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.