arithmetic series

Find the value of $a_2 + a_4 + \ldots + a_{98}$

if $a_1, a_2, a_3, \ldots$ is an arithmetic progression with common difference 1

and $a_1 + a_2 + ... + a_{98} = 99$.

$\begin{array}{|rcll|} \hline a_1 + a_2 + ... + a_{98} &=& 99 \\\\ \left(\dfrac{a_1 + a_{98}}{2}\right)*98 &=& 99 \\ \mathbf{a_1 + a_{98}} &=& \mathbf{\dfrac{ 2*99 }{98} } \\ \hline \end{array}$

$\text{Let $x=a_2 + a_4 + \ldots + a_{98}$}\\ \begin{array}{|rcll|} \hline x &=& a_2 + a_4 + \ldots + a_{98} \\\\\ x &=& \left(\dfrac{a_2 + a_{98}}{2}\right)*\dfrac{98}{2} \quad | \quad a_2 = a_1 +1 \\\\ x &=& \left(\dfrac{1 + a_1 + a_{98}}{2}\right)*\dfrac{98}{2} \\\\ x &=& (1 + a_1 + a_{98})*\dfrac{98}{4} \\\\ x &=& \dfrac{98}{4} + (a_1 + a_{98})*\dfrac{98}{4} \quad | \quad \mathbf{a_1 + a_{98}=\dfrac{ 2*99 }{98} } \\\\ x &=& \dfrac{98}{4} + \dfrac{ 2*99 }{98}*\dfrac{98}{4} \\\\ x &=& \dfrac{98}{4} + \dfrac{ 2*99 }{4} \\\\ x &=& \dfrac{98+2*99}{4} \\\\ \mathbf{x} &=& \mathbf{74} \\ \hline \end{array}$

$a_2 + a_4 + \ldots + a_{98} = \mathbf{74}$