What number must be added to each of the numbers 5, 11, 19 and 37 so that they are in proportion

Let the number to be added be x. So, new numbers are 5 + x, 11 + x, 19 + x, 37 + x

Since, these numbers are in proportion,

∴5+x11+x=19+x37+x⇒(5+x)(37+x)=(19+x)(11+x)(185+5x+37x+x2)=209+19x+11x+x2⇒x2−x2+42x−30x+185−209=0⇒12x−24=0⇒12x=24⇒x=2412=2.\therefore \dfrac{5 + x}{11 + x} = \dfrac{19 + x}{37 + x} \\[0.5em] \Rightarrow (5 + x)(37 + x) = (19 + x)(11 + x) \\[0.5em] (185 + 5x + 37x + x^2) = 209 + 19x + 11x + x^2 \\[0.5em] \Rightarrow x^2 - x^2 + 42x - 30x + 185 - 209 = 0 \\[0.5em] \Rightarrow 12x - 24 = 0 \\[0.5em] \Rightarrow 12x = 24 \\[0.5em] \Rightarrow x = \dfrac{24}{12} = 2.∴11+x5+x​=37+x19+x​⇒(5+x)(37+x)=(19+x)(11+x)(185+5x+37x+x2)=209+19x+11x+x2⇒x2−x2+42x−30x+185−209=0⇒12x−24=0⇒12x=24⇒x=1224​=2.

Hence, the number that must be added to make the numbers in proportion is 2.