Question -
Answer -
(i)┬аa2, b2,c2
Given that a, b, c arein GP.
By using the propertyof geometric mean,
b2┬а=ac
on squaring both thesides we get,
(b2)2┬а=(ac)2
(b2)2┬а=a2c2
тИ┤ a2, b2,c2┬аare in G.P.
(ii)┬аa3, b3,c3
Given that a, b, c arein GP.
By using the propertyof geometric mean,
b2┬а=ac
on squaring both thesides we get,
(b2)3┬а=(ac)3
(b2)3┬а=a3c3
(b3)2┬а=a3c3
тИ┤ a3, b3,c3┬аare in G.P.
(iii)┬аa2┬а+ b2,ab + bc, b2┬а+ c2
Given that a, b, c arein GP.
By using the propertyof geometric mean,
b2┬а=ac
a2┬а+ b2,ab + bc, b2┬а+ c2┬аor (ab + bc)2┬а=(a2┬а+ b2) (b2┬а+ c2) [byusing the property of GM]
Let us consider LHS:(ab + bc)2
Upon expansion we get,
(ab + bc)2┬а=a2b2┬а+ 2ab2c + b2c2
= a2b2┬а+2b2(b2) + b2c2┬а[Since, ac = b2]
= a2b2┬а+2b4┬а+ b2c2
= a2b2┬а+b4┬а+ a2c2┬а+ b2c2┬а{againusing b2┬а= ac }
= b2(b2┬а+a2) + c2(a2┬а+ b2)
= (a2┬а+b2)(b2┬а+ c2)
= RHS
тИ┤┬аLHS = RHS
Hence a2┬а+b2, ab + bc, b2┬а+ c2┬аare in GP.