Answer
$\Rightarrow[t=39.7]$
Work Step by Step
The rate of the growth $r(t)=k t$ gram/week.
$\therefore$ the mass of the tumor is: $m=\int k t d t=\frac{k t^{2}}{2}+C$
The average mass of the tumor during the second 26 weeks is:
\[
\begin{aligned}
\frac{1}{-26+52} \int_{26}^{52} \frac{k t^{2}}{2}+C d x &=\left.\frac{1}{26}\left(Ct+\frac{k t^{3}}{6}\right)\right|_{26} ^{52} \\
&=\left(26C+\frac{123032}{6} k\right) \frac{1}{26}\\
&=\frac{4732}{6} k+C
\end{aligned}
\]
To find the time, we will set equal the average of the mass of the tumor during the second 26 weeks and the mass of the tumor:
$\frac{k t^{2}}{2}+C=\frac{4732}{6} k+C$
$\Rightarrow \frac{t^{2}}{2} k=\frac{4732}{6} k$
$\Rightarrow 1577.3=\frac{9464}{6}= t^{2}$
$\Rightarrow[t=39.7]$ weeks
