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in Geometric Progressions by (15.7k points)
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If a, b, c are in GP, prove that 

(i) a(b2 + c2) = c(a2 + b2)

(ii) \(\frac{1}{(a^2 - b^2)} + \frac{1}{b^2} = \frac{1}{(b^2 - c^2)}\)

(iii) (a + 2b + 2c)(a – 2b + 2c) = a2 + 4c2

(iv) a2b2c2 \(\left(\frac{1}{a^3} + \frac{1}{b^3}+\frac{1}{c^3}\right)\) =  a3 + b3 + c3

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(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

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