Question

Find the number of natural numbers between 102&998which are divisible by2&5both​

Answer 1

Step-by-step explanation:

First term is a=110

The common difference is d=10

The last term is l=990

The last term formula is given by,

l=a+(n-1)d

990=110+(n-1)10

990-110=(n-1)10

880=(n-1)10

\frac{880}{10}=n-1

n-1=88

n=88+1

n=89

The total number of natural numbers is 89.

Answer 2

Answer:

Step-by-step explanation:

As the numbers are divisible by both 2 and 5,so their last digit will always be 0. hence the series are 110,120,130,....,990

here a=110,d=10,an=990

an=a+(n-1)d

990=110+(n-1)10

(n-1)10=880

n-1=88

n=88+1

n=89.

thus the number of natural numbers between 102 and 998 which are divisible by both 2 and 5 is 89

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