Question

Find the natural numbers between 101 and 999 which are divisible by both 2 and 5.


Class 10


Arithmetic Progressions

Answer 1

Hey there !!


➡ Given :-

→ The number should be divisible ( multiples ) by both 2 and 5.


➡ To find :-

→ The natural numbers between 101 and 999 which are divisible by both 2 and 5. [ n ]


➡ Solution :-

→ The number that is divisible by 2 and 5 both , it means that number is the divisible of 10.

→ So, the first number divisible by 10 after 101 = 110.

And, the last number divisible by 10 till 999 = 990.

→ And, the common difference is 10.

▶ So, these numbers are 110, 120, 130, ... , 990.


→ a = 110.

→ d = 10.

→ a$\tiny n$ = 990.

→ n = ?


▶ So, the nth term is given by,

a$\tiny n$ = a + ( n - 1 )d.

=> 990 = 110 + ( n - 1 ) × 10.

=> 990 - 110 = 10n - 10.

=> 880 + 10 = 10n

=> 10n = 890.

=> n = $\frac{890}{10} .$

$\huge \boxed{ \boxed{ \bf => n = 89. }}$


✔✔ Hence, it is solved ✅✅.

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THANKS


#BeBrainly.

Answer 2

$\huge{hey}$


$\huge{ \mathfrak{here \: is \: answer}}$

$\bf{see \: in \: pic}$

$\huge \boxed{\ulcorner{hope \: it \: helps}}$


$\huge{ \mathbb{THANKS}}$