Find the sum of the numbers between 500 and 7 | vedclass

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(D) The least number which is divisible by $6, 8$ and $12$ is the $LCM$ of $6, 8$ and $12$,which is $24$.We need to find the numbers between $500$ and

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$5400$

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(D) The least number which is divisible by $6, 8$ and $12$ is the $LCM$ of $6, 8$ and $12$,which is $24$.We need to find the numbers between $500$ and $700$ that are multiples of $24$.Dividing $500$ by $24$: $500 = 24 	imes 20 + 20$. The first multiple of $24$ greater than $500$ is $24 	imes 21 = 504$.Dividing $700$ by $24$: $700 = 24 	imes 29 + 4$. The last multiple of $24$ less than $700$ is $24 	imes 29 = 696$.These numbers form an arithmetic progression: $504, 528, 552, \dots, 696$.Here,the first term $a = 504$,the last term $l = 696$,and the common difference $d = 24$.The number of terms $n$ is given by $l = a + (n - 1)d$,so $696 = 504 + (n - 1)24$.$192 = (n - 1)24 \implies n - 1 = 8 \implies n = 9$.The sum $S_n = \frac{n}{2}(a + l) = \frac{9}{2}(504 + 696) = \frac{9}{2}(1200) = 9 	imes 600 = 5400$.

Find the sum of the numbers between $500$ and $700$ such that when they are divided by $6, 8$ and $12$ it leaves no remainder.

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