$\left(a_n\right)=\left(a_1,a_1q,a_1q^2,...,a_1q^{98},a_1q^{99}\right)$, gdzie $q$ jest ilorazem ciągu geometrycznego $\left(a_n\right)$.

$\begin{split}
&a_1+a_3+\dots+a_{99}=100\cdot \left(a_2+a_4+\dots+a_{100}\right)\\
&a_1+a_1 q^2+\dots+a_1 q^{98}=100\left(a_1q+a_1 q^3+\dots+a_1\cdot q^{99}\right)\\
&a_1\left(1+q^2+\dots+q^{98}\right)=100a_1\left(q+q^3+\dots+q^{99}\right)\Big/:a_1\\
&1\cdot \frac{1-\left(q^2\right)^{50}}{1-q^2}=100\cdot q\cdot\frac{1-\left(q^2\right)^{50}}{1-q^2}\Big/:\frac{1-\left(q^2\right)^{50}}{1-q^2}\\
&1=100q\Big/:100\\
&q=\frac{1}{100}.
\end{split}$

$\begin{split}
&\log a_1+\log a_2+\log a_3+\dots+\log a_{100}=100\\
&\log a_1+\log a_1q+\log a_1q^2+\dots+\log a_1q^{99}=100\\
&\log a_1\cdot a_1q\cdot a_1q^2\dots a_1q^{99}=100\\
&\log a_1^{100}\cdot q^{1+2+...+99}=100\\
&\log a_1^{100}\cdot q^{\frac{1+99}{2}\cdot 99}=100\\
&\log a_1^{100} +\log q^{50\cdot 99}=100\\
&100\log a_1+\log \left(\frac{1}{100}\right)^{4950}=100\\
&100\log a_1+\log 10^{-9900}=100\\
&100\log a_1-9900=100\Big/:100\\
&\log a_1-99=1\\
&\log a_1=100\\
&a_1=10^{100}.
\end{split}$