The sum of the common terms of the following | vedclass

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(B) Let the three arithmetic progressions be $A_1, A_2, A_3$.$A_1: 3, 7, 11, 15, \ldots, 399$ with common difference $d_1 = 4$.$A_2: 2, 5, 8, 11, \ldo

Vedclass · Accepted Answer

$321$

Vedclass · Answer

(B) Let the three arithmetic progressions be $A_1, A_2, A_3$.$A_1: 3, 7, 11, 15, \ldots, 399$ with common difference $d_1 = 4$.$A_2: 2, 5, 8, 11, \ldots, 359$ with common difference $d_2 = 3$.$A_3: 2, 7, 12, 17, \ldots, 197$ with common difference $d_3 = 5$.The common difference of the sequence of common terms is $L = \operatorname{LCM}(4, 3, 5) = 60$.To find the first common term $a$,we check the intersection of the sequences:$A_1: a_n = 3 + (n-1)4 = 4n - 1$$A_2: a_m = 2 + (m-1)3 = 3m - 1$$A_3: a_k = 2 + (k-1)5 = 5k - 3$Equating $A_1$ and $A_2$: $4n - 1 = 3m - 1 \implies 4n = 3m$. The smallest common term is $11$.Now check $11$ in $A_3$: $5k - 3 = 11 \implies 5k = 14$ (No).Next common term for $A_1$ and $A_2$ is $11 + 12 = 23$. Check in $A_3$: $5k - 3 = 23 \implies 5k = 26$ (No).Next is $23 + 12 = 35$. Check in $A_3$: $5k - 3 = 35 \implies 5k = 38$ (No).Next is $35 + 12 = 47$. Check in $A_3$: $5k - 3 = 47 \implies 5k = 50 \implies k = 10$. So,$47$ is the first common term.The common terms are $47, 47+60=107, 107+60=167$.The next term $167+60=227$ exceeds the limit of $A_3$ $(197)$.Sum $= 47 + 107 + 167 = 321$.

The sum of the common terms of the following three arithmetic progressions:
$3, 7, 11, 15, \ldots, 399$
$2, 5, 8, 11, \ldots, 359$ and
$2, 7, 12, 17, \ldots, 197$,is equal to $................$.

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The sum of the common terms of the following three arithmetic progressions:$3, 7, 11, 15, \ldots, 399$$2, 5, 8, 11, \ldots, 359$ and$2, 7, 12, 17, \ldots, 197$,is equal to $................$.