What number must be added to each of the number 4, 6, 8, 11 in order to get the four numbers in proportion ?
Let x be added to each number.
So, 4 + x, 6 + x, 8 + x and 11 + x must be in proportion.
∴4+x6+x=8+x11+x⇒(4+x)×(11+x)=(8+x)×(6+x)⇒44+4x+11x+x2=48+8x+6x+x2⇒44+15x+x2=48+14x+x2⇒15x−14x=48−44⇒x=4.\therefore \dfrac{4 + x}{6 + x} = \dfrac{8 + x}{11 + x} \\[1em] \Rightarrow (4 + x) \times (11 + x) = (8 + x) \times (6 + x) \\[1em] \Rightarrow 44 + 4x + 11x + x^2 = 48 + 8x + 6x + x^2\\[1em] \Rightarrow 44 + 15x + x^2 = 48 + 14x + x^2 \\[1em] \Rightarrow 15x - 14x = 48 - 44 \\[1em] \Rightarrow x = 4.∴6+x4+x=11+x8+x⇒(4+x)×(11+x)=(8+x)×(6+x)⇒44+4x+11x+x2=48+8x+6x+x2⇒44+15x+x2=48+14x+x2⇒15x−14x=48−44⇒x=4.
Hence, required number is 4.
