How many three digit numbers are divisible by 7?

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Hint: An arithmetic progression can be given by a, (a+d), (a+2d), (a+3d), ……
a, (a+d), (a+2d), (a+3d),….. where a = first term, d = common difference.
a,b,c are said to be in AP if the common difference between any two consecutive number of the series is same ie ba=cb2b=a+c
Formula to consider for solving these questions
Where d -> common difference
A -> first term
n-> term
an>nth term

Complete step-by-step answer:
We have AP starting from 105 because it is the first three digit number divisible by 7.
AP will end at 994 because it is the last three digit number divisible by 7.
Therefore, we have AP of the form 105, 112, 119…, 994
994 is the nth term of AP.
We need to find n here.
First term = a = 105, Common difference = d = 112 – 105 = 7
Using formula , to find nth term of arithmetic progression,
994 = 105 + (n  1) (7)994 = 105 + 7n7896 = 7nn = 128
It means 994 is the 128th term of AP.
Therefore, there are 128 terms in AP.
Hence, 128 three digit numbers are divisible by 7.

Note: To solve most of the problems related to AP, the terms can be conveniently taken as
3 terms: (a−d),a,(a+d)
4 terms: (a−3d),(a−d),(a+d),(a+3d)
5 terms: (a−2d),(a−d),a,(a+d),(a+2d)
If each term of an AP is increased, decreased, multiplied or divided by the same non-zero constant, the resulting sequence also will be in AP.
In an AP, the sum of terms equidistant from beginning and end will be constant.