a – b)
= 2d [from eqn. (i)]
Taking (c + a – b) and (a + b – c)
Common Difference = (a + b – c) - (c + a – b)
= a + b – c – c – a + b
= 2b – 2c
= 2(b – c)
= 2d [from eqn. (i)]
Here we can see that we have obtained a common difference between numbers i.e. 2d
Hence, (b + c – a), (c + a – b), (a + b – c) are in AP.
(ii) (bc – a2), (ca – b2), (ab – c2) are in AP.
To prove: (bc – a2), (ca – b2), (ab – c2) are in AP.
Given: a, b, c are in A.P.
Proof: Let d be the common difference for the A.P. a,b,c
Since a, b, c are in A.P.
⇒ b – a = c – b = common differnce
⇒ a – b = b – c = d … (i)
Considering series (bc – a2), (ca – b2), (ab – c2)
For numbers to be in A.P. there must be a common difference between them
Taking (bc – a2) and (ca – b2)
Common Difference = (ca – b2) – (bc – a2)
= [ca – b2 – bc + a2]
= [ca – bc + a2 – b2]
= [c (a – b) + (a + b) (a – b)]
= [(a – b ) (a + b + c)]
a – b = d, from eqn. (i)
⇒ [(d) (a + b + c)]
Taking (ca – b2) and (ab – c2)
Common Difference = (ab – c2) – (ca – b2)
= [ab – c2 – ca + b2]
= [ab – ca + b2 – c2]
= [a (b – c) + (b – c) (b + c)]
= [(b – c) (a + b + c)]
b – c = d, from eqn. (i)
⇒ [(d) (a + b + c)]
Here we can see that we have obtained a common difference between numbers i.e. [(d) (a + b + c)]
Hence, (bc – a2), (ca – b2), (ab – c2) are in AP.