Mean delay for a M tap harmonically weighted filter I am trying to derive the mean delay for $M$ filter taps harmonically weighted:

$$y[n] = \frac{M\,x[n] + (M-1)x[n−1]... + 1 \cdot x[n−M+1]}{\tfrac12 M (M + 1)}$$.

For a uniformly weighted filter, I used group delay formula and solved it but I am not able to solve this question in a similar way. If I find frequency response $H(e^{j\omega})$ for the above mentioned equation, I end with a arithmetic-geometric progression 

$$H(e^{j\omega})=M+(M-1)e^{-j\omega}+...e^{-j\omega(M-1)}$$.

I don't know whether my approach is right or not. 

Also, if this is a non-linear phase filter, how should I solve it?

Thanks in advance.
The coefficients of this FIR filter do not posses any type of symmetry, hence it cannot be a linear phase filter. However, you can still compute its phase response either analytically (or numerically) as:

$$\phi(\omega) = \tan^{-1}( \frac{ \mathcal{Im} \{ H(e^{j\omega}) \} }{\mathcal{Re} \{ H(e^{j\omega}) \} } ) $$

Then you can compute its derivative (again either analytically or numerically) to obtain the group delay function

$$ \tau(\omega) = -\frac{d\phi(\omega)}{d\omega} $$

from which you can compute the mean delay according to its definition.