Question

The sum of the squares of three positive numbers that are consecutive multiples of 5 is 725. Find the three numbers.​

Answer 1

Given:

The sum of the squares of three positive numbers that are consecutive     multiples of 5 is 725.

To Find:

Find the three numbers.​

Step-by-step explanation:

• Let three consecutive numbers which are multiple of 5 are

              5x, 5(x+1) , 5(x+2)

• Now it is given that sum of there square is 725.

           $(5x)^2+(5(x+1))^2+(5(x+2))^2=725$

• Use the fomula of whole square to solve this equation

          [tex]25x^2+25(x^2+1+2x)+25(x^2+4+4x)=725\\\\ 25x^2+25x^2+25+50x+25x^2+100+100x=725\\\\ 75x^2+150x+125=725\\\\ 75x^2+150x=600\\\\ 75x^2+150x-600=0\\\\ x^2+2x-8=0[/tex]

• Now factorise the equation to get value of x

            [tex]x^2+2x-8=0\\\\ x^2+4x-2x-8=0\\\\ x(x+4)-2(x+4)=0\\\\ (x+4)(x-2)=0\\\\ THUS x=-4,x=2[/tex]

Since it is given number are positive hence x= 2

Thus consecutive number will be 10,15,20.

Answer 2

$\Huge\textrm{10, 15, and 20.}$

$\Large\textrm{Arithmetic Sequence}$

"The set of numbers that increases or decreases constantly is called the arithmetic sequence."

$\large\textrm{ \bigstar General form of A.P. \bigstar }$

$\cdots\longrightarrow\boxed{a,\ a+d,\ a+2d,\ \cdots}$

$\Large\textrm{General form of A.P.: -}$

$\textrm{Given that all of the numbers are positive}$

$\cdots\longrightarrow\boxed{5a-5,\ 5a,\ 5a+5\ (a>1)}$

$\Large\textrm{Then: -}$

$(5a-5)^{2}+(5a)^{2}+(5a+5)^{2}=725$

$\small\text{ 25\times\left\{(a^{2}-2a+1)+a^{2}+(a^{2}+2a+1)\right\}=725 }$

$3a^{2}+2=29$

$a^{2}=9,\ \boxed{a>1}$

$a=3$

$\therefore 5a-5=10, 5a=15, 5a+5=20$

Hence, the three multiples of 5 are 10, 15, and 20.

$\Large\textrm{Verification}$

$\text{ 10^{2}=100 , 15^{2}=225 , 20^{2}=400 }$

$10^{2}+15^{2}+20^{2}=725$

$\textrm{Hence all three numbers are verified.}$