If first term is 3 and common ratio is 3 then find the 6th term of G.P

Explanation

Let `P(n): tan n alpha gt n tan alpha`
Step I For `n=2, tan 2 alpha gt 2 tan alpha `
`rArr (2 tan alpha)/(1-tan^2alpha)-2tan alpha gt 0`
`rArr 2 tan alpha ((1-(1-tan^2alpha)/(1-tan^2alpha))gt0`
`rArr tan^2alpha.tan 2alpha gt 0 [because 0 lt alpha lt (pi)/(4) "for"n=2]`
`rArr tan 2 alpha gt 0 [because 0 lt 2 alpha lt (pi)/(2)]`
Which is true (`because` in first quadrant , `tan2alpha ` is positive)
Therefore , P(2) is true.
Step II Assume that P(k) is true , then `P(k): tan k alpha gt tan alpha`
Step III For `n=k+1`, we shall prove that `tan(k+1)alpha gt (k+1)tan alpha`
`because tan (k+1)alpha=(tan k alpha +tan alpha)/(1-tan k alpha tan alpha )` .......(i)
When `0 lt alpha lt (pi)/(4k)or 0 lt kalpha lt (pi)/(4)`
i.e., `0 lt tan k alpha lt1`, also `0 lt tan alpha lt 1`
`therefore tan k alpha tan alpha lt1`
`1-tan k alpha tan alpha gt 0 and 1-tan k alpha tan alpha lt 1`......(ii)
From Eqs. (i) and (ii) , we get
`tan(k+1alpha gt (tan kalpha +tanalpha)/(1) gt tan kalpha +tanalpha gt k tan alpha +tan alpha ` [ by assumption step ]
`therefore tan (k+1)alpha gt (k+1)tan alpha`
Therefore , `P(k+1)` is true , Hence by the principle of mathematical induction P(n) is true for all `n in N`.