Fifth term of a GP is 2, then the product of its 9 terms is (A) 256 (B) 512 (C) 1024 (D) none of the

Q: Why does the product of all terms depend only on the middle term?

In a Geometric Progression, the product of terms equidistant from the center is constant. For 9 terms, the middle term is $T_5$. The product is $T_1 \times T_9 = T_2 \times T_8 = T_3 \times T_7 = T_4 \times T_6 = T_5^2$. Thus, the product of all 9 terms is $(T_1 \times T_9) \times (T_2 \times T_8) \times (T_3 \times T_7) \times (T_4 \times T_6) \times T_5 = T_5^2 \times T_5^2 \times T_5^2 \times T_5^2 \times T_5 = T_5^9$. It is a beautiful property of symmetry!

Q: Is there a general formula for the product of $n$ terms of a GP?

Yes! The product of the first $n$ terms of a GP is given by $P = (a \cdot T_n)^{n/2}$, where $a$ is the first term and $T_n$ is the $n^{th}$ term. In our case, $n=9$, so $P = (a \cdot T_9)^{9/2}$. Since $a \cdot T_9 = a \cdot ar^8 = a^2r^8 = (ar^4)^2 = T_5^2$, this simplifies to $(T_5^2)^{9/2} = T_5^9$. It works perfectly.

Q: What if the number of terms was even, say 8? Would this trick still work?

Absolutely! If you had 8 terms, the product would be $(T_1 \times T_8) \times (T_2 \times T_7) \times (T_3 \times T_6) \times (T_4 \times T_5)$. Since $T_1 \times T_8 = T_2 \times T_7 = T_3 \times T_6 = T_4 \times T_5 = ar^7$, the product would be $(ar^7)^4$. The logic remains the same: group the terms symmetrically!

Question

Fifth term of a GP is $2$, then the product of its $9$ terms is

Options:

(A) 256

(B) 512

(C) 1024

(D) none of these

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow traveler on the JEE journey. Today, we are not just solving a problem; we are uncovering the hidden symmetry within a Geometric Progression.

Imagine you are standing at the start of a sequence, looking at nine distinct numbers, $T_1, T_2, \dots, T_9$, all dancing to the rhythm of a common ratio $r$. The question asks for their product.

At first glance, this might seem like a tedious task of multiplying nine terms, but in the world of JEE, we look for the shortcut, the elegant path.

### The Anatomy of the Sequence

Let us define our terms. The $n^{th}$ term of a Geometric Progression is defined as $T_n = ar^{n-1}$. This is our fundamental building block.

We have nine terms, and we need their product, $P = T_1 	imes T_2 	imes \dots 	imes T_9$. If we write these out, we get:

$$P = (a) \cdot (ar) \cdot (ar^2) \dots (ar^8)$$

Notice how the first term $a$ appears in every single one of the nine terms. When we multiply them all together, we are essentially multiplying $a$ by itself nine times, which gives us $a^9$.

Now, consider the common ratio $r$. We have $r^0 \cdot r^1 \cdot r^2 \dots r^8$. When multiplying terms with the same base, we add the exponents.

The exponent of $r$ becomes the sum $1 + 2 + 3 + 4 + 5 + 6 + 7 + 8$.

### The Exponent Magic

This is where the precision of a mathematician is required. The sum of the first $n$ natural numbers is given by the formula:

$$\frac{n(n+1)}{2}$$

Here, $n=8$, so the sum is:

$$\frac{8 	imes 9}{2} = 36$$

Thus, our product $P$ simplifies beautifully to $P = a^9 r^{36}$.

Now, look at this expression with the eyes of a strategist. We are given that the fifth term $T_5 = 2$. Using our general formula:

$$T_5 = ar^{5-1} = ar^4 = 2$$

### The Final Revelation

We have $P = a^9 r^{36}$ and we know $ar^4 = 2$. Notice that $36 = 4 	imes 9$.

This means we can rewrite $r^{36}$ as $(r^4)^9$. Therefore, we can express the product as:

$$P = a^9 (r^4)^9$$

By the laws of exponents, this simplifies to:

$$P = (ar^4)^9$$

Substituting the known value $ar^4 = 2$ into our expression, we get $P = (2)^9$.

The calculation is straightforward:

$$2^9 = 512$$

There it is. We did not need to know the individual values of $a$ or $r$. We only needed to understand the structure of the sequence.

The final answer is **512**.

Animated Solution for Mathematics - Sequence and Series: Fifth term of a GP is $2$ , then the product of its $9$ terms is

Select Answer:

Visualized Solution

Visualizing the G.P. Sequence

General Term of a G.P.

The Middle Term $T_{5}$

Given Value of $T_{5}$

Setting up the Product $P$

Expanding the Product

Grouping the First Term $a$

Grouping the Common Ratio $r$

Summing the Exponents

Connecting to the Known Value

Rewriting the Product

Substituting the Value

Final Calculation

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

Analyzing the Setup

The Anatomy of the Sequence

The Exponent Magic

The Final Revelation

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Animated Solution for Mathematics - Sequence and Series: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Fifth term of a GP is 2, then the product of its 9 terms is