Q. 7

# 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

⇒ A_{2} = a+d

⇒ A_{3} = a+2d

⇒ A_{6} = a+5d

If a,b,c are consecutive terms of GP then we can write b^{2} =a.c

∴ We can write (a+2d)^{2} = (a+d).(a+5d)

⇒ a^{2}+4d^{2}+4ad = a^{2}+6ad+5d^{2}

⇒ d^{2}+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.

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