Sequences & Series Question - None

Question

For Problems 25–27: Two arithmetic progressions have the same numbers. The ratio of the last term of the first progression to the first term of the second progression is equal to the ratio of the last term of the second progression to the first term of the first progression and is equal to 4. The ratio of the sum of the $n$ terms of the first progression to the sum of the $n$ terms of the second progression is equal to 2.

The ratio of their common difference is

Accepted Answer

Step 1: Set up the progressions.

Let the first AP have first term $a_1$, common difference $d_1$, and $n$ terms. Last term: $l_1 = a_1 + (n-1)d_1$

Let the second AP have first term $a_2$, common difference $d_2$, and $n$ terms. Last term: $l_2 = a_2 + (n-1)d_2$

Step 2: Apply the ratio condition.

Given: $\frac{l_1}{a_2} = \frac{l_2}{a_1} = 4$

This gives us:

$l_1 = 4a_2$ ... (1)

$l_2 = 4a_1$ ... (2)

Step 3: Apply the sum ratio condition.

Sum of $n$ terms: $S_n = \frac{n}{2}(a + l) = \frac{n}{2}(2a + (n-1)d)$

Given: $\frac{S_1}{S_2} = 2$

$\frac{\frac{n}{2}(a_1 + l_1)}{\frac{n}{2}(a_2 + l_2)} = 2$

$\frac{a_1 + l_1}{a_2 + l_2} = 2$ ... (3)

Step 4: Substitute from equations (1) and (2) into (3).

$\frac{a_1 + 4a_2}{a_2 + 4a_1} = 2$

$a_1 + 4a_2 = 2(a_2 + 4a_1)$

$a_1 + 4a_2 = 2a_2 + 8a_1$

$2a_2 = 7a_1$

$a_2 = \frac{7a_1}{2}$ ... (4)

Step 5: Express the last terms using the AP formula.

From (1): $a_1 + (n-1)d_1 = 4 \cdot \frac{7a_1}{2} = 14a_1$

$(n-1)d_1 = 13a_1$ ... (5)

From (2): $\frac{7a_1}{2} + (n-1)d_2 = 4a_1$

$(n-1)d_2 = 4a_1 - \frac{7a_1}{2} = \frac{a_1}{2}$ ... (6)

Step 6: Find the ratio of common differences.

$\frac{d_1}{d_2} = \frac{\frac{13a_1}{n-1}}{\frac{a_1}{2(n-1)}} = \frac{13a_1}{n-1} \cdot \frac{2(n-1)}{a_1} = 13 \cdot 2 = 26$

∴ Answer: A

Question:

Step-by-Step Solution

Master Sequences & Series with Mathbee