Find fourier series of my signal
First, your period is $T = 1/25s=0.04s$ which means you do not need to take integral of 3 periods for getting fourier series. A plot of the DFT is given below when I took only first $T=0.04s$ is given by 


As you can see, all even multiples of $25\text{Hz}$ is zero, while odd multiples of $25\text{Hz}$ is decreasing with increasing frequency. From the plot $|X[5]|^2 = 0.002914$, while $|X[7]|^2=0.0008386$. Even multiples of $25\text{Hz}$ have very low values ($\lt 10^{-12}$). Taking DFT is not the correct method, so read on.

The above values are expected because fourier series of your even symmetric signal with non-zero DC has all $b_k =0$. For computing $a_k$, your signal $x(t)=2t/T$ for $0 \le t \le T/2$ and $x(t)=-2t/T-2$ for $T/2 \le t \le T$. I am not going into details of how to compute fourier series coefficients but I will give hints
$$
a_0=\frac{2}{T}\int_0^T x(t)dt\\
a_k=\frac{2}{T}\int_0^T x(t)\cos(2\pi kt/T)dt \\
=\frac{2}{T}\int_0^{T/2} 2t/T\cos(2\pi kt/T)dt + \frac{2}{T}\int_{T/2}^T (-2t/T-2)\cos(2\pi kt/T)dt\\
=\frac{2((-1)^k-1)}{\pi^2k^2}\\
a_k=\frac{-4}{\pi^2k^2}\text{  for k odd}\\
a_k=0\text{ for k even}
$$
So as you can see, the even harmonics are all zeros, while odd harmonics decrease by a factor of $1/k^2$. So
$$a_5=\frac{-4}{\pi^2 25}\\
a_7=\frac{-4}{\pi^2 49}
$$