smc

The $40\text{th}$ term of an arithmetic progression with a first term $p$ and a common difference $2p-1$ is $p+39(2p-1)=79p-39$. Therefore, the sum of the first $40$ terms of such a progression is $\frac{40}{2}(79p-39+p)=1600p-780$. We now want to evaluate ${\sum }_{p=1}^{10}(1600p-780)$. $$\sum _{p=1}^{10}(1600p-780)=1600\sum _{p=1}^{10}(p)-\sum _{p=1}^{10}(780)$$ $$=(1600)\left(\frac{10\cdot 11}{2}\right)-(780)(10)=88000-7800=80200,\boxed{80200}.$$