Respuesta :

ANSWER

65

EXPLANATION

This will form a sequence with first term:

[tex]a_{1} = 102[/tex]

with constant difference d=6 and last term

[tex]l = 486[/tex]

The last term is a term in the sequence, therefore;

[tex]l=a_1+d(n-1)[/tex]

[tex]486=102+6(n-1)[/tex]

[tex]486 - 102 = 6(n-1)[/tex]

Simplify the LHS

[tex]384= 6(n-1)[/tex]

Divide both sides by 6.

[tex]64=n-1[/tex]

[tex]64 + 1 = n[/tex]

Therefore

[tex]n = 65[/tex]

Hence there are 65 multiples of 6.

Answer:

67

Step-by-step explanation:

First number between 100 to 500 which is divisible by 6 is 102

Multiples of 6 between 100 to 500 :

102,102+6,102+6+6,....

This Forms an AP

a= first term = 102

d = common difference = 6

Last number between 100 to 500 which is divisible by 6 is 498

So, [tex]a_n=498[/tex]

Formula of nth term = [tex]a_n=a+(n-1)d[/tex]

[tex]498=102+(n-1)6[/tex]

[tex]498-102=(n-1)6[/tex]

[tex]396=(n-1)6[/tex]

[tex]\frac{396}{6}=n-1[/tex]

[tex]66=n-1[/tex]

[tex]66+1=n[/tex]

[tex]67=n[/tex]

Hence there are 67 multiples of 6 between 100 to 500.