If a, b, c are in A.P., then show that:
(i) a² (b + c), b² (c + a), c² (a + b) are also in A.P.
(ii) b + c − a, c + a − b, a + b − c are in A.P.
(iii) bc − a², ca − b², ab − c² are in A.P.

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Solution

$Since a, b, c are in A . P ., we have : 2 b = a + c (i) We have to prove the following : 2 b^{2} (a + c) = a^{2} (b + c) + c^{2} (a + b) LHS : 2 b^{2} \times 2 b (Given) = 4 b^{3} RHS : a^{2} b + a^{2} c + a c^{2} + c^{2} b = a c (a + c) + b (a^{2} + c^{2}) = a c (a + c) + b [(a + c)^{2} - 2 a c] = a c (2 b) + b [{(2 b)}^{2} - 2 a c] = 2 a b c + 4 b^{3} - 2 a b c = 4 b^{3} LHS = RHS Hence, proved .$

$(ii) We have to prove the following : 2 (c + a - b) = (b + c - a) + (a + b - c) LHS : 2 (c + a - b) = 2 (2 b - b) (∵ 2 b = a + c) = 2 b RHS : (b + c - a) + (a + b - c) = 2 b LHS = RHS Hence, proved .$

$(iii) We have to prove t h e f o l l o w i n g : 2 (c a - b^{2}) = (b c - a^{2} + a b - c^{2}) RHS : b c - a^{2} + a b - c^{2} = c (b - c) + a (b - a) = c (\frac{a + c}{2} - c) + a (\frac{a + c}{2} - a) (∵ 2 b = a + c) = c (\frac{a + c - 2 c}{2}) + a (\frac{a + c - 2 a}{2}) = \frac{c (a - c)}{2} + a (\frac{c - a}{2}) = \frac{c a}{2} - \frac{c^{2}}{2} + \frac{a c}{2} - \frac{a^{2}}{2} = a c - \frac{1}{2} (c^{2} + a^{2}) = a c - \frac{1}{2} (4 b^{2} - 2 a c) (∵ a^{2} + c^{2} + 2 a c = 4 b^{2} \Rightarrow a^{2} + c^{2} = 4 b^{2} - 2 a c) = a c - 2 b^{2} + a c = 2 a c - 2 b^{2} = 2 (a c - b^{2}) = LHS Hence, proved .$

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If a, b, c are in A.P., then show that: (i) a2 (b + c), b2 (c + a), c2 (a + b) are also in A.P. (ii) b + c − a, c + a − b, a + b − c are in A.P. (iii) bc − a2, ca − b2, ab − c2 are in A.P.

If a, b, c are in A.P., then show that:
(i) a² (b + c), b² (c + a), c² (a + b) are also in A.P.
(ii) b + c − a, c + a − b, a + b − c are in A.P.
(iii) bc − a², ca − b², ab − c² are in A.P.