Question

How many three digit numbers divisible by 7

Answer 1

The first three digit number which is divisible by 7 is 105 and last three digit number which is divisible by 7 is 994.

This is an A.P. in which a = 105, d = 7 and l = 994.

Let the number of terms be n . Then tn = 994.

nth term of A.P = tn = a + (n - 1)d.

⇒ 994 = 105 + (n -1)7.

⇒ 889 = 7(n-1)

⇒ n -1 = 127

∴ n = 128.

∴ There are128 three digit numbers which are divisible by 7.

Answer 2

✌️✌️ hey frnds.......


_______________________________________⬇️⬇️⬇️


solution :- Three digit no. , divisible by 7 are :

105 , 112 , 119,.........................., 994

it is an AP as 112 -105 =119-112 = 7

(i.e. , the common difference is always same )

Here
$a \: = 105 \: and \: d = 7$

Also if there are n no. in the list , then ...


$a_{n} \: = 994$


→
$a + (n - 1)d = 994$

$105 + (n - 1)7 = 994$

$105 + 7n - 7 = 994$

$98 + 7n = 994$


$7n = 994 - 98$

$7n = 896$

$n = \frac{896}{7} = 128$

•°• There are 128 , three digit no. divisible by 7.





I hope it's helpful ☺️✅...