If the sum of the second, fourth and sixth terms of a G.P. of positive terms is 21 and the sum of it

Q: Wait, why do we divide the equations instead of substituting $a$ from the first one into the second?

That is a great question! While substitution is a valid algebraic method, it would lead to a very messy expression involving $a$ and $r$ that would be difficult to solve. By dividing the equations, we exploit the symmetry of the G.P. terms, allowing us to cancel out the complex polynomial $(1 + r^2 + r^4)$ instantly. In JEE Advanced, recognizing these symmetries is often the key to saving time and avoiding calculation errors.

Q: Why do we reject the negative root for $r$ when solving $r^6 = 729$?

The problem statement explicitly mentions that the G.P. consists of 'positive terms'. If $r$ were negative, the terms of the G.P. would alternate in sign (positive, negative, positive, negative...). Since all terms must be positive, $r$ must be positive. Always keep an eye out for these constraints—they are often the 'trap' that catches students who rush the calculation!

Q: Is there a trick to calculating $3^9$ without doing long multiplication?

Absolutely! You don't need to calculate $3^9$ from scratch. We already know $3^6 = 729$. Since $3^9 = 3^6 \times 3^3$, you can simply calculate $729 \times 27$. It is much safer and faster to break down large powers into smaller, known components.

Question

If the sum of the second, fourth and sixth terms of a G.P. of positive terms is 21 and the sum of its eighth, tenth and twelfth terms is 15309, then the sum of its first nine terms is :

Options:

(A) 760

(B) 755

(C) 750

(D) 757

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing before a sequence of numbers, a Geometric Progression (G.P.), where every term is born from the previous one by a constant multiplier, $r$. In the world of JEE Advanced, problems like this are not just about calculation; they are about pattern recognition.

We are given two conditions: the sum of the second, fourth, and sixth terms is $21$, and the sum of the eighth, tenth, and twelfth terms is $15309$.

At first glance, this looks like a system of equations that might lead to a nightmare of algebra. But pause. Take a breath. Look at the indices.

The terms are $T_2, T_4, T_6$ and $T_8, T_{10}, T_{12}$. Each term in the second set is exactly $r^6$ times the corresponding term in the first set. This is the hidden geometric reality.

### Decoding the Conditions

Let us formalize this. We define our G.P. with first term $a$ and common ratio $r$. The first condition is $T_2 + T_4 + T_6 = 21$.

Expanding this using $T_n = ar^{n-1}$, we get $ar + ar^3 + ar^5 = 21$. If we factor out $ar$, we are left with:

$$ar(1 + r^2 + r^4) = 21 \quad 	ext{(Equation 1)}$$

Now, look at the second condition: $T_8 + T_{10} + T_{12} = 15309$. Expanding this, we get $ar^7 + ar^9 + ar^{11} = 15309$.

If we factor out $ar^7$, we get:

$$ar^7(1 + r^2 + r^4) = 15309 \quad 	ext{(Equation 2)}$$

Notice the beauty? The term $(1 + r^2 + r^4)$ is identical in both equations. This is not a coincidence; it is the problem designer's gift to you.

### The Power of Division

Now, we perform the masterstroke. We divide Equation 2 by Equation 1:

$$\frac{ar^7(1 + r^2 + r^4)}{ar(1 + r^2 + r^4)} = \frac{15309}{21}$$

The $a$ cancels. The entire bracket $(1 + r^2 + r^4)$ cancels. We are left with:

$$\frac{r^7}{r} = \frac{15309}{21} \Rightarrow r^6 = 729$$

We have successfully isolated the common ratio. Since $3^6 = 729$, we find $r = 3$. We reject $r = -3$ because the problem implies positive terms.

With $r=3$ in hand, we return to Equation 1:

$$a(3)(1 + 3^2 + 3^4) = 21$$

This becomes $3a(1 + 9 + 81) = 21$, or $3a(91) = 21$. Solving for $a$, we get:

$$a = \frac{21}{273} = \frac{1}{13}$$

### The Final Summation

We have our foundation: $a = \frac{1}{13}$ and $r = 3$. The question asks for the sum of the first nine terms, $S_9$.

We use the formula $S_n = \frac{a(r^n - 1)}{r - 1}$. Substituting our values:

$$S_9 = \frac{\frac{1}{13}(3^9 - 1)}{3 - 1} = \frac{3^9 - 1}{13 	imes 2} = \frac{3^9 - 1}{26}$$

We know $3^6 = 729$, so $3^9 = 729 	imes 27 = 19683$. Thus:

$$S_9 = \frac{19683 - 1}{26} = \frac{19682}{26} = 757$$

The journey is complete. We navigated the algebra, respected the constraints, and arrived at the elegant solution of **757**. Keep this mindset—look for the symmetry, and the math will always reveal its secrets.

Animated Solution for Mathematics - Sequence and Series: If the sum of the second, fourth and sixth terms of a G.P. of positive terms is 21 and the sum of its eighth, tenth and twelfth terms is 15309, then the sum of its first nine terms is :

Select Answer:

Visualized Solution

Define the G.P. and First Condition

Establish the Second Condition

Divide Equations to Isolate $r$

Solve for Common Ratio $r$

Find the First Term $a$

Calculate Sum of First Nine Terms

Final Computation

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

Analyzing the Setup

Decoding the Conditions

The Power of Division

The Final Summation

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