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 …

Let four terms are \(a−3d,a−d,a+d,a+3d\)

Where \(d=\frac{l}{2}\)

\(∵{a}_{1}+{a}_{2}+{a}_{3}+{a}_{4}=48\)

\(\Rightarrow 4a=48\Rightarrow a=12\)

\({a}_{1}{a}_{2}{a}_{3}{a}_{4}+{l}^{4}=361\)

\(\Rightarrow \left({a}^{2}−9{d}^{2}\right)\left({a}^{2}−{d}^{2}\right)+16{d}^{4}=361\)

\(\Rightarrow \left(144−9{d}^{2}\right)\left(144−{d}^{2}\right)+16{d}^{4}=361\)

\(\Rightarrow 25{d}^{4}−1440{d}^{2}+{(144)}^{2}=361\)

\(\Rightarrow {\left(5{d}^{2}−144\right)}^{2}={19}^{2}\)

\(\Rightarrow 5{d}^{2}−144=19,−19\)

\(\Rightarrow {d}^{2}=\frac{163}{5},25\)

\(\Rightarrow d=\sqrt{\frac{163}{5}},5\)

\(∴l=2\sqrt{\frac{163}{5}},10\)

Common difference is an integer therefore \(l=10\)

Largest term \(= 12 + 15 = 27\)