\(\frac{A}{7}\cdot3=\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+...+\frac{3}{43\cdot46}\)
\(\frac{A}{7}\cdot3=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{43}-\frac{1}{46}\)
\(\frac{A}{7}\cdot3=\frac{45}{46}\)
\(\frac{A}{7}=\frac{15}{46}\)
\(A=\frac{105}{46}\)
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