Q12 of 104 Page 468

If a, b, c are in GP, prove that are in AP.

To prove: are in AP

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 and

⇒

⇒ [ b² = ac ]

⇒

We can see that

Hence we can say that are in AP.

If a, b, c are in GP, prove that

(i) a(b² + c²) = c(a² + b²)

(ii)

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

If a, b, c, d are in GP, prove that

(i) (b + c)(b + d) = (c + a)(c + a)

(ii)

(iii) (a + b + c + d)² = (a + b)² + 2(b + c)² + (c + d)²

If a, b, c are in GP, prove that a², b², c² are in GP.

If a, b, c are in GP, prove that a³, b³, c³ are in GP