Question -
Answer -
(i) a2(b + c),b2(c + a), c2(a + b) are also in A.P.
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
And since a, b, c arein AP,
c(b2 –a2 ) + ab(b – a) = a(c2 – b2 )+ bc(c – b)
(b – a) (ab + bc + ca)= (c – b) (ab + bc + ca)
Upon cancelling, ab +bc + ca from both sides
b – a = c – b
2b = c + a [which istrue]
Hence, given terms arein AP
(ii) b + c – a, c + a– b, a + b – c are in A.P.
If (c + a – b) – (b +c – a) = (a + b – c) – (c + a – b)
Then, b + c – a, c + a– b, a + b – c are in A.P.
Let us consider LHSand RHS
(c + a – b) – (b + c –a) = (a + b – c) – (c + a – b)
2a – 2b = 2b – 2c
b – a = c – b
And since a, b, c arein AP,
b – a = c – b
Hence, given terms arein AP.
(iii) bc – a2,ca – b2, ab – c2 are in A.P.
If (ca – b2)– (bc – a2) = (ab – c2) – (ca – b2)
Then, bc – a2,ca – b2, ab – c2 are in A.P.
Let us consider LHSand RHS
(ca – b2) –(bc – a2) = (ab – c2) – (ca – b2)
(a – b2 –bc + a2) = (ab – c2 – ca + b2)
(a – b) (a + b + c) =(b – c) (a + b + c)
a – b = b – c
And since a, b, c arein AP,
b – c = a – b
Hence, given terms arein AP