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From the given information, we have that the three numbers $\sqrt{36+k},\sqrt{300+k},\sqrt{596+k}$ form an arithmetic progression, in that order. Therefore, we have $$2\sqrt{300+k}=\sqrt{36+k}+\sqrt{596+k}.$$Squaring both sides of the equation, we get $$4(300+k)=(36+k)+2\sqrt{(36+k)(596+k)}+(596+k)$$or $$568+2k=2\sqrt{(36+k)(596+k)}.$$Dividing by $2$ and then squaring again, we have $$(284+k{)}^2=(36+k)(596+k),$$or $$284^2+2\cdot 284k+k^2=36\cdot 596+632k+k^2.$$Thus, $$k=\frac{284^2-36\cdot 596}{632-2\cdot 284}=\frac{284^2-36\cdot 596}{64}=\boxed{925}.$$