$\text{ Let }{\mathrm{S}}_n\text{ denote the sum of first n-terms of an A.P. If }{\mathrm{S}}_4=16,{\mathrm{S}}_6=-48\text{ then}\left|S_{10}\right|\text{is}$

${\text{S}}_4=2\left[\text{2a+3d}\right]=16 \left(S_n=\frac{n}{2}\left(2a+\left(n-1\right)d\right)\right)$

$\begin{array}{l}\to 2\mathrm{a}+3\mathrm{d}=8--\left(1\right)\\ {\mathrm{S}}_6=3[2\mathrm{a}+5\mathrm{d}]=-48\\ 2\mathrm{a}+5\mathrm{d}=-16---\left(2\right)\\ \left(2\right)-\left(1\right)\Rightarrow \mathrm{d}=-12\\ \Rightarrow 2\mathrm{a}=44\\ \mathrm{a}=22\\ {\mathrm{S}}_{10}=\frac{10}{2}[2\mathrm{a}+9\mathrm{d}]\\ =5[44+9(-12\left)\right]=5(44-108)\\ =-320,\left|{\mathrm{S}}_{10}\right|=320\end{array}$