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

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 \(-3\).