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

Let us denote the three consecutive terms of the required arithmetic progression (A.P.) by $$a-d,\; a,\; a+d$$ where $$a$$ is the middle term and $$d$$ is the common difference. Choosing the terms in this symmetric form keeps the algebra simple because their sum depends only on $$a$$ while their product separates neatly into a factor of $$a$$ and a quadratic expression in $$d$$.

We are told that the sum of these three terms equals 33. Using the definition of sum we write

$$ (a-d)+a+(a+d)=33. $$

On the left-hand side the terms $$-d$$ and $$+d$$ cancel each other, leaving

$$ 3a = 33. $$

Dividing both sides by 3 gives

$$ a = \frac{33}{3}=11. $$

Next we use the information about the product of the same three terms, which is given as 1155. Writing the product explicitly we have

$$ (a-d)\,a\,(a+d)=1155. $$

First we substitute the value $$a=11$$ obtained above:

$$ (11-d)\,11\,(11+d)=1155. $$

Because multiplication is associative and commutative, we can take the constant factor 11 outside and recognise the remaining product as a difference of squares:

$$ 11\bigl[(11-d)(11+d)\bigr]=1155. $$

Using the algebraic identity $$ (x-y)(x+y)=x^{2}-y^{2} $$ with $$x=11$$ and $$y=d$$, we rewrite the bracketed term:

$$ 11\bigl[\,11^{2}-d^{2}\bigr]=1155. $$

We now compute the square of 11:

$$ 11\bigl[121-d^{2}\bigr]=1155. $$

To isolate the bracket we divide both sides by 11:

$$ 121-d^{2} = \frac{1155}{11}=105. $$

Rearranging gives a simple quadratic equation in $$d$$:

$$ d^{2}=121-105=16. $$

Taking square roots yields two possible real values for the common difference:

$$ d = \pm4. $$

So the required A.P. can rise by 4 each step (when $$d=+4$$) or fall by 4 each step (when $$d=-4$$). Both progressions satisfy the given sum and product conditions. We must now find the eleventh term and then see which value matches the options provided.

The general formula for the $$n^{\text{th}}$$ term of an A.P. whose first term is $$A_1$$ and common difference $$d$$ is

$$ A_n = A_1 + (n-1)d. $$

In our symmetric notation the first term is $$A_1 = a-d$$. Therefore the eleventh term is

$$ A_{11} = (a-d) + (11-1)d = (a-d) + 10d. $$

We already know $$a=11$$, so we substitute this value:

$$ A_{11} = 11 - d + 10d = 11 + 9d. $$

Now we consider the two possible values of $$d$$ separately.

Case 1: $$d=+4$$

$$ A_{11} = 11 + 9(4) = 11 + 36 = 47. $$

Case 2: $$d=-4$$

$$ A_{11} = 11 + 9(-4) = 11 - 36 = -25. $$

The question asks for “a value of its 11th term”, and among the four options given <-25> is the only one that appears in our calculation. The positive value <47> does not figure in the list. Therefore the admissible eleventh term consistent with the options is $$-25$$.

Hence, the correct answer is Option A.