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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 of these terms is $51$:$(a -

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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 of these terms is $51$:$(a - d) + a + (a + d) = 51$$3a = 51$$a = 17$The product of the first term $(a - d)$ and the last term $(a + d)$ is $273$:$(a - d)(a + d) = 273$$a^2 - d^2 = 273$Substituting $a = 17$:$17^2 - d^2 = 273$$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 $21, 17, 13$.Both sets represent the same sequence. Thus,the numbers are $21, 17, 13$.

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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