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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`
Posted by justaguide on July 3, 2012 at 12:33 PM (Answer #1)
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