Question

Prove that the arithmetic sequence with first term 1/3 and common difference 1/6 contain all natural numbers

Answer 1

Answer:

First term(a)=1/3

Common difference (d)=1/6

a=1/3

a2=⅓+1/6=(2+1)/6=3/6=1/2

a3=½+1/6=(3+1)/6=4/6=⅔

the AP is ⅓, ½,⅔......................

No it does not contain any natural number

Answer 2

Answer:

a=1/3

d=1/6

xn=dn+f-d

=1/6n+1/3-1/6

=1/6n+1/6

=(n+1)/6

The value of n should be a multiple of 6-1

(for example:12 is a multiple of 6.The value of n should be 12-1=11)

If n=5

Then,

(n+1)/6=(5+1)/6

=6/6

=1

If n=11

Then,

(n+1)/6=(11+1)/6

=12/6

=2

If n=17

Then,

(n+1)/6=(17+1)/6

=18/6

=3

Hence, The arithmetic sequence with first term 1/3 and common difference 1/6 contains all natural numbers.