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.
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.
⇒ An = 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.