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\(A=\frac{3a}{4.1}+\frac{3a}{7.4}+\frac{3a}{10.7}+\frac{3a}{13.10}+..+\frac{3a}{22.19}+\frac{3a}{25.22}=\frac{48}{25}\)
\(a.\left(\frac{3}{4.1}+\frac{3}{7.4}+\frac{3}{10.7}+\frac{3}{13.10}+..+\frac{3}{22.19}+\frac{3}{25.22}\right)=\frac{48}{25}\)
\(B=\left(\frac{3}{4.1}+\frac{3}{7.4}+\frac{3}{10.7}+\frac{3}{13.10}+..+\frac{3}{22.19}+\frac{3}{25.22}\right)\)
\(B=\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+..+\frac{1}{22}-\frac{1}{25}\)
\(B=\frac{1}{1}-\frac{1}{25}=\frac{24}{25}\)
\(A=a.B=\frac{24a}{25}=\frac{48}{25}\Rightarrow a=2\)
\(\frac{3a}{4}+\frac{3a}{28}+\frac{3a}{70}+...+\frac{3a}{418}+\frac{3a}{550}=\frac{48}{25}\)
\(\Rightarrow a\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{19.22}+\frac{3}{22.25}\right)=\frac{48}{25}\)
\(\Rightarrow a\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{19}-\frac{1}{22}+\frac{1}{22}-\frac{1}{25}\right)=\frac{48}{25}\)
\(\Rightarrow a\left(1-\frac{1}{25}\right)=\frac{48}{25}\)
\(\Rightarrow a.\frac{24}{25}=\frac{48}{25}\)
\(\Rightarrow a=2\)
\(\frac{3a}{4}+\frac{3a}{28}+\frac{3a}{70}+\frac{3a}{130}+...+\frac{3a}{418}+\frac{3a}{550}=\frac{48}{25}\)
\(\Leftrightarrow a\left(\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+\frac{3}{7\cdot10}+\frac{3}{10\cdot13}+...+\frac{3}{19\cdot22}+\frac{3}{22\cdot25}\right)=\frac{48}{25}\)
\(\Leftrightarrow a\left(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+...+\frac{1}{19}-\frac{1}{22}+\frac{1}{22}-\frac{1}{25}\right)=\frac{48}{25}\)
\(\Leftrightarrow a\left(\frac{1}{1}-\frac{1}{25}\right)=\frac{48}{25}\)
\(\Leftrightarrow a\cdot\frac{24}{25}=\frac{48}{25}\)
\(\Leftrightarrow a=2\)