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