Let 3, 7, 11, 15, ldots, 403 and 2, 5, 8, 11, | vedclass

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(D) The first arithmetic progression is $A_1 = 3, 7, 11, 15, \ldots, 403$ with common difference $d_1 = 4$.The second arithmetic progression is $A_2 =

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$6699$

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(D) The first arithmetic progression is $A_1 = 3, 7, 11, 15, \ldots, 403$ with common difference $d_1 = 4$.The second arithmetic progression is $A_2 = 2, 5, 8, 11, \ldots, 404$ with common difference $d_2 = 3$.The common terms form a new arithmetic progression with a common difference equal to $	ext{LCM}(4, 3) = 12$.The first common term is $11$.Let the common terms be $11, 23, 35, \ldots, L$,where $L \leq 403$.The $n$-th term is given by $a_n = 11 + (n - 1) 	imes 12$.We need $11 + (n - 1) 	imes 12 \leq 403$,which implies $(n - 1) 	imes 12 \leq 392$,so $n - 1 \leq 32.66$.Thus,$n - 1 = 32$,which gives $n = 33$.The last term is $L = 11 + 32 	imes 12 = 11 + 384 = 395$.The sum of these $33$ terms is $S_{33} = \frac{n}{2}(a + L) = \frac{33}{2}(11 + 395) = \frac{33}{2}(406) = 33 	imes 203 = 6699$.

Let $3, 7, 11, 15, \ldots, 403$ and $2, 5, 8, 11, \ldots, 404$ be two arithmetic progressions. Then the sum of the common terms in them is equal to:

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