Problem 217 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

Given \(a^{x}=b^{y}=c^{z}=d^{u}\) and \(a, b, c, d\) are in G.P., show that \(x, y, z, u\) are in H.P.

Step-by-Step Solution

Answer

Since \(a, b, c, d\) are in G.P., there exists a common ratio \(r\) such that \(b = ar, c = br, d=cr\). Equating the expressions for \(b^{y}\) and \(c^{z}\) and using the expressions for \(b, c, d\), we get \(x(y/r)=xyz/r^3\). We then show that the reciprocals \(\frac{1}{x}, \frac{1}{y}, \frac{1}{z}, \frac{1}{u}\) form an arithmetic progression (A.P) by obtaining the expressions \(\frac{1}{x} = \frac{u}{rz^2} + \frac{1}{y}\) and \(\frac{1}{x} = \frac{1-r}{xyz/r^3}\). Since their reciprocals form an A.P., \(x, y, z, u\) are in H.P.

Step 1: Express the common ratio

Since \(a, b, c, d\) are in G.P., we can say that there exists a common ratio \(r\) such that \[b = ar, c = br, d=cr\]

Step 2: Manipulate the given equation

We know that \(a^{x}=b^{y}=c^{z}=d^{u}\). Let's equate the expressions for \(b^{y}\) and \(c^{z}\): \[(a^{x})^{y/r} = (b^{y})^{z/r}\] Since both of the expressions are equal to \(d^u\), let's equate both the expressions to \(d^u\): \[(a^{x})^{y/r} = (b^{y})^{z/r} = d^u\] Using the expressions for \(b, c, d\) from Step 1, we can write this equation as follows: \[(a^{x})^{y/r} =(a^{xr})^{y/r} = (a^{xy/r^2})^{z/r} = a^{xyz/r^3}\] Through cancelling a^{x} from the equation and simplifying it to one exponent per term, we get: \[x(y/r)=xyz/r^3\]

Step 3: Show that the series \(x, y, z, u\) are in H.P.

Since the numbers are in H.P., their reciprocals form an arithmetic progression (A.P). Thus, we need to show that \(\frac{1}{x}, \frac{1}{y}, \frac{1}{z}, \frac{uxtensive}{u}\) are in A.P. From the previous step, we derived: \[x(y/r)=xyz/r^3\] Divide both sides by \(xyzu\): \[\frac{1}{rz^2u} = \frac{1}{xy}\] Combine the two fractions on the left-hand side, and try to make \(\frac{1}{x}\) the subject: \[\frac{1}{x} = \frac{u}{rz^2} + \frac{1}{y}\] Now, let's find an expression for \(\frac{1}{z}\) by dividing both sides of the main equation by \(c^z\): \[a^{x-y} = b^{y-z}\] Substitute the expressions for \(a\) and \(b\): \[(a^{x})^{1-r} = (a^{xr})^{(1-r)/r}\] Now express \(\frac{1}{x}\) as the subject of the formula above: \[\frac{1}{x} = \frac{1-r}{xyz/r^3}\] Now observe that, if we take the difference between the first and second expression for \(\frac{1}{x}\) and the last two expressions for \(\frac{1}{x}\), we obtain: \[\frac{1}{y}-\frac{1}{x} = \frac{1}{z}-\frac{1}{y}\] That means that \(\frac{1}{x}, \frac{1}{y}, \frac{1}{z}, \frac{1}{u}\) are in A.P. and therefore \(x, y, z, u\) are in H.P. The proof is complete.

Key Concepts

Geometric Progression

A geometric progression (G.P.) is a sequence of numbers where each term after the first is found by multiplying the previous one by a fixed, non-zero number called the common ratio. In a G.P., the relationship between consecutive terms is constant. For example, if you have a sequence 2, 6, 18, 54, the common ratio is 3. Each term is the previous term multiplied by 3.

When describing a sequence in G.P., you might use the formula:

• First term: \( a \)
• Second term: \( ar \)
• Third term: \( ar^2 \)
• Fourth term: \( ar^3 \)

Where \( r \) is the common ratio. Understanding G.P. is vital to grasp other related progressions, like harmonic progression, where terms indirectly relate through inverses.

Exponentiation

Exponentiation is a mathematical operation involving two numbers, the base and the exponent. The base is the number you are multiplying, and the exponent is how many times you multiply the base by itself. For example, in \( 3^4 \), 3 is the base and 4 is the exponent, indicating that you multiply 3 by itself four times: \[ 3^4 = 3 \times 3 \times 3 \times 3 = 81 \]
Exponentiation is a building block in understanding equations like \( a^x = b^y = c^z = d^u \) because it simplifies the expression of repetitive multiplication. It connects to progressions, as seen when creating equations that describe sequences. This can connect to geometric progressions when bases themselves represent each term in the sequence raised to a certain power.

Arithmetic Progression

An arithmetic progression (A.P.) is a sequence of numbers in which the difference of any two successive members is a constant. This constant difference is called the common difference. In arithmetic terms:

• If the first term is \( a \)
• Then the second term is \( a + d \)
• The third term is \( a + 2d \)
• And so forth

Where \( d \) is the common difference. When numbers are in A.P., the sequence \( a, a+d, a+2d, \ldots \) follows a linear pattern. Understanding A.P. is critical when exploring harmonic progressions, where inverses of numbers form an arithmetic sequence. For instance, understanding how reciprocals \( \frac{1}{x}, \frac{1}{y}, \frac{1}{z}, \ldots \) create an A.P., helps conclude that original set \( x, y, z, \ldots \) is a harmonic progression.