Question

If both the sequences x, a1, a2, y and x, b1, b2, z are in a.P. and it is given that \(y > x\) and \(z < x\), then which of the following values can \(\frac{a_1 - a_2}{b_1 - b_2}\)possibly take?

• A. 2
• B. 0
• C. -3
• D. 5
• E. 1

Answer 1

The Correct Option is C

To solve this problem, we need to analyze the sequences provided. Both sequences are in Arithmetic Progression (A.P.). Let's examine the sequences:

1. Sequence: \(x, a_1, a_2, y\)

2. Sequence: \(x, b_1, b_2, z\)

Given that these sequences are in A.P., we know the common difference for sequence 1 is:

\(d_1 = a_1 - x = a_2 - a_1 = y - a_2\)

Similarly, the common difference for sequence 2 is:

\(d_2 = b_1 - x = b_2 - b_1 = z - b_2\)

Given:

• \(y > x\)
• \(z < x\)

We want to find the possible value of \(\frac{a_1 - a_2}{b_1 - b_2}\).

Using the relations of A.P., we substitute:

• In the first sequence:
• In the second sequence:

Thus, we have:

\(\frac{a_1 - a_2}{b_1 - b_2} = \frac{(x + d_1) - (x + 2d_1)}{(x + d_2) - (x + 2d_2)} = \frac{-d_1}{-d_2} = \frac{d_1}{d_2}\)

Given \(y > x\) implies \(d_1 > 0\) and \(z < x\) implies \(d_2 < 0\).

Therefore, \(\frac{d_1}{d_2}\) is a negative number.

Checking the provided options, the only negative value is -3.

Thus, the possible value for \(\frac{a_1 - a_2}{b_1 - b_2}\) is -3.

Answer 2

Given that both sequences \(x, a_1, a_2, y\) and \(x, b_1, b_2, z\) are in arithmetic progression (A.P), it implies that for these sequences the common differences are consistent across their terms. Therefore:

For sequence \(x, a_1, a_2, y\): 

\(\frac{a_2-a_1}{a_1-x} = \frac{y-a_2}{a_2-a_1}\)

For the second sequence \(x, b_1, b_2, z\):

\(\frac{b_2-b_1}{b_1-x} = \frac{z-b_2}{b_2-b_1}\)

These imply:

\(y = x + 3(a_1-x)\)

\(z = x + 3(b_1-x)\)

From the problem, \(y > x\) and \(z < x\), indicating:

1. \(a_1 > x\)

2. \(b_1 < x\)

We calculate the ratio \(\frac{a_1 - a_2}{b_1 - b_2}\):

Since both sequences are A.P, we have:

\(a_2 = \frac{x+y}{2} = 2a_1-x\)

\(b_2 = \frac{x+z}{2} = 2b_1-x\)

The differences become:

\(a_1-a_2 = a_1-(2a_1-x) = x-a_1\)

\(b_1-b_2 = b_1-(2b_1-x) = x-b_1\)

Thus, the desired ratio is:

\(\frac{a_1-a_2}{b_1-b_2} = \frac{x-a_1}{x-b_1}\)

Given \(a_1 > x\) and \(b_1 < x\), both terms are negative, simplifying as:

\(\frac{x - a_1}{x - b_1} = \frac{|x - a_1|}{|x - b_1|}\)

Substitute possible known values from constraints:

Simplifies to \(-3\) based on options provided and constraints \((y > x\), \(z < x)\).

Thus, the value \(\frac{a_1 - a_2}{b_1 - b_2}\) possibly takes is \(-3\).