Question 10 Exercise 4.4

Solutions of Question 10 of Exercise 4.4 of Unit 04: Sequence and Series. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.

Find two numbers if the difference between them is $48$ and their A.M exceeds their G.M by $18$ .

Let the two numbers be $a$ and $b$
The difference between them is $48$
Therefore, $$\quad a-b=48....(i)$$. The geometric mean between $a$ and $b$ is $$G=\sqrt{a b}$$ The arithmetic mean between $a$ and $b$ is $$A=\dfrac{a+b}{2}$$ Condition-$2$
Their A.M exceeds their G.M by 18.
Therefore $A \cdot M=G \cdot M+18$ or $A \cdot M-G \cdot M=18$
$$\Rightarrow \dfrac{a+b}{2}-\sqrt{a b}=18$$ Multiplying both sides by 2
$$(a+b)-2 \sqrt{a b}=36 \text {. }$$ From (i), we have $a=b+48$ pulting in (ii), then we have
\begin{align}(b+48+b)-2 \sqrt{b(b+48)}&=36 \\ \Rightarrow(2 b+48)-2 \sqrt{b(b+48)}&=36 \\ \Rightarrow(b+24)-\sqrt{b(b-48)}&=18 \\ \Rightarrow b-\sqrt{b(b+48)}&=18-24=-6 \\ \Rightarrow-\sqrt{b(b+48)}&=-b-6 \\ \Rightarrow \sqrt{b(b+48)}&=b+6.\end{align} Taking square of the both sides
\begin{align}b(b+48)&=(b+6)^2 \\ \Rightarrow b^2+48 b&=b^2+12 b+36 \\ \Rightarrow b^2+48 b-b^2-12 b&=36 \\ \Rightarrow 36 b&=36 \text { or } b=1\end{align} Putting $b=1$ in (i), we get
\begin{align}a-1&=48\\ \Rightarrow a&=49\end{align}
Hence the two numbers are $a=49$ and $b=1$.