Question

If \(a_1, a_2, a_3\) ….. and \(b_1, b_2, b_3\) ….. are A.P., and \(a_1 = 2, a_{10} = 3, a_1b_1 = 1 = a_{10}b_{10}\), then \(a_4b_4\) is equal to

• A. \(\frac{35}{27}\)
• B. 1
• C. \(\frac{27}{28}\)
• D. \(\frac{28}{27}\)

Answer 1

The Correct Option is D

To solve the given problem, we need to find the value of \(a_4b_4\) where \(a_n\) and \(b_n\) are terms of two arithmetic progressions (A.P.) with given conditions.

First, let's consider the sequence \(a_n\):

1. We know \(a_1 = 2\) and \(a_{10} = 3\).
2. The nth term of an A.P. is given by the formula: \(a_n = a_1 + (n-1)d\).
3. Substitute \(n = 10\): \(a_{10} = a_1 + 9d = 2 + 9d = 3\).
4. This gives: \(9d = 1 \Rightarrow d = \frac{1}{9}\).
5. Thus, the general term of the sequence is: \(a_n = 2 + (n-1)\times\frac{1}{9}\).

Now, let's compute \(b_4\):

1. The condition given is \(a_1b_1 = 1\) and \(a_{10}b_{10} = 1\).
2. Substitute: \(\Rightarrow 2b_1 = 1 \Rightarrow b_1 = \frac{1}{2}\).
3. Substitute for \(a_{10}\): \(3b_{10} = 1 \Rightarrow b_{10} = \frac{1}{3}\).

Since \(b_n\) is also an A.P., the nth term can be written as: \(b_n = b_1 + (n-1)d_b\)

4. Substitute into the equation for \(b_n\): \(b_{10} = b_1 + 9d_b = \frac{1}{2} + 9d_b = \frac{1}{3}\).
5. This gives: \(9d_b = \frac{1}{3} - \frac{1}{2} = -\frac{1}{6}\).
6. Simplifying: \(\Rightarrow d_b = -\frac{1}{54}\).
7. General term for \(b_n\) is: \(b_n = \frac{1}{2} + (n-1)\left(-\frac{1}{54}\right)\).

Next, let's find \(a_4b_4\):

1. Find \(a_4\): \[ a_4 = 2 + (4-1)\times\frac{1}{9} = 2 + \frac{1}{3} = \frac{7}{3} \].
2. Find \(b_4\): \[ b_4 = \frac{1}{2} + 3\times\left(-\frac{1}{54}\right) = \frac{1}{2} - \frac{1}{18} = \frac{9}{18} - \frac{1}{18} = \frac{8}{18} = \frac{4}{9} \].

Now calculate \(a_4 \times b_4\):

\[ a_4b_4 = \left(\frac{7}{3}\right)\left(\frac{4}{9}\right) = \frac{28}{27} \]

Thus, the correct answer is \(\frac{28}{27}\).