If S=a ∈ R:|2 a-1|=3[a]+2a, where [t] denotes the greatest integer less than or equal to t and t represents the fractional part of t, then 72 Σ_a ∈ S…

$\begin{aligned} & |2 \mathrm{a}-1|=3[\mathrm{a}]+2\{\mathrm{a}\} \\ & |2 \mathrm{a}-1|=[\mathrm{a}]+2 \mathrm{a}\end{aligned}$​∣2a−1∣=3[a]+2{a}∣2a−1∣=[a]+2a​ $\displaystyle \begin{aligned} & \text { Case-1 }: \mathrm{a}>\frac{1}{2} \\ & 2 a-1=[a]+2 a \\ & {[a]=-1 \quad \therefore a \in[-1,0) \text { Reject }} \\ & \text { Case-2 }: \mathrm{a} < \frac{1}{2} \\ & -2 \mathrm{a}+1=[\mathrm{a}]+2 \mathrm{a} \\ & \mathrm{a}=\mathrm{I}+\mathrm{f} \\ & -2(\mathrm{I}+\mathrm{f})+1=\mathrm{I}+2 \mathrm{I}+2 \mathrm{f} \\ & \mathrm{I}=0, \mathrm{f}=\frac{1}{4} \quad \therefore \mathrm{a}=\frac{1}{4} \\ & \text { Hence } \mathrm{a}=\frac{1}{4} \\ & 72 \sum_{\mathrm{a} \in \mathrm{S}} \mathrm{a}=72 \times \frac{1}{4}=18\end{aligned}$​ Case-1 :a>21​2a−1=[a]+2a[a]=−1∴a∈[−1,0) Reject  Case-2 :a<21​−2a+1=[a]+2aa=I+f−2(I+f)+1=I+2I+2fI=0,f=41​∴a=41​ Hence a=41​72a∈S∑​a=72×41​=18​