Let $$a_{1},a_{2},a_{3},a_{4}$$ be an A.P. of four terms such that each term of the A.P. and its common difference $$l$$ are integers. If $$a_{1} +a_{2}+a_{3}+a_{4}= 48$$ and $$a_{1} a_{2}a_{3}a_{4} + l^{4} = 361,$$ then the largest term of the A.P. is equal to

We are to determine the largest term of a four‐term arithmetic progression when the sum of its terms satisfies $$a_1 + a_2 + a_3 + a_4 = 48$$ and the product condition $$a_1 a_2 a_3 a_4 + l^4 = 361$$ holds, where $$l$$ denotes the common difference.

Denote the four terms by $$a,\,a+l,\,a+2l,\,a+3l$$. Their sum gives $$4a + 6l = 48$$, which simplifies to $$2a + 3l = 24$$ and hence $$a = \frac{24 - 3l}{2}\,. $$

The product condition can be written as $$a(a+l)(a+2l)(a+3l) + l^4 = 361\,. $$ Notice that $$a(a+3l)=a^2 + 3al\quad\text{and}\quad (a+l)(a+2l)=a^2 + 3al +2l^2\,, $$ so if we set $$u = a^2 + 3al$$, then the product becomes $$u\,(u + 2l^2)$$ and $$a_1 a_2 a_3 a_4 + l^4 = u^2 + 2u l^2 + l^4 = (u + l^2)^2 = 361 = 19^2\,. $$ It follows that $$u + l^2 = \pm 19$$, or $$a^2 + 3al + l^2 = \pm 19\,. $$

Substituting $$a = \frac{24 - 3l}{2}$$ into the equation $$\left(\frac{24-3l}{2}\right)^2 + 3l\left(\frac{24-3l}{2}\right) + l^2 = \pm 19$$ and multiplying by 4 yields $$(24 - 3l)^2 + 6l(24 - 3l) + 4l^2 = \pm 76\,. $$ Expanding gives $$576 - 144l + 9l^2 + 144l - 18l^2 + 4l^2 = \pm 76\,, $$ which simplifies to $$576 - 5l^2 = \pm 76\,. $$

In the case $$576 - 5l^2 = 76$$ one finds $$5l^2 = 500\implies l^2=100\implies l=\pm10\,. $$ If $$l=10$$ then $$a = \frac{24 - 30}{2} = -3$$ and the terms are $$-3,\,7,\,17,\,27$$, so the largest term is $$27$$. If $$l=-10$$ then $$a = \frac{24 + 30}{2} = 27$$ and the terms are $$27,\,17,\,7,\,-3$$, again giving the largest term as $$27$$. In the alternative case $$576 - 5l^2 = -76$$ one obtains $$l^2 = 130.4$$, which is not an integer and thus inadmissible.

Therefore, the largest term is Option 2: 27.