What number must be added to each of the numbers 7, 11 and 19 so that the resulting numbers may be in continued proportion?

### Concept & Continued Proportion Three numbers $a, b,$ and $c$ are in continued proportion if the ratio of the first to the second equals the ratio of the second to the third, which means $\frac{a}{b} = \frac{b}{c}$, or $b^2 = ac$. ### Step-by-Step Solution * Let the number to be added be $x$. * The resulting numbers will be $(7 + x)$, $(11 + x)$, and $(19 + x)$. * For these numbers to be in continued proportion, the condition is: $$ \frac{7 + x}{11 + x} = \frac{11 + x}{19 + x} $$ * Cross-multiply to solve for $x$: $$ (7 + x)(19 + x) = (11 + x)^2 $$ * Expand both sides of the equation: $$ 133 + 7x + 19x + x^2 = 121 + 22x + x^2 $$ $$ 133 + 26x + x^2 = 121 + 22x + x^2 $$ * Cancel out $x^2$ from both sides: $$ 133 + 26x = 121 + 22x $$ * Rearrange to solve for $x$: $$ 26x - 22x = 121 - 133 $$ $$ 4x = -12 $$ $$ x = -3 $$ ### Exam Strategy & Shortcut Instead of expanding the algebraic expression, test the options directly into the condition $b^2 = ac$. Test (a) $-3$: The numbers become $4, 8, 16$. Is $8^2 = 4 \times 16$? Yes, $64 = 64$. Testing the first option provides the answer in seconds without any complex algebra. ### Common Pitfall A common error is confusing "continued proportion" with standard proportion of four numbers, or incorrectly expanding the binomial squared $(11 + x)^2$ as just $121 + x^2$. ### Final Answer Therefore, the correct answer is **-3**.