Question

3. Show that (a - b)^2, (?a^2+ b^2) and (a + b)^2 are in AP​

Answer 1

$let \: (a - b) ^{2} be \: term \: a \\ Consider \: all \: of \: them \: to \: be \: in \: AP \\ AP= {(a - b)}^{2} ,({a}^{2} + {b}^{2} ),( {a + b})^{2} \\ Therefore, \\ d = {a}^{2} + {b}^{2} - ( {a - b})^{2} = ( {a + b})^{2} - ( {a}^{2} + {b}^{2} ) \\ d = {a}^{2} + {b}^{2} - {a}^{2} - {b}^{2} + 2ab = {a}^{2} + {b}^{2} + 2ab - {a}^{2} - {b}^{2} \\ d = 2ab = 2ab \\ Since \: d \: is \: equal \\ Therefore, \: the \: given \: series \: is \: in \: AP$