Let a and b be two distinct positive real numbers. Let 11th term of a GP, whose first term is a and

Q: Wait, why can we just equate the exponents? What if the base is 1?

That is a brilliant observation! The problem explicitly states that $a$ and $b$ are distinct positive real numbers. If $a = b$, then $\frac{b}{a} = 1$, and the equation $1^5 = 1^{\frac{p-1}{4}}$ would be true for any $p$. Because they are distinct, $\frac{b}{a} \neq 1$, which guarantees that the exponents must be identical for the equality to hold.

Q: Why did we use $r^2$ and $R^4$ instead of solving for $r$ and $R$ directly?

Solving for $r$ would introduce square roots or fourth roots, which makes the algebra messy and prone to error. By keeping the terms as powers of $(\frac{b}{a})$, we maintain the structural symmetry of the problem, allowing the math to 'cancel out' elegantly.

Q: Is there a shortcut to visualize the relationship between the indices?

Yes! Think of the 'distance' covered in terms of the number of common ratio multiplications. In the first GP, to get from the 1st term to the 11th, you multiply by $r$ ten times. Since $r^2 = \frac{b}{a}$, you are essentially multiplying by the factor $(\frac{b}{a})$ five times. You are looking for the same 'total growth' in the second GP.

Q: What if the first term was not 'a'? Would the answer change?

If the first terms were different, the ratio $\frac{b}{a}$ would not be the common base for both sequences, and the terms would not cancel out so cleanly. The fact that they share the same starting point is the 'anchor' that makes this comparison possible.

Q: I got $p=20$. Where did I miss the +1?

This is the most common trap in sequence problems! Remember that the $n^{th}$ term involves $(n-1)$ multiplications. When you equate the exponents, you are equating the number of 'steps' taken. Always double-check if you are counting the terms or the jumps between them.

Question

Let $a$ and $b$ be two distinct positive real numbers. Let $11^{th}$ term of a GP, whose first term is $a$ and third term is $b$, is equal to $p^{th}$ term of another GP, whose first term is $a$ and fifth term is $b$. Then $p$ is equal to

Options:

(A) 20

(B) 25

(C) 21

(D) 24

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing at the starting line of a race. You have two runners, let's call them GP 1 and GP 2. Both start at the same position, $a$.

Their goal is to reach the same finish line, $b$. GP 1 reaches the finish line in 3 steps, while GP 2 takes 5 steps to reach the same destination. We need to find the position of GP 2 when GP 1 reaches its 11th step.

### Decoding the Strides

To solve this, we must first understand the 'stride' or the common ratio of each runner. For GP 1, the first term is $a$ and the third term is $b$.

Using the standard formula for the $n^{th}$ term, $T_n = ar^{n-1}$, we find that the third term is $ar^2$. Thus, we have the relationship:

$$ar^2 = b \implies r^2 = \frac{b}{a}$$

Now, consider GP 2. It also starts at $a$, but its fifth term is $b$. Following the same logic, $aR^4 = b$, where $R$ is the common ratio of the second sequence. This implies:

$$R^4 = \frac{b}{a}$$

### The 11th Milestone

Let's calculate the 11th term of GP 1. Using our formula, $t_{11} = ar^{10}$.

Since we know $r^2 = \frac{b}{a}$, we can rewrite $r^{10}$ as $(r^2)^5$. Substituting our known value, we get:

$$t_{11} = a\left(\frac{b}{a}ight)^5$$

This expression represents the 'value' of the milestone. It tells us exactly how many times the ratio $\frac{b}{a}$ has been applied to the starting value $a$.

### The Search for p

Now, we turn our attention to GP 2. We want to find the $p^{th}$ term, $T_p$, such that it equals $t_{11}$.

The formula for the $p^{th}$ term is $T_p = aR^{p-1}$. Given $R^4 = \frac{b}{a}$, we find $R = (\frac{b}{a})^{\frac{1}{4}}$. Plugging this into our expression for $T_p$, we get:

$$T_p = a\left(\left(\frac{b}{a}ight)^{\frac{1}{4}}ight)^{p-1} = a\left(\frac{b}{a}ight)^{\frac{p-1}{4}}$$

### The Grand Equivalence

We are now at the climax of our journey. We set the two terms equal:

$$a\left(\frac{b}{a}ight)^5 = a\left(\frac{b}{a}ight)^{\frac{p-1}{4}}$$

Since $a$ and $b$ are distinct positive real numbers, we know $\frac{b}{a} 
eq 1$. This allows us to equate the exponents:

$$5 = \frac{p-1}{4}$$

Multiplying both sides by 4, we get $20 = p - 1$. Adding 1 to both sides, we arrive at our destination:

**$p = 21$**

### Reflection

It is truly fascinating how complex-looking sequences can be reduced to simple algebraic relationships. We didn't need to know the specific values of $a$ or $b$; we only needed to understand the ratio of their growth.

By focusing on the structure of the exponents, we bypassed the need for tedious calculations. Whenever you see a problem involving powers and sequences, look for the underlying ratio. You will find that the math often solves itself if you let it breathe.

Animated Solution for Mathematics - Sequence and Series: Let $a$ and $b$ be two distinct positive real numbers. Let $1 1^{t h}$ term of a GP, whose first term is $a$ and third term is $b$ , is equal to $p^{t h}$ term of another GP, whose first term is $a$ and fifth term is $b$ . Then $p$ is equal to

Select Answer:

Visualized Solution

Understanding the Problem Setup

Analyzing the First G.P.

Finding the $1 1^{t h}$ Term of G.P. 1

Analyzing the Second G.P.

Expressing the $p^{t h}$ Term of G.P. 2

Equating the Two Terms

Solving for $p$

The Sigma Insight: Geometric Progression (G.P.)

Analyzing the Setup

Decoding the Strides

The 11th Milestone

The Search for p

The Grand Equivalence

Reflection

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