Arithmetic Means B3444C

1. **State the problem:** We need to find the value of $n$ when $n$ arithmetic means are inserted between $a$ and $100$, the ratio of the first mean to the last mean is $1:7$, and $a + n = 33$. 2. **Recall the formula for arithmetic means:** If $n$ arithmetic means are inserted between $a$ and $b$, the sequence is $a, A_1, A_2, \ldots, A_n, b$ where the common difference $d = \frac{b - a}{n+1}$. 3. **Express the first and last means:** - First mean: $A_1 = a + d$ - Last mean: $A_n = a + n d$ 4. **Given ratio:** $$\frac{A_1}{A_n} = \frac{1}{7}$$ Substitute $A_1$ and $A_n$: $$\frac{a + d}{a + n d} = \frac{1}{7}$$ 5. **Express $d$ in terms of $a$, $n$, and $b=100$:** $$d = \frac{100 - a}{n+1}$$ 6. **Substitute $d$ into the ratio equation:** $$\frac{a + \frac{100 - a}{n+1}}{a + n \cdot \frac{100 - a}{n+1}} = \frac{1}{7}$$ 7. **Multiply numerator and denominator by $n+1$ to clear fractions:** $$\frac{a(n+1) + 100 - a}{a(n+1) + n(100 - a)} = \frac{1}{7}$$ Simplify numerator: $$a(n+1) + 100 - a = a n + a + 100 - a = a n + 100$$ Simplify denominator: $$a(n+1) + n(100 - a) = a n + a + 100 n - a n = a + 100 n$$ So the ratio becomes: $$\frac{a n + 100}{a + 100 n} = \frac{1}{7}$$ 8. **Cross multiply:** $$7(a n + 100) = 1(a + 100 n)$$ $$7 a n + 700 = a + 100 n$$ 9. **Rearrange terms:** $$7 a n - 100 n = a - 700$$ $$n(7 a - 100) = a - 700$$ 10. **Recall $a + n = 33$, so $a = 33 - n$. Substitute into the equation:** $$n(7(33 - n) - 100) = (33 - n) - 700$$ $$n(231 - 7 n - 100) = 33 - n - 700$$ $$n(131 - 7 n) = -667 - n$$ 11. **Expand left side:** $$131 n - 7 n^2 = -667 - n$$ 12. **Bring all terms to one side:** $$131 n - 7 n^2 + n + 667 = 0$$ $$-7 n^2 + 132 n + 667 = 0$$ Multiply entire equation by $-1$ for simplicity: $$7 n^2 - 132 n - 667 = 0$$ 13. **Solve quadratic equation:** $$n = \frac{132 \pm \sqrt{132^2 - 4 \cdot 7 \cdot (-667)}}{2 \cdot 7}$$ Calculate discriminant: $$132^2 = 17424$$ $$4 \cdot 7 \cdot 667 = 18676$$ $$\Delta = 17424 + 18676 = 36100$$ $$\sqrt{36100} = 190$$ 14. **Calculate roots:** $$n = \frac{132 \pm 190}{14}$$ - For $+$: $$n = \frac{132 + 190}{14} = \frac{322}{14} = 23$$ - For $-$: $$n = \frac{132 - 190}{14} = \frac{-58}{14} = -\frac{29}{7}$$ (not valid since $n$ must be positive) 15. **Final answer:** $$\boxed{23}$$ Thus, the value of $n$ is $23$.