Let S_n denote the sum of the first 'n' terms of an A.P. If S₄ = 16 and S₆ = –48 then S₁₀ is equal to

$\begin{array}{l}{S}_{4=}\frac{4}{2}(2\mathrm{a}+3\mathrm{d})=16\Rightarrow 2\mathrm{a}+3\mathrm{d}=8\dots \dots .\left(1\right)\\ {\mathrm{S}}_6=\frac{6}{2}(2\mathrm{a}+5\mathrm{d})=-48\Rightarrow 2\mathrm{a}+5\mathrm{d}=-16...\left(2\right)\end{array}$
Solving (1) and (2) we get a = 22 and d = –12
${\mathrm{S}}_{10}=\frac{10}{2}\{2\times 22+(10-1\left)\right(-12\left)\right\}=-320$