For the A.P. let $a$ be the first term and $d$ be the common difference

Given $T_{p+1}=2T_{q+1}$
$\Rightarrow a+pd=2\left(a+qd\right)\Rightarrow a=\left(p-2q\right)d$
Also,
$T_{3p+1}=\lambda T_{p+q+1}$
$a+3pd=\lambda \left(a+\left(p+q\right)d\right)$
$a\left(\lambda -1\right)=d\left(3p-\lambda p-\lambda q\right)$
So, $\left(p-2q\right)d=\frac{\left(\lambda p+\lambda q-3p\right)d}{1-\lambda }$
$\Rightarrow p-p\lambda -2q+2q\lambda =\lambda p+\lambda q-3p$
$\Rightarrow \lambda \left(p+q+p-2q\right)=p-2q+3p$
$\Rightarrow \lambda =\frac{4p-2q}{2p-q}=\frac{2\left(2p-q\right)}{2p-q}=2$