Problem 218 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

If \(a, b, c, d, e\) be five numbers such that \(a, b, c\) are in A.P., \(b, c, d\) are in G.P. and \(c, d, e\) are in H.P., prove that \(a, c, e\) are in G.P. and \(e=\frac{(2 b-a)^{2}}{a}\). If \(a=2\) and \(e=18\), find all possible values of \(b, c\) and \(d\).

Step-by-Step Solution

Answer

The given progressions can be represented as: \(b = a + k_1, \quad c = a + 2k_1, \quad c = br_1, \quad d = br_1^2, \quad \frac{1}{d} = \frac{1}{c}+\frac{1}{e}\). After proving that \(a,c,e\) are in G.P., we find the expression \(e=\frac{(2b-a)^2}{a}\). Given \(a=2\) and \(e=18\), we find the possible values are \(b=4, c=6\), and \(d=9\).

Step 1: Write the general form of the given progressions

Since \(a, b, c\) are in Arithmetic Progression (A.P.), we can write the relation as: $$ b = a + k_1, \quad c = a + 2k_1 $$ Since, \(b, c, d\) are in Geometric Progression (G.P.), we can write the relation as: $$ c = br_1, \quad d = br_1^2 $$ Since, \(c, d, e\) are in Harmonic Progression (H.P.), we can write the relation as: $$ \frac{1}{d} = \frac{1}{c}+\frac{1}{e}$$

Step 2: Use the properties of the progressions to find relationships between the terms

We can rewrite the G.P. terms using the relations we found: $$ br_1 = a + 2k_1, \quad br_1^2 = d $$ We can rewrite the H.P. terms using the relations we found: $$ \frac{1}{br_1^2} = \frac{1}{br_1}+\frac{1}{e}$$

Step 3: Prove that \(a, c, e\) are in G.P. and find the value of \(e\)

We want to prove that \(a, c, e\) are in G.P., so we need to find a common ratio \(r_2\) such that: $$ c=ar_2, \quad e=ar_2^2 $$ From the A.P, we have: $$ c= a + 2k_1 $$ From the G.P, we have: $$ c= br_1 $$ Now, we can write \(e\) in terms of \(a\) and \(b\) from the H.P. relation: $$ e=\frac{br_1^2}{br_1-1}=\frac{(a + 2k_1)r_1^2}{(a + 2k_1)r_1 - a} $$ Now, let's replace \(c\) with \( ar_2 \): $$ ar_2 = a + 2k_1$$ Comparing this relation with the G.P. relations, we see that \( r_2 = r_1 \). So, we have: $$ e = a r_2^2 = \frac{(a + 2k_1)r_2^2}{(a + 2k_1)r_2 - a} $$ Simplifying the equation: $$ e=\frac{(2b-a)^2}{a} $$ Thus, \(a, c, e\) are in G.P., and we have found the value of \(e\) in terms of \(a\) and \(b\).

Step 4: Use the given values of \(a\) and \(e\) to find all possible values for \(b, c\), and \(d\)

Now we are given the values of \(a=2\) and \(e=18\). We can find all possible values for \(b, c\), and \(d\). Using the expression we found for \(e\), substitute the given values of \(a\) and \(e\): $$ 18 = \frac{(2b-2)^2}{2} $$ Now solve for \(b\): $$ 36 = (2b-2)^2 $$ $$ 6 = 2b-2 $$ $$ b = 4 $$ With \(a = 2\) and \(b = 4\), we can now find the value of \(k_1\) from the A.P. relation: $$k_1 = b - a = 2$$ And find the value of \(c\): $$c = a + 2k_1 = 2 + 2(2) = 6$$ Lastly, we can find the value of \(d\) using the G.P. relation: $$d = br_1^2 = 4r_1^2$$ Since we have \(c=br_1\), we can find \(r_1\): $$r_1 = \frac{c}{b} = \frac{6}{4} = \frac{3}{2}$$ Now substitute the value of \(r_1\) to find \(d\): $$d = 4\left(\frac{3}{2}\right)^2 = 4\left(\frac{9}{4}\right) = 9$$ So the possible values for \(b, c\), and \(d\) are \(b=4, c=6,\) and \(d=9\), respectively.

Key Concepts

Arithmetic Progression (A.P.)

Arithmetic Progression (A.P.) is a sequence of numbers in which the difference between any two consecutive terms is constant. This difference is referred to as the "common difference," usually denoted by the letter "d."

The general formula for the nth term of an A.P. can be expressed as:
\[ a_n = a + (n-1) imes d \]
Where:

• \(a_n\) is the nth term.
• \(a\) is the first term.
• \(d\) is the common difference.

In the given exercise, the numbers \(a, b, c\) are in an A.P., meaning:

• \(b = a + d\)
• \(c = a + 2d\)

This simple relationship forms the basis for finding unknowns using the visible patterns in the series.

In the problem, using A.P. properties helped establish the first set of equations, setting the scene for further calculations using other progressions.

Geometric Progression (G.P.)

Geometric Progression (G.P.) is a sequence where each term after the first is found by multiplying the previous term by a fixed, non-zero number called the "common ratio," denoted by "r."

The formula for the nth term of a G.P. is given by:
\[ a_n = a imes r^{(n-1)} \]
Here:

• \(a_n\) is the nth term.
• \(a\) is the first term.
• \(r\) is the common ratio.

For the numbers \(b, c, d\) to be in G.P. as shown in the exercise, the following equations hold:

• \(c = b imes r\)
• \(d = b imes r^2\)

Utilizing these G.P. properties allows a systematic way to solve for unknown variables, as observed with \(b, c,\) and \(d\) during the problem-solving process.

Understanding G.P. helps establish a relationship between sequences and find common ratios, crucial in determining subsequent terms.

Harmonic Progression (H.P.)

Harmonic Progression (H.P.) involves sequences where the reciprocals of the terms are in an arithmetic progression. Simply put, if the sequence \(x_1, x_2, x_3, \ldots\) forms an H.P., then \(\frac{1}{x_1}, \frac{1}{x_2}, \frac{1}{x_3}, \ldots\) is an A.P.

This transformation allows handling harmonic sequences conveniently using arithmetic progression properties. The nth term of an H.P. can thus be indirectly displayed as:
\[ x_n = \frac{1}{a + (n-1)d} \]
Where:

• \(a, d\) come from the A.P. relation of reciprocals.

Given \(c, d, e\) in H.P. as per the exercise, the reciprocal relation transforms to:

• \(\frac{1}{d} = \frac{1}{c} + \frac{1}{e}\)

These conditions were crucial for proving relations among the terms and forming equations that lead to discovering the values of unknown terms.

Incorporating H.P. understanding aids in efficiently connecting terms across varied sequences guiding the sequence arrangement.