Question

If the A.M of two numbers is $10$ and their product is $91,$ then find the numbers.

Answer 1

We have A.M of an A.P$=\frac{a+b}{2}=10$

$\Rightarrow a+b=20$ .......$\left(1\right)$

Product$ab=91$ ........$\left(2\right)$

or $b=\frac{91}{a}$

Put $b=\frac{91}{a}$ in $\left(1\right)$ we get

$a+\frac{91}{a}=20$

$\Rightarrow a^2-20a+91=0$

$\Rightarrow a^2-7a-13a+91=0$

$\Rightarrow a(a-7)-13(a-7)=0$

$\Rightarrow (a-13)(a-7)=0$

$\therefore a=13,7$

Put $a=13$ in $\left(2\right)$

$b=\frac{91}{a}=\frac{91}{13}=7$

Put $a=7$ in $\left(2\right)$

$b=\frac{91}{a}=\frac{91}{7}=13$

$\therefore a=13,b=7$ or $b=13,a=7$ are the two numbers.