Answer:
4,5,6 are the three consecutive numbers. 16, 25 and 36 are their squares.
Step-by-step explanation:
Let the three consecutive numbers be x, (x+1), (x+2)
Now, the squares of these three numbers are [tex]x^{2} ,(x+1)^{2} ,(x+2)^{2}[/tex]
Sum = 77
∴by the problem ,
[tex]x^{2} +(x+1)^{2}+(x+2)^{2} = 77\\x^{2} +(x^{2} +2x+1)+(x^{2}+4x +4) = 77\\x^{2} +x^{2} +2x+1+x^{2} +4x +4 = 77\\3x^{2} +6x+5 = 77\\3x^{2} +6x = 77-5\\3x^{2} + 6x = 72\\3x^{2} +6x-72= 0\\[/tex]
{Taking 3 common }
[tex]x^{2} +2x- 24 = 0\\[/tex]
{By factorization}
[tex]x^{2} +2x- 24 =0\\ x^{2} +6x-4x-24=0\\x(x+6)-4(x+6)=0\\(x+6)(x-4)=0\\[/tex]
Therfore,
[tex]x= -6,4\\[/tex]
X can't be negetive
∴ [tex]x =4\\x+1=5\\x+2=6[/tex]
The squares of the three consecutive numbers are 16, 25, 36
The three consecutive numbers whose sum is 77 are 4, 5, 6
