Question

Find the sum of first all integers between 100 and 1000 which are divisible by 7.

Answer 1

Hint: Find the first number that is divisible by 7 between 100 and 1000. Now, form an A.P. with the first term being the first number we have found and a common difference as 7. Similarly, find the last term divisible by 7 between 100 and 1000. Find the number of terms in the A.P. by using the formula: - \[{{T}_{n}}=a+\left( n-1 \right)d\], where ‘a’ is the first term, ‘\[{{T}_{n}}\]’ is the last term, ‘d’ is a common difference and ‘n’ is the number of terms. Once ‘n’ is found, apply the formula: - \[{{S}_{n}}=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]\] to find the sum. Here, ‘\[{{S}_{n}}\]’ is the sum of ‘n’ terms.

Complete step by step answer:
Here, we have been asked to find the sum of all the numbers between 100 and 1000 that are divisible by 7.
Now, first, we need to determine the first number between 100 and 1000 that will be divisible by 7. So, we have,
\[\begin{align}
  & \Rightarrow 100=7\times 14+2 \\
 & \Rightarrow \left( 100-2 \right)=7\times 14 \\
 & \Rightarrow 98=7\times 14 \\
\end{align}\]
So, we can see that 98 is the number just below 100 which is divisible by 7. So, the next number that will be divisible by 7 will be 98 + 7 = 105. Hence, the first number between 100 and 1000 divisible by 7 is 105. Similarly, considering a difference of 7 we get that: - 112, 119, 126, 133, …. and so on will be the successive numbers divisible by 7.
Now, let us determine the last number between 100 and 1000 that will be divisible by 7. So, we have,
\[\begin{align}
  & \Rightarrow 1000=7\times 142+6 \\
 & \Rightarrow \left( 1000-6 \right)=7\times 142 \\
 & \Rightarrow 994=7\times 142 \\
\end{align}\]
Therefore, 994 will be the last number that will be divisible by 7. So, the terms between 100 and 1000 divisible by 7 can be given as: - 105, 112, 119, 126, ….., 994.
Now, clearly we can see that these terms are forming an A.P. with first term (a) as 105 and last term (\[{{T}_{n}}\]) as 994. So, applying the formula for \[{{n}^{th}}\] term of an A.P. given as: - \[{{T}_{n}}=a+\left( n-1 \right)d\], we have,
\[\Rightarrow {{T}_{n}}=a+\left( n-1 \right)d\]
Here, a = first term = 105
d = common difference = 7
n = number of terms
\[{{T}_{n}}\] = last term = 994
\[\begin{align}
  & \Rightarrow 994=105+\left( n-1 \right)7 \\
 & \Rightarrow 7\left( n-1 \right)=889 \\
 & \Rightarrow \left( n-1 \right)=127 \\
 & \Rightarrow n=128 \\
\end{align}\]
Therefore, the total number of terms between 100 and 1000 that are divisible by 7 is 128.
Now, applying the formula of sum of ‘n’ terms of A.P., we get,
\[\Rightarrow {{S}_{n}}=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]\], where \[{{S}_{n}}\] denotes the sum of ‘n’ terms.
\[\begin{align}
  & \Rightarrow {{S}_{128}}=\dfrac{128}{2}\left[ 2\times 105+\left( 128-1 \right)\times 7 \right] \\
 & \Rightarrow {{S}_{128}}=64\left[ 210+127\times 7 \right] \\
 & \Rightarrow {{S}_{128}}=64\left[ 210+889 \right] \\
 & \Rightarrow {{S}_{128}}=64\times 1099 \\
 & \Rightarrow {{S}_{128}}=70336 \\
\end{align}\]
Hence, our answer is 70336.

Note:
 One may note that is not possible for us to consider each individual term and take their sum one – by – one. This is the reason we needed to form a series in A.P. and apply the formula for the sum of ‘n’ terms. You may see that while finding the last term that is divisible by 7 we have not added 7 to 994, this is because we needed the number below 1000. You must remember the formulas of \[{{n}^{th}}\] term and the sum of ‘n’ terms to solve the question.