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.


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.


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