If a, b, c are in continued proportion then prove that (ab + bc + ac)² = ac(a + b + c)²

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Solution

$If a, b a n d c are in continued proportion, b^{2} = a c . . . (i) We have to show {(a b + b c + a c)}^{2} = a c {(a + b + c)}^{2} . F o r t h a t, c onsider the LHS . {(a b + b c + a c)}^{2} = a^{2} b^{2} + b^{2} c^{2} + a^{2} c^{2} + 2 a b^{2} c + 2 b c^{2} a + 2 a^{2} b c From (i), w e g e t : {(a b + b c + a c)}^{2} = a^{3} c + a c^{3} + a^{2} c^{2} + 2 a^{2} c^{2} + 2 b c^{2} a + 2 a^{2} b c . . . (ii)$

$Now consider t h e RHS . a c {(a + b + c)}^{2} = a c (a^{2} + b^{2} + c^{2} + 2 a b + 2 b c + 2 c a) = a^{3} c + a b^{2} c + a c^{3} + 2 a^{2} b c + 2 a b c^{2} + 2 a^{2} c^{2} From (i), w e g e t : a c {(a + b + c)}^{2} = a^{3} c + a^{2} c^{2} + a c^{3} + 2 a^{2} b c + 2 a b c^{2} + 2 a^{2} c^{2} . . . (iii) From (ii) & (iii), w e g e t : {(a b + b c + a c)}^{2} = a c {(a + b + c)}^{2}$

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