The 5th, 8th and 11th terms of a G.P. are a,b and c respectively. Show that b^2=ac.

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The nth term of a geometric progression is `T_n = x*r^(n - 1)`

It is given that `T_5 = x*r^4 = a` , `T_8 = x*r^7 = b` and `T_11 = x*r^10 = c`

`a*c = x*r^4*x*r^10 = x^2*r^(10 + 4) = x^2*r^14` ...(1)

`b^2 = (x*r^7)^2 = x^2*r^14` ...(2)

As the result in (1) and (2) is the same it shows that `b^2 = a*c`

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