If second, third and sixth terms of an A.P. are consecutive terms of a

1.	If second, third and sixth terms of an A.P. are consecutive terms of a G.P., write the common ratio of the G.P.
Answer» Given: Second, third and sixth terms of an A.P. are consecutive terms of a G.P. Let the first term of AP be a and the common difference be d. ⇒ A_n = a+(n-1)d ⇒ A₂ = a+d ⇒ A₃ = a+2d ⇒ A₆ = a+5d If a,b,c are consecutive terms of GP then we can write b² = a.c ∴ We can write (a+2d)² = (a+d).(a+5d) ⇒ a²+4d²+4ad = a²+6ad+5d² ⇒ d²+2ad = 0 ⇒ d(d+2a) = 0 ∴ d = 0 or d = - 2a When d = 0 then the GP becomes a, a, a. ∴ The common ration becomes 1. When d = - 2a then the GP becomes – a, - 3a,- 9a ∴ The common ratio becomes 3.

If second, third and sixth terms of an A.P. are consecutive terms of a G.P., write the common ratio of the G.P.

Answer»

Given: Second, third and sixth terms of an A.P. are consecutive terms of a G.P.

Let the first term of AP be a and the common difference be d.

⇒ A_n = a+(n-1)d

⇒ A₂ = a+d

⇒ A₃ = a+2d

⇒ A₆ = a+5d

If a,b,c are consecutive terms of GP then we can write b² = a.c

∴ We can write (a+2d)² = (a+d).(a+5d)

⇒ a²+4d²+4ad = a²+6ad+5d²

⇒ d²+2ad = 0

⇒ d(d+2a) = 0

∴ d = 0 or d = - 2a

When d = 0 then the GP becomes a, a, a.

∴ The common ration becomes 1.

When d = - 2a then the GP becomes – a, - 3a,- 9a

∴ The common ratio becomes 3.