The total spring constant of the system as shown in the figure will be
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$\frac{1}{\mathrm{k}_{\mathrm{eq}}}=\frac{1}{2 \mathrm{k}_{1}}+\frac{1}{\mathrm{k}_{2}}$
$\mathrm{k}_{\mathrm{eq}}=\left[\frac{1}{2 \mathrm{k}_{1}}+\frac{1}{\mathrm{k}_{2}}\right]^{-1}$
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