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# How many three digit numbers are divisible by 7?  Verified
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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 $b-a=c-b⇒2b=a+c$$b - a = c - b \Rightarrow 2b = a + c$
Formula to consider for solving these questions
${a}_{n}=a+\left(n-1\right)d$${a_n}=a+ (n− 1)d$
Where d -> common difference
A -> first term
n-> term
${a}_{n}->{n}^{th}$${a_n} -&amp;amp;amp;amp;gt; {n^{th}}$ term

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 ${n}^{th}$${n^{th}}$ term of AP.
We need to find n here.
First term = a = 105, Common difference = d = 112 – 105 = 7
Using formula , to find ${n}^{th}$${n^{th}}$ term of arithmetic progression,
$\begin{array}{*{20}{l}} { \Rightarrow 994{\text{ }} = {\text{ }}105{\text{ }} + {\text{ }}\left( {n{\text{ }} - {\text{ }}1} \right){\text{ }}\left( 7 \right)} \\ { \Rightarrow 994{\text{ }} = {\text{ }}105{\text{ }} + {\text{ }}7n - 7} \\ \Rightarrow 896{\text{ }} = {\text{ }}7n \\ \Rightarrow n{\text{ }} = {\text{ }}128 \\ \end{array}$
It means 994 is the ${128}^{th}$${128^{th}}$ 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)
${t}_{n}={S}_{n}-{S}_{n-1}$${t_n}={S_n}-{S_{n-1}}$
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.