Question

Prove that the arithmetic sequence with first term 1/5
and common difference
1/10
contains all natural numbers ?

Answer 1

Step-by-step explanation:

a=1/5

d=1/10

so nth term is = a+(n-1)d

= 1/5 + (n-1).1/10

= 1/5 + n/10 -1/10

= n/10 + 1/10

= (n+1)/10

so the values of n must be 1 less than the multiples of 10..

like 10-1=9,20-1=19 and so on..

putting the value 9 we get (9+1)/10=1

(19+1)/10=2......

so this sequence contains all natural numbers