Question

plzzz ans....Thnk u....Third part plxz ans

Answer 1

HELLO DEAR,

given that:-

If a,b, c,d are in continued proportion, a/b = b/c = c/d i.e a,b,c and d are in GP
$= > \frac{a}{b} = \frac{b}{c} = \frac{c}{d} \\ = > \frac{a}{b} = \frac{b}{c} \\ = > {b}^{2} = ac.....(1) \\ = >\frac{b}{c} = \frac{c}{d} \\ = > {c}^{2} = bd.....(2) \\ \\ = > \frac{a}{b} = \frac{c}{d} \\ = > bc = ad.....(3)$

=> a × d = b × c

=> ad=bc----------(1)

(a+d) (b+c) - (a+c) (b+d) = (b-c)²

=> a (b+c)+d (b+c) -[ a(b+d)+c(b+d) ]

=> ab+ac+db+dc -[ab+ad+cb+cd]

=> ab + ac + db + dc - ab - ad - cb - cd

=> ab - ab + ac + db + cd - cd - ad - cb

=> ac + bd - ad - bc

from--(1) and--(2) and--(3)

we get,

we know that;-

a²+b²-2ab=(a-b)²

=> b²+ c² -2bc

=> (b-c)²

hence,

R.H.S = L.H.S.


I HOPE ITS HELP YOU DEAR,
THANKS

Answer 2

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

=> a / b = b / c = c / d

=> a / b = c / d

=> ad = bc

and , b / c = c / d

=> bd = c²

and a / b = b / c

=> ac = b²

LHS => ( a + d ) ( b + c ) - ( a + c ) ( b + d )

=> ab + ac + bd +cd - [ ab + ad + bc + cd ]

=> ab + ac + bd + cd - ab - ad - bc - cd

=> ab and - ab & cd and - cd got cancelled out

=> ac + bd - ad - bc

=> b² + c² - bc - bc

( as , ac = b² , bd = c² , ad = bc )

=> b² + c² - 2 bc

=> ( b - c )²


RHS = ( b - c ) ²

=> since , LHS = RHS


hence , PROVED


hope this helps