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(C) The even natural numbers divisible by $5$ are multiples of $10$,which are $10, 20, 30, 40, \ldots$This sequence forms an Arithmetic Progression $(

Vedclass · Accepted Answer

$50500$

Vedclass · Answer

(C) The even natural numbers divisible by $5$ are multiples of $10$,which are $10, 20, 30, 40, \ldots$This sequence forms an Arithmetic Progression $(A.P.)$ where the first term $a = 10$ and the common difference $d = 10$.The sum of the first $n$ terms of an $A.P.$ is given by the formula $S_n = \frac{n}{2}[2a + (n - 1)d]$.For $n = 100$,we have:$S_{100} = \frac{100}{2}[2 	imes 10 + (100 - 1) 	imes 10]$$S_{100} = 50[20 + 99 	imes 10]$$S_{100} = 50[20 + 990]$$S_{100} = 50[1010]$$S_{100} = 50500$.

Find the sum of the first hundred even natural numbers divisible by $5$.

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