Let a_1, a_2, a_3, a_4 be an A.P.

Question

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

Accepted Answer

It is given that ${a}_{1},{a}_{2},{a}_{3},{a}_{4}$ are in A.P with common difference $l$.

four terms in A.P are $(a-3d),(a-d),(a+d),(a+3d)$ common difference $l=2d\Rightarrow d=\frac{l}{2}$

$\begin{gathered} {a}_{1}=a-\frac{3l}{2},{a}_{2}=(a-\frac{l}{2}) \ {a}_{3}=a+\frac{l}{2},{a}_{4}=(a+\frac{3l}{2}) \end{gathered}$

It is given that ${a}_{1}+{a}_{2}+{a}_{3}+{a}_{4}=48$

$\begin{gathered} a-\frac{3l}{2}+a-\frac{l}{2}+a+\frac{l}{2}+a+\frac{3l}{2}=48 \ \Rightarrow 4a=48\Rightarrow a=12 \ {a}_{1}{a}_{2}{a}_{3}{a}_{4}+{l}^{4}=361 \ \Rightarrow (12-\frac{3l}{2})(12-\frac{l}{2})(12+\frac{l}{2})(12+\frac{3l}{2})+{l}^{4}=361 \ \Rightarrow (144-\frac{{l}^{2}}{4})(144-\frac{9{l}^{2}}{4})+{l}^{4}=361 \ \Rightarrow 20736-324{l}^{2}-36{l}^{2}+\frac{9}{16}{l}^{4}+{l}^{4}=361 \end{gathered}$

$\Rightarrow$ $\frac{25}{16}{l}^{4}-360{l}^{2}+20375=0$

multiply both side by $\frac{16}{5}$.

$5{l}^{4}-1152{l}^{2}+65200=0$

using quadratic formula

${l}^{2}=\frac{1152\pm \sqrt{(1152{)}^{2}-4	imes 5	imes 65200}}{2	imes 5}$

$\begin{gathered} \Rightarrow {l}^{2}=\frac{1152\pm \sqrt{23104}}{10}=\frac{1152\pm 152}{10} \ \Rightarrow {l}^{2}=\frac{1152+152}{10},\frac{1152-152}{10} \ \Rightarrow {l}^{2}=\frac{1304}{10},\frac{1000}{10} \ \Rightarrow {l}^{2}=\frac{1304}{10},100 \end{gathered}$

Since $l$ is integer so $l=\pm 10$

$\begin{gathered} l=10 \ \begin{gathered} {a}_{1}=12-\frac{3}{2}	imes 10=-3 \ \, {a}_{2}=12-\frac{10}{2}=7 \ \, {a}_{3}=12+\frac{10}{2}=17 \ {a}_{4}=12+\frac{3}{2}	imes 10=27 \end{gathered}$

$\begin{gathered} l=-10 \ {a}_{1}=12+\frac{3}{2}	imes 10=27 \ {a}_{2}=12+\frac{10}{2}=17 \ {a}_{3}=12-\frac{10}{2}=7 \ {a}_{4}=12-\frac{3}{2}	imes 10=-3 \end{gathered} \end{gathered}$

So, numbers are same but in reverse order therefore, largest term of A.P is 27.

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