Question

Which is value of `n` knowing that the 6th term of `(sqrt(2^(log_10(10-3x)))+root(3)(2^((x-2)log_10 3)))^n` is `21`, and `C_n^1,C_n^2,C_n^3` are 1th,3th and 5th terms of an arithmetic progression?

Answer 1

`n=7`

Explanation:

We have

`{(C_n^1=n=a_0+a),(C_n^2=(n(n-1))/(2!)=a_0+3a),(C_n^3 = (n(n-1)(n-2))/(3!)=a_0+5a):}`

Solving for `n,a_0,a` we obtain

`a_0=0,a=7,n=7` the feasible solution. So `n=7`