The sum of the first three terms of an AP is | vedclass

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(A) Let the three terms in $AP$ be $a-d, a, a+d$.According to the problem,the sum of these terms is $33$:$(a-d) + a + (a+d) = 33$$3a = 33$$a = 11$Also

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$2, 11, 20, \dots$ and $20, 11, 2, \dots$

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(A) Let the three terms in $AP$ be $a-d, a, a+d$.According to the problem,the sum of these terms is $33$:$(a-d) + a + (a+d) = 33$$3a = 33$$a = 11$Also,the product of the first and the third term exceeds the second term by $29$:$(a-d)(a+d) = a + 29$$a^2 - d^2 = a + 29$Substituting $a = 11$:$11^2 - d^2 = 11 + 29$$121 - d^2 = 40$$d^2 = 121 - 40 = 81$$d = \pm 9$Case $1$: If $d = 9$,the terms are $11-9, 11, 11+9$,which are $2, 11, 20$.Case $2$: If $d = -9$,the terms are $11-(-9), 11, 11+(-9)$,which are $20, 11, 2$.Thus,the $AP$ is $2, 11, 20, \dots$ or $20, 11, 2, \dots$.

The sum of the first three terms of an $AP$ is $33$. If the product of the first and the third term exceeds the second term by $29$,find the $AP$.

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