Three numbers are in A.P. whose sum is 33 and | vedclass

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Question

(A) Let the three numbers in $A.P.$ be $(a - d), a, (a + d)$.According to the problem,the sum of these numbers is $33$:$(a - d) + a + (a + d) = 33$$3a

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(A) Let the three numbers in $A.P.$ be $(a - d), a, (a + d)$.According to the problem,the sum of these numbers is $33$:$(a - d) + a + (a + d) = 33$$3a = 33$$a = 11$The product of these numbers is $792$:$(a - d) \cdot a \cdot (a + d) = 792$$a(a^2 - d^2) = 792$$11(11^2 - d^2) = 792$$121 - d^2 = 72$$d^2 = 121 - 72 = 49$$d = 7$ (taking the positive value for the sequence).The numbers are $(11 - 7), 11, (11 + 7)$,which are $4, 11, 18$.Thus,the smallest number is $4$.

Three numbers are in $A.P.$ whose sum is $33$ and product is $792$. The smallest number among these is:

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