1.

If a, b, c are in G.P., prove that the following are also in G.P.:(i) a2, b2, c2(ii) a3, b3, c3(iii) a2 + b2, ab + bc, b2 + c2

Answer»

(i) a2, b2, c2

Given that a, b, c are in GP.

By using the property of geometric mean,

b2 = ac

on squaring both the sides 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 are in GP.

By using the property of geometric mean,

b2 = ac

on squaring both the sides 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 are in GP.

By using the property of geometric mean,

b2 = ac

a2 + b2, ab + bc, b2 + c2 or (ab + bc)2 = (a2 + b2) (b2 + c2) [by using 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 {again using 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.



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