If a , b , c are in G.P prove that: 1 a 2 − b 2 + 1 b 2 = 1 b 2 − c 2

Given: a , b , c are in G.P ⇒ b 2 = a c Consider, L . H . S : 1 a 2 − b 2 + 1 b 2 = b 2 + a 2 − b 2 a 2 − b 2 b 2 = a 2 a 2 b 2 − b 4 = a 2 a 2 b 2 − b 2 2 = a 2 a 2 b 2 − ( a c ) 2 using b 2 = a c = a 2 a 2 b 2 − a 2 c 2 = a 2 a 2 b 2 − c 2 = 1 b 2 − c 2 = R . H . S Hence proved.

If a , b , c are in G.P prove that: a 2 b 2 c 2 1 a 3 + 1 b 3 + 1 c 3 = a 3 + b 3 + c 3

Given a , b , c are in G.P ⇒ b 2 = a c Consider, L . H . S : a 2 b 2 c 2 1 a 3 + 1 b 3 + 1 c 3 = a 2 b 2 c 2 a 3 + a 2 b 2 c 2 b 3 + a 2 b 2 c 2 c 3 = b 2 c 2 a + a 2 c 2 b + a 2 b 2 c = ( a c ) c 2 a + b 2 2 b + a 2 ( a c ) c using b 2 = a c = a c 3 a + b 4 b + a 3 c c = a 3 + b 3 + c 3 = R . H . S Hence proved.

If a , b , c are in G.P, prove that 1 ( a + b ) , 1 2 b , 1 ( b + c ) are in A.P.

Given: a , b , c in G.P. Let b = a r and c = a r 2 where r is common ratio. Then, 1 a + b = 1 a + a r = 1 a ( 1 + r ) and 1 2 b = 1 2 a r and 1 ( b + c ) = 1 ar ⁡ ( 1 + r ) ∴ 1 ( a + b ) + 1 ( b + c ) = 1 a ( 1 + r ) + 1 a r ( 1 + r ) = ( 1 + r ) a r ( 1 + r ) = 1 a r = 2 × 1 2 b [multiply and divide by 2 ] Hence, 1 ( a + b ) , 1 2 b , 1 ( b + c ) are in A.P.