(i) a(b2 + c2 ) = c(a2 + b2 )
To prove: a(b2 + c2 ) = c(a2 + b2 )
Given: a, b, c are in GP
Formula used: When a,b,c are in GP, b2 = ac
When a,b,c are in GP, b2 = ac
Taking LHS = a(b2 + c2 )
= a(ac + c2 ) [b2 = ac]
= (a2c + ac2 )
= c(a2 + ac)
= c(a2 + b2 ) [b2 = ac]
= RHS
Hence Proved
(ii) \(\cfrac{1}{(a^2 - b^2)} + \cfrac{1}{b^2} = \cfrac{1}{(b^2 - c^2)}\)
To prove: a(b2 + c2 ) = c(a2 + b2 )
Given: a, b, c are in GP Formula used:
When a,b,c are in GP, b2 = ac
Proof: When a,b,c are in GP, b2 = ac
Taking LHS = \(\frac{1}{(a^2 - b^2)} + \frac{1}{b^2}\)
Hence Proved
(iii) (a + 2b + 2c)(a – 2b + 2c) = a2 + 4c2
To prove: (a + 2b + 2c)(a – 2b + 2c) = a2 + 4c2
Given: a, b, c are in GP
Formula used: When a,b,c are in GP, b2 = ac
Proof: When a,b,c are in GP, b2 = ac
Taking LHS = (a + 2b + 2c)(a – 2b + 2c)
⇒ [(a + 2c) + 2b] [(a + 2c) – 2b]
⇒ [(a + 2c)2 – (2b)2 ] [(a + b) (a – b) = a2 – b2 ]
⇒ [(a2 + 4ac + 4c2 ) – 4b2 ]
⇒ [(a2 + 4ac + 4c2 ) – 4b2 ] [ b2 = ac]
⇒ [(a2 + 4ac + 4c2 – 4ac]
⇒ a2 + 4c2 = RHS
Hence Proved
(iv) a2b2c2 \(\left(\frac{1}{a^3} + \frac{1}{b^3}+\frac{1}{c^3}\right)\) = a3 + b3 + c3
To prove: a2b2c2 \(\left(\frac{1}{a^3} + \frac{1}{b^3}+\frac{1}{c^3}\right)\) = a3 + b3 + c3
Given: a, b, c are in GP
Formula used: When a,b,c are in GP, b2 = ac
Proof: When a,b,c are in GP, b2 = ac
Hence Proved