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(A) Let the three numbers in $A.P.$ be $a-d, a, a+d$.Given that the sum is $(a-d) + a + (a+d) = 33$.$3a = 33 \Rightarrow a = 11$.Given that the produc

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$-25$

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(A) Let the three numbers in $A.P.$ be $a-d, a, a+d$.Given that the sum is $(a-d) + a + (a+d) = 33$.$3a = 33 \Rightarrow a = 11$.Given that the product is $(a-d)(a)(a+d) = 1155$.$a(a^2 - d^2) = 1155$.$11(121 - d^2) = 1155$.$121 - d^2 = 105$.$d^2 = 16 \Rightarrow d = \pm 4$.Case $1$: If $d = 4$,the first term $A = a-d = 11-4 = 7$. The $11^{th}$ term $T_{11} = A + 10d = 7 + 10(4) = 47$.Case $2$: If $d = -4$,the first term $A = a-d = 11 - (-4) = 15$. The $11^{th}$ term $T_{11} = A + 10d = 15 + 10(-4) = 15 - 40 = -25$.Thus,one possible value for the $11^{th}$ term is $-25$.

If the sum and product of the first three terms in an $A.P.$ are $33$ and $1155$,respectively,then a value of its $11^{th}$ term is

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