Q. 7

# If second, third

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)

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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