find two numbers whose A.M exceeds their G.M by 30 and their H.M by 48

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$$\begin{gathered} \mathrm{Let} \mathrm{A}.\mathrm{M}. \mathrm{be} \mathrm{A}, \mathrm{G}.\mathrm{M}. \mathrm{be} \mathrm{G} \mathrm{and} \mathrm{H}.\mathrm{M}. \mathrm{be} \mathrm{H} \mathrm{and} \mathrm{two} \mathrm{numbers} \mathrm{be} \mathrm{a} \mathrm{and} \mathrm{b}. \\ \mathrm{Now} \mathrm{given} \mathrm{that} \\ \mathrm{A}-\mathrm{G}=30 \\ \Rightarrow \mathrm{G}=\mathrm{A}-30 \\ \mathrm{A}-\mathrm{H}=48 \\ \Rightarrow \mathrm{H}=\mathrm{A}-48 \\ \mathrm{We} \mathrm{know} \mathrm{that} \\ {\mathrm{G}}^2=\mathrm{AH} \\ \Rightarrow {\left(\mathrm{A}-30\right)}^2=\mathrm{A}\left(\mathrm{A}-48\right) \\ \Rightarrow {\mathrm{A}}^2+900-60\mathrm{A}={\mathrm{A}}^2-48\mathrm{A} \\ \Rightarrow 60\mathrm{A}-48\mathrm{A}=900 \\ \Rightarrow 12\mathrm{A}=900 \\ \Rightarrow \mathrm{A}=\frac{900}{12} \\ \Rightarrow \mathrm{A}=75 \\ \Rightarrow \frac{\mathrm{a}+\mathrm{b}}{2}=75 \\ \Rightarrow \mathrm{a}+\mathrm{b}=150 ;\left(\mathrm{i}\right) \\ \mathrm{G}=\mathrm{A}-30=75-30 \\ \Rightarrow \mathrm{G}=45 \\ \Rightarrow \sqrt{\mathrm{ab}}=45 \\ \Rightarrow \mathrm{ab}=2025 \\ \Rightarrow \mathrm{a}=\frac{2025}{\mathrm{b}} \\ \mathrm{Hence} \left(\mathrm{i}\right) \mathrm{becomes} \\ \frac{2025}{\mathrm{b}}+\mathrm{b}=150 \\ \Rightarrow 2025+{\mathrm{b}}^2=150\mathrm{b} \\ \Rightarrow {\mathrm{b}}^2-150\mathrm{b}+2025=0 \\ \Rightarrow {\mathrm{b}}^2-150\mathrm{b}+2025=0 \\ \Rightarrow {\mathrm{b}}^2-135\mathrm{b}-15\mathrm{b}+2025=0 \\ \Rightarrow \mathrm{b}\left(\mathrm{b}-135\right)-15\left(\mathrm{b}-135\right)=0 \\ \Rightarrow \left(\mathrm{b}-15\right)\left(\mathrm{b}-135\right)=0 \\ \Rightarrow \mathrm{b}=15 \mathrm{and} \mathrm{b}=135 \\ \mathrm{When} \mathrm{b}=15 \\ \mathrm{a}=\frac{2025}{15}=135 \\ \mathrm{When} \mathrm{b}=135 \\ \mathrm{a}=\frac{2025}{135}=15 \\ \mathbf{Hence}\mathbf{ }\mathbf{numbers}\mathbf{ }\mathbf{are}\mathbf{ }\mathbf{15}\mathbf{ }\mathbf{and}\mathbf{ }\mathbf{135}\mathbf{.} \end{gathered}$$

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