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Find the sum of all multiples of 7 lying between 100 and 1000.

Last updated date: 26th Feb 2024
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IVSAT 2024
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Hint: According to the question, we need to find the sum of the multiples of 7 that lie between 100 and 1000. Now, here if we observe we can see that difference between any two consecutive numbers would be the same as 7. Therefore, we get the idea that we need to use the concept of arithmetic progression.

Complete step-by-step solution:
In this question, we want the sum of all multiples of 7 from 100 to 1000 and for this we know that the first multiple of 7 above 100 is 105 and the multiple of 7 which is just before 1000 is 994.
Now, if we can clearly see that since all the multiples of 7 are taken therefore, the common difference between them is going to be the same which is 7.
Therefore, the multiples are 105, 112, …, 994 and this is an arithmetic progression where we have the first and last term and common difference.
Now, the sum of the progression is given by $S=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]$ .
And in order to find the number of multiples we know that the formula is ${{a}_{n}}=a+\left( n-1 \right)d$
Now, substituting the values we get
  & {{a}_{n}}=a+\left( n-1 \right)d \\
 & \Rightarrow 994=105+\left( n-1 \right)7 \\
 & \Rightarrow 889=7n-7 \\
 & \Rightarrow 896=7n \\
 & \Rightarrow n=128 \\
Now, substituting the values in the sum formula we get,
  & S=\dfrac{128}{2}\left[ 2\times 105+\left( 128-1 \right)7 \right] \\
 & \Rightarrow S=64\left[ 210+127\times 7 \right] \\
 & \Rightarrow S=64\left[ 210+889 \right] \\
 & \Rightarrow S=70336 \\
Therefore, the sum of the multiples of 7 from 100 to 1000 is 70336.

Note: In such questions, we must be clear that some or the other concept is used and for that we need to observe the similarity or relation between the numbers amongst each other. Also, we need to be clear with the different kinds of series and progressions in order to solve this.
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