Question

plzzzz ans....Thnk u

Answer 1

since , a , b , c , d are in continued proportion

=> a / b = b / c = c / d
=> ac = b²
=> bd = c²
=> ad = bc


LHS = ( 5a + 7b )( 2c - 3d )

=> [ 5b²/c + 7c²/d ] (2c - 3d )

=> [ ( 5b²d + 7c³ ) / cd ] × ( 2c - 3d )

=> 2c ( 5b²d +7c³/ cd ) - 3d ( 5b²d + 7c³/cd )

=> c in denominator got cancelled out and d in denominator got cancelled out

=> 10b²d / d + 14c³ / d - 15b²d / c - 21c³ / c

=> 10b² + 14 (c²)(c) / d - 15(bd)(b) / c - 21c²

=> 10b² + 14 ( bd ) (c) / d - 15 ( c² ) (b) / c - 21c²

=> 10b² + 14bc - 15bc - 21 c²

=> 10b²-bc-21c²


Now , RHS = ( 5c+7d ) ( 2a-3b )

=> 5c [ 2b²/c - 3c²/d ] + 7d [ 2b²/c - 3c²/d ]

=> 5c [ 2b²d - 3c³ / cd ] + 7d [ 2b²d - 3c³ / cd ]

=> c in denominator got cancelled out and d in denominator got cancelled out

=> 10b²d / d - 15c³/d + 14b²d/c - 21c³/c

=> 10b² - 15 (c²)(c) / d + 14 (bd)(b) / c - 21c²

=> 10b² - 15 (bd)(c) / d + 14 c²b/c - 21c³

=> 10b² - 15bc + 14bc - 21c³


=> 10b² - bc - 21c³


since , LHS= RHS = 10b² - bc - 21c³

hence , PROVED

hope this helps