Find the sum of two-digit natural numbers div | vedclass

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(B) The two-digit natural numbers divisible by $6$ are $12, 18, 24, \dots, 96$.This forms an arithmetic progression where the first term $a = 12$,the

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$810$

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(B) The two-digit natural numbers divisible by $6$ are $12, 18, 24, \dots, 96$.This forms an arithmetic progression where the first term $a = 12$,the common difference $d = 6$,and the last term $l = 96$.To find the number of terms $n$,we use the formula $a_n = a + (n - 1)d$.$96 = 12 + (n - 1)6$$84 = (n - 1)6$$14 = n - 1$$n = 15$.The sum of $n$ terms is given by $S_n = \frac{n}{2}(a + l)$.$S_{15} = \frac{15}{2}(12 + 96)$$S_{15} = \frac{15}{2}(108)$$S_{15} = 15 	imes 54 = 810$.

Find the sum of two-digit natural numbers divisible by $6$.

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