Question

plzzz ans....Thnk u....plzzz ans EIGHTH PART

Answer 1

This is your answer. Hope it helps.

Answer 2

=> TO PROVE :- abcd ( 1 / a² + 1 / b² + 1 / c² + 1 / d² ) = a² + b² + c² + d²


a , b , c , d are in continued proportion

hence , a / b = b / c = c / d

=> a / b = c / d

=> ad = bc


LHS =>

abcd ( 1 / a² + 1 / b² + 1 / c² + 1 / d² )

=> a × a × d × d ( b²c²d² + a²c²d² + a²b²d² + a²b²c² ) / a²b²c²d²

[ as , ad = bc ]

=> a²d² × ( b²c²d² + a²c²d² + a²b²d² + a²b²c² ) / a²b²c²d²


=> a²d² gets cancelled out from both numerator and denominator

=> ( b²c²d² + a²c²d² + a²b²d² + a²b²c² ) / b²c²

=> ( b²c²d² + b²c²c² + b²c²b² + b²c²a² ) / b²c²


( as , bc = ad

=> ( bc ) ² = ( ad ) ²

=> b²c² = a²d² )


=> taking b²c² common


=> b²c² ( d² + c² + b² + a² ) / b²c²

=> b²c² gets cancelled out


=> d² + c² + b² + a²

=> a² + b² + c² + d²


Since , LHS = RHS


hence , PROVED


hope this helps