Question

In a G.P of 6 terms, the first and last terms are \(\frac{\mathrm{x}^{3}}{\mathrm{y}^{2}}\) and \(\frac{\mathrm{y}^{3}}{\mathrm{x}^{2}}\) respectively. Find the ratio of \(3^{\text {rd }}\) and \(4^{\text {th }}\) terms of that G P. (1) \(x^{2}: 1\) (2) \(\mathrm{y}^{2}: \mathrm{x}\) (3) \(\mathrm{y}: \mathrm{x}\) (4) \(\mathrm{x}: \mathrm{y}\)

Short answer

Answer: The ratio between the 3rd and 4th term is \(y^2:x^2\).

Step-by-step solution

Identifying the G.P. terms

The geometric progression (G.P.) consists of 6 terms. The first term is \(a_1 = \frac{x^3}{y^2}\), and the last term is \(a_6 = \frac{y^3}{x^2}\).

Finding the common ratio (r)

As we know the first (a_1) and last term (a_6) of the G.P., we can find the common ratio (r) using the formula: \(a_{n} = a_1 * r^{n - 1}\). Here, we have: $$a_6 = a_1 * r^{6 - 1} \\ \frac{y^3}{x^2} = \frac{x^3}{y^2} * r^5$$ Now, we solve for r: $$r^5 = \frac{y^3}{x^2}*\frac{y^2}{x^3} = \frac{y^5}{x^5}$$ $$r = \sqrt[5]{\frac{y^5}{x^5}} = \frac{y}{x}$$

Calculate the 3rd and 4th term

Now that we have the common ratio, we can calculate the 3rd (a_3) and 4th (a_4) term using the general formula: \(a_{n} = a_1 * r^{n - 1}\). For the 3rd term: $$a_3 = a_1 * r^2 = \frac{x^3}{y^2}*\left(\frac{y}{x}\right)^2 = \frac{x^3}{y^2}*\frac{y^2}{x^2}$$ For the 4th term: $$a_4 = a_1 * r^3 = \frac{x^3}{y^2}*\left(\frac{y}{x}\right)^3 = \frac{x^3}{y^2}*\frac{y^3}{x^3}$$

Calculate the ratio of the 3rd and 4th term

Finally, we can find the ratio between the 3rd and 4th term: $$\frac{a_3}{a_4} = \frac{\frac{x^3}{y^2}*\frac{y^2}{x^2}}{\frac{x^3}{y^2}*\frac{y^3}{x^3}}$$ $$\frac{a_3}{a_4}=\frac{\cancel{x^3}\cancel{y^2}\cancel{x^2}y^2}{\cancel{x^3}\cancel{y^2}y^3\cancel{x^3}}= \frac{y^2}{x^2}$$ The required ratio is \(y^2:x^2\), which corresponds to option (2).

Key concepts

Common Ratio

In any geometric progression (G.P.), the common ratio is a pillar concept. It defines how each term is related to the previous one. To identify this ratio, you divide any term by its preceding term in the sequence.
This relationship holds constant across the entire sequence.

For example, in the given exercise where the first term is \(a_1 = \frac{x^3}{y^2}\) and the last term is \(a_6 = \frac{y^3}{x^2}\), we use the formula for the n-th term of a G.P.:
\[\ a_n = a_1 \times r^{n-1} \]
Applying it between the first and last terms allows us to express:
\[\ a_6 = a_1 \times r^5 \]
Substituting the given terms, we solve for \(r\):
\[\ \frac{y^3}{x^2} = \frac{x^3}{y^2} \cdot r^5 \]
Resulting in:
\[\ r^5 = \frac{y^5}{x^5} \]
Finally:
\[\ r = \frac{y}{x} \]
This statement tells us every term is \(\frac{y}{x}\) times the previous term.
The common ratio enables prediction of any term in the series and is instrumental in understanding progression.

Third and Fourth Terms

To derive specific terms within a geometric progression like the third and fourth terms, we utilize the common ratio along with the known first term. This lies at the heart of calculating any term in G.P.

Starting with the third term, \(a_3\), use the formula:
\[\ a_3 = a_1 \times r^2 \]
Substituting known values provides:
\[\ \frac{x^3}{y^2} \times \left(\frac{y}{x}\right)^2 = \frac{x^3}{y^2} \times \frac{y^2}{x^2} \]
which simplifies to \(1\), showing a neat balance in geometric terms.

Similarly, for the fourth term, \(a_4\):
\[\ a_4 = a_1 \times r^3 \]
Substituting yields:
\[\ \frac{x^3}{y^2} \times \left(\frac{y}{x}\right)^3 = \frac{x^3}{y^2} \times \frac{y^3}{x^3} \]
Results in \(\frac{y}{x}\).

Each term calculation exploits power laws impacting proportions in G.P.
This emphasis on systematic simplification pinpoints term behavior through algebraic manipulation.

Sequence Formulas

Sequence formulas are the blueprint for navigating through a geometric progression, offering a structured path to find any term or relationship therein. These formulas reveal hidden sequences through consistent multiplication by a common ratio.
Given \(a_1\) and a known common ratio \(r\), any term \(a_n\) can be expressed as:
\[\ a_n = a_1 \times r^{n-1} \]
This central relation applies equally to building up or breaking down sequences.

• Example Use: In understanding the G.P. structure for finding the ratio of 3rd and 4th terms, the formula was repeatedly used to allocate positions:
• \(a_3 = a_1 \times r^2\)
• \(a_4 = a_1 \times r^3\)

These enable quick location of term position and interaction within any progression.
Ultimately, given required values, these formulas unlock clarity and structure data towards meaningful sequence relationships, such as comparative ratios in terms.