Sequences & Series Question - None

Question

n arithmetic means are inserted between x and 2y and then between 2x and y. If the rth means in each case be equal, then find the ratio x/y.

Accepted Answer

Step 1: Set up the first AP sequence. When n arithmetic means are inserted between x and 2y, we have an AP with (n+2) terms total: x, A₁, A₂, ..., Aₙ, 2y. The common difference is:

$d_1 = \frac{2y - x}{n+1}$

Step 2: Find the rth mean in the first case. The rth mean is the (r+1)th term of this AP:

$A_r^{(1)} = x + r \cdot d_1 = x + r \cdot \frac{2y - x}{n+1}$

Step 3: Set up the second AP sequence. When n arithmetic means are inserted between 2x and y, we have an AP with (n+2) terms: 2x, B₁, B₂, ..., Bₙ, y. The common difference is:

$d_2 = \frac{y - 2x}{n+1}$

Step 4: Find the rth mean in the second case. The rth mean is the (r+1)th term of this AP:

$B_r^{(1)} = 2x + r \cdot d_2 = 2x + r \cdot \frac{y - 2x}{n+1}$

Step 5: Equate the rth means. Given that the rth means are equal:

$x + r \cdot \frac{2y - x}{n+1} = 2x + r \cdot \frac{y - 2x}{n+1}$

Step 6: Simplify the equation. Multiply both sides by (n+1):

$x(n+1) + r(2y - x) = 2x(n+1) + r(y - 2x)$

$x(n+1) + 2ry - rx = 2x(n+1) + ry - 2rx$

$xn + x + 2ry - rx = 2xn + 2x + ry - 2rx$

Step 7: Collect terms. Rearranging:

$xn + x - 2xn - 2x = ry - 2rx - 2ry + rx$

$-xn - x = -ry - rx$

$-x(n + 1) = -r(y + x)$

$x(n + 1) = r(y + x)$

Step 8: Solve for x/y. Dividing both sides by y(n+1):

$\frac{x}{y} = \frac{r(y + x)}{(n+1)y} = \frac{r}{n+1} \cdot \frac{y+x}{y}$

Let $\frac{x}{y} = k$. Then:

$k(n+1) = r(1 + k)$

$kn + k = r + rk$

$kn - rk = r - k$

$k(n - r) = r - k$

$k(n - r + 1) = r$

$k = \frac{r}{n - r + 1}$

∴ Answer: $\dfrac{x}{y} = \dfrac{r}{n-r+1}$

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