Question

Four integers are consecutive terms of an arithmetic sequence. The sum of these four numbers is 24 and the product is 945.Find the largest of these four integers.

Answer 1

When common difference, $d=\sqrt{39}$ then the largest term of these four terms will be $6+3\sqrt{39}$

Let four consecutive terms of AP is,

     $a-3d, a-d, a+d, a+3d$

Since, sum of all four terms is 24

So,   $a-3d+a-d+a+d+a+3d=24\\\\a=6$

And product is 945

So,  $(a-3d)(a+3d)(a-d)(a+d)=945\\\\9d^4-360d^2+351=0\\\\d^4-40d^2+39=0\\\\(d^2-1)(d^2-39)=0\\\\d=1,-1\\d=\sqrt{39}, -\sqrt{39}$

When d= 1, -1 then largest term is 9

When $d=\sqrt{39}, -\sqrt{39}$ then largest term is $6+3\sqrt{39}$

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