Question

If the nth term of the geometric progression, $5, -\frac{5}{2}, \frac{5}{4}, -\frac{5}{8}, \ldots$ is $\frac{5}{1024}$, then the value of n is

• A. $11$
• B. $10$
• C. $9$
• D. $4$

Hint:

For GP, $T_n = ar^{n-1}$. Sign depends on whether $n-1$ is even/odd.

Answer 1

The Correct Option is A

To find the value of \(n\) for the given geometric progression (G.P.), we first need to understand the properties of geometric progressions. In a G.P., each term after the first is the product of the previous term and a constant known as the common ratio.

Given the sequence: \(5, -\frac{5}{2}, \frac{5}{4}, -\frac{5}{8}, \ldots\)

We can calculate the common ratio (\(r\)) as follows:

Common ratio, \(r = \frac{\text{second term}}{\text{first term}} = \frac{-\frac{5}{2}}{5} = -\frac{1}{2}\)

The general formula for the \(n\)th term of a geometric progression is:

\(a_n = a \cdot r^{n-1}\)

where:

• \(a\) is the first term,
• \(r\) is the common ratio,
• \(n\) is the term number.

In our sequence, \(a = 5\) and \(r = -\frac{1}{2}\). We are given that the \(n\)th term is \(\frac{5}{1024}\). Therefore:

\(5 \cdot \left(-\frac{1}{2}\right)^{n-1} = \frac{5}{1024}\)

Dividing both sides by 5, we get:

\(\left(-\frac{1}{2}\right)^{n-1} = \frac{1}{1024}\)

This implies:

\((-1)^{n-1} \cdot \left(\frac{1}{2}\right)^{n-1} = (-1)^0 \cdot \left(\frac{1}{2}\right)^{10}\)

For the equality to hold, \((-1)^{n-1} = 1\), and \(\left(\frac{1}{2}\right)^{n-1} = \left(\frac{1}{2}\right)^{10}\).

This gives us two equations:

• \(n-1\) should be even for \((-1)^{n-1} = 1\).
• \(n-1 = 10\) from the second equation.

The only value that satisfies both is \(n-1 = 10\), consequently \(n = 11\).

Therefore, the value of \(n\) is 11.