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Question

(A) Let the three consecutive terms of an $A.P.$ be $(a - d)$,$a$,and $(a + d)$.According to the given condition,the sum is $(a - d) + a + (a + d) = 5

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$21, 17, 13$

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(A) Let the three consecutive terms of an $A.P.$ be $(a - d)$,$a$,and $(a + d)$.According to the given condition,the sum is $(a - d) + a + (a + d) = 51$.$3a = 51 \Rightarrow a = 17$.The product of the first and last term is $(a - d)(a + d) = 273$.$a^2 - d^2 = 273$.Substituting $a = 17$,we get $17^2 - d^2 = 273$.$289 - d^2 = 273$.$d^2 = 289 - 273 = 16$.$d = \pm 4$.If $d = 4$,the terms are $(17 - 4), 17, (17 + 4)$,which are $13, 17, 21$.If $d = -4$,the terms are $(17 - (-4)), 17, (17 + (-4))$,which are $21, 17, 13$.Thus,the numbers are $21, 17, 13$ or $13, 17, 21$.

If the sum of three consecutive terms of an $A.P.$ is $51$ and the product of the last and first term is $273$,then the numbers are

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