how many three digit number divided by 7
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Answer:
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.
I Hope it will help!
^_^
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.
I Hope it will help!
^_^
Answered by
1
Sol:
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.
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.
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