Let a, b, c be positive integers such that b/a is an integer. If a, b, c are in geometric progressio

Q: Wait, why is $r$ an integer? The problem only says $b/a$ is an integer.

That is a fantastic observation! In a Geometric Progression, the common ratio $r$ is defined as $b/a$. Since the problem explicitly states that $b/a$ is an integer, it directly follows that $r$ must be an integer. This is the crucial first step that unlocks the entire problem!

Q: Why can't $r=1$?

If $r=1$, then $a=b=c$. The arithmetic mean of three identical numbers is simply $a$. The equation would then become $a = a + 2$, which simplifies to $0 = 2$. This is a contradiction, so $r$ cannot be 1.

Q: How do we know $(r-1)^2$ must be 1?

We derived the equation $a(r-1)^2 = 6$. Since $a$ is a positive integer, $(r-1)^2$ must be a factor of 6. Furthermore, $(r-1)^2$ is a perfect square. The factors of 6 are 1, 2, 3, and 6. Among these, only 1 is a perfect square. Therefore, $(r-1)^2$ must be 1.

Question

Let $a, b, c$ be positive integers such that $b/a$ is an integer. If $a, b, c$ are in geometric progression and the arithmetic mean of $a, b, c$ is $b + 2$, then the value of $\frac{a^2 + a - 14}{a + 1}$ is \dots.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow traveler, to the elegant world of sequences! Today, we are going to unravel a problem that is a beautiful dance between Geometric Progressions (G.P.) and Arithmetic Means (A.M.).

Imagine three positive integers, $a, b,$ and $c$, forming a G.P. This means there is a constant ratio, $r$, such that $b = ar$ and $c = ar^2$.

Since $b/a$ is an integer, our common ratio $r$ must also be an integer. This is our first foothold on this mountain.

### The Bridge: The Arithmetic Mean Equation

Now, let us look at the second condition: the arithmetic mean of $a, b,$ and $c$ is $b + 2$. Mathematically, this is written as:

$$\frac{a + b + c}{3} = b + 2$$

This equation is our bridge, connecting the geometric nature of the sequence to the arithmetic property of its mean. Let us substitute our G.P. terms into this equation by replacing $b$ with $ar$ and $c$ with $ar^2$:

$$\frac{a + ar + ar^2}{3} = ar + 2$$

### The Algebraic Transformation: The "Aha!" Moment

Let us simplify this by multiplying both sides by 3 to clear the fraction:

$$a + ar + ar^2 = 3(ar + 2)$$

Expanding the right side gives us $a + ar + ar^2 = 3ar + 6$. Now, let us bring all the terms involving $a$ and $r$ to one side:

$$ar^2 - 2ar + a = 6$$

Every term on the left side has an $a$ in it. Let us factor out $a$:

$$a(r^2 - 2r + 1) = 6$$

The expression inside the parentheses is a perfect square, $(r-1)^2$. Thus, our equation becomes:

$$a(r-1)^2 = 6$$

### The Detective Work: Integer Constraints

We know that $a$ and $r$ are positive integers. This means $(r-1)^2$ must be a perfect square factor of 6.

The factors of 6 are 1, 2, 3, and 6. The only perfect square among these is 1.

Therefore, $(r-1)^2 = 1$. This implies $r-1 = 1$ (since $r$ must be a positive integer, $r-1 = -1$ would result in $r=0$, which is invalid).

Thus, $r = 2$. Substituting $r=2$ back into our equation:

$$a(2-1)^2 = 6 \Rightarrow a(1)^2 = 6 \Rightarrow a = 6$$

### The Final Calculation: The Reward

We have found our value for $a = 6$. The problem asks us to evaluate the expression:

$$\frac{a^2 + a - 14}{a + 1}$$

Substituting $a = 6$ into the expression:

$$\frac{6^2 + 6 - 14}{6 + 1} = \frac{36 + 6 - 14}{7}$$

$$\frac{28}{7} = 4$$

We have reached the summit! The final answer is **4**.

Animated Solution for Mathematics - Sequence and Series: Let $a, b, c$ be positive integers such that $b / a$ is an integer. If $a, b, c$ are in geometric progression and the arithmetic mean of $a, b, c$ is $b + 2$ , then the value of $\frac{a ^{2} + a - 14}{a + 1}$ is \dots.

Enter Numerical Value:

Visualized Solution

Understanding the Given Conditions

Defining the G.P. Terms

The Arithmetic Mean Equation

Substituting G.P. Terms

Simplifying the Equation

Rearranging Terms

Factoring the Expression

Integer Constraint Analysis

Finding the Values of $r$ and $a$

The Final Expression

Calculating the Result

The Sigma Insight: Relation Between A.M., G.M., and H.M.

Analyzing the Setup

The Bridge

The Algebraic Transformation

The Detective Work

The Final Calculation

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