Let the first three terms 2, p and q, with q ≠ 2, of a G.P. be respectively the 7th, 8th and 13th te

Q: Wait, why do we reject the root $p=2$ in the quadratic equation? Is it just a random choice?

Great catch! In JEE problems, always look at the constraints. The problem explicitly states $q \neq 2$. If we set $p=2$, our relationship $q = 6p - 10$ forces $q$ to be $6(2) - 10 = 2$. This violates the condition $q \neq 2$. We aren't choosing; we are obeying the constraints provided by the problem setter!

Q: Why can't we just use the sum of terms formula for the A.P.?

The sum formula is for adding terms together. Here, we are dealing with specific positions (the 7th, 8th, and 13th terms). We need the general term formula $T_n = a + (n-1)d$ to pinpoint the exact value at those specific positions. Think of it as finding a specific seat in a theater rather than counting the total number of people.

Q: Is there a faster way to find $n$ without calculating $a$ and $d$?

While you might find shortcuts, calculating $a$ and $d$ is the safest path. It builds a solid foundation. Once you have $a = -46$ and $d = 8$, the final step becomes a simple linear equation. Don't rush the foundation; the clarity you gain from finding $a$ and $d$ prevents silly calculation errors later.

Question

Let the first three terms $2, p$ and $q$, with $q 
eq 2$, of a G.P. be respectively the $7^{	ext{th}}, 8^{	ext{th}}$ and $13^{	ext{th}}$ terms of an A.P. If the $5^{	ext{th}}$ term of the G.P. is the $n^{	ext{th}}$ term of the A.P., then $n$ is equal to:

Options:

(A) 163

(B) 151

(C) 177

(D) 169

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, future engineer. Today, we are not just solving a problem; we are embarking on a journey through the architecture of numbers. We have two distinct sequences—an Arithmetic Progression (A.P.) and a Geometric Progression (G.P.)—and they are intertwined in a beautiful, logical dance.

Our goal is to find the position $n$ where these two worlds collide. Let’s break this down step by step.

### The A.P. Landscape

Imagine you are standing on a staircase. An A.P. is like a staircase where every step has the exact same height, $d$. If we define the first step as $a$, then the $n^{	ext{th}}$ step is simply $a + (n-1)d$.

The problem gives us three specific markers on this staircase:

1. The $7^{	ext{th}}$ term: $T_7 = a + 6d = 2$
2. The $8^{	ext{th}}$ term: $T_8 = a + 7d = p$
3. The $13^{	ext{th}}$ term: $T_{13} = a + 12d = q$

These three markers are the foundation of our G.P. By looking at these equations, we can see the relationship between $p$, $q$, and the common difference $d$.

Subtracting the first equation from the second yields:
$$(a + 7d) - (a + 6d) = p - 2 \Rightarrow d = p - 2$$

Similarly, subtracting the second from the third gives:
$$(a + 12d) - (a + 7d) = q - p \Rightarrow 5d = q - p$$

### The G.P. Bridge

The problem states that $2, p$, and $q$ are the first three terms of a G.P. In a G.P., the ratio between consecutive terms is constant, leading to the property that the square of the middle term equals the product of its neighbors:
$$p^2 = 2q$$

We now have a system of equations. Substituting $d = p - 2$ into the second A.P. relation $5d = q - p$:
$$5(p - 2) = q - p$$
$$5p - 10 = q - p \Rightarrow q = 6p - 10$$

### The Quadratic Conflict

We substitute our expression for $q$ into the G.P. property $p^2 = 2q$:
$$p^2 = 2(6p - 10)$$
$$p^2 - 12p + 20 = 0$$

Factoring this quadratic equation gives:
$$(p - 10)(p - 2) = 0$$

We have two potential candidates for $p$: $10$ and $2$. However, the problem explicitly states $q 
eq 2$. If we test $p = 2$, then $q = 6(2) - 10 = 2$, which violates the condition. Therefore, we must reject $p = 2$ and accept **$p = 10$**.

### The Final Convergence

With $p = 10$, we find the common difference $d$:
$$d = p - 2 = 10 - 2 = 8$$

Using the first A.P. equation $a + 6d = 2$:
$$a + 6(8) = 2 \Rightarrow a = 2 - 48 = -46$$

The G.P. is defined by the first term $2$ and common ratio $r = 10/2 = 5$. The $5^{	ext{th}}$ term of this G.P. is:
$$T_5 = a_{gp} \cdot r^4 = 2 \cdot 5^4 = 2 \cdot 625 = 1250$$

Finally, we equate this to the $n^{	ext{th}}$ term of our A.P.:
$$1250 = a + (n - 1)d$$
$$1250 = -46 + (n - 1)8$$
$$1296 = (n - 1)8$$
$$162 = n - 1 \Rightarrow n = 163$$

The final answer is **$n = 163$**.

Animated Solution for Mathematics - Sequence and Series: Let the first three terms $2, p$ and $q$ , with $q \neq = 2$ , of a G.P. be respectively the $7^{th}, 8^{th}$ and $1 3^{th}$ terms of an A.P. If the $5^{th}$ term of the G.P. is the $n^{th}$ term of the A.P., then $n$ is equal to:

Select Answer:

Visualized Solution

Defining the Sequences

Setting up A.P. Equations

Finding the Common Difference $d$

Relating $p$ and $q$

Applying G.P. Property

Solving the Quadratic Equation

Filtering the Value of $p$

Finding $a$ and $d$ for A.P.

Finding the $5^{t h}$ Term of G.P.

Equating G.P. and A.P. Terms

Solving for $n$

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

Analyzing the Setup

The A.P

The G.P

The Quadratic Conflict

The Final Convergence

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