If the sum of the second, third and fourth terms of a positive term G.P. is 3 and the sum of its six

Q: Why did we divide the equations instead of solving for 'a' in terms of 'r' and substituting?

Great question! While substitution works, it often leads to messy algebraic expressions. In a Geometric Progression, terms are defined by multiplication. Dividing one equation by another is a powerful 'shortcut' because it allows the common factors—like the entire $(1 + r + r^2)$ block—to cancel out instantly, leaving you with a clean power of $r$.

Q: How do I know if I should take the positive or negative root for $r^4 = 81$?

Always look back at the problem statement! The problem explicitly states it is a 'positive term G.P.' If $r$ were negative, the terms would oscillate between positive and negative values. Since all terms must be positive, $r$ must be positive. Thus, we reject $r = -3$.

Q: Is there a faster way to see that $r^4 = 81$?

Yes! Notice the indices of the terms. The second term is $ar^1$ and the sixth is $ar^5$. The ratio of the sixth term to the second term is $r^4$. Because the sums are just scaled versions of these terms, the ratio of the sums must also be $r^4$. It's a beautiful property of G.P.s!

Q: What if the common ratio was 1? Would the formula still work?

If $r = 1$, the formula $S_n = \frac{a(r^n - 1)}{r - 1}$ would result in a division by zero. In that case, the G.P. is just $a, a, a, \dots$ and the sum is simply $n \times a$. Always check your $r$ value before plugging it into the formula!

Q: Why is the final answer so large? Should I expand $3^{50}$?

In JEE Advanced, you are almost never expected to calculate massive powers like $3^{50}$. Leaving the answer in exponential form is standard practice and actually preferred, as it shows you understand the structure of the sequence rather than just performing arithmetic.

Question

If the sum of the second, third and fourth terms of a positive term G.P. is 3 and the sum of its sixth, seventh and.eighth terms is 243, then the sum of the first 50 terms of this G.P. is :

Options:

(A) $\frac{1}{26}(3^{50}-1)$

(B) $\frac{1}{13}(3^{50}-1)$

(C) $\frac{1}{26}(3^{49}-1)$

(D) $\frac{1}{26}(3^{50}-1)$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing at the beginning of a path that stretches infinitely forward. This path is a Geometric Progression (G.P.), where every step you take is a constant multiple of the one before it.

We are given two snapshots of this journey: the sum of the second, third, and fourth terms, and the sum of the sixth, seventh, and eighth terms. Our goal is to map the entire terrain by identifying the sequence's parameters.

### Defining the DNA of the Sequence

Every G.P. is defined by its DNA: the first term $a$ and the common ratio $r$. We are told this is a 'positive term G.P.', which implies $a > 0$ and $r > 0$.

The second, third, and fourth terms are $ar$, $ar^2$, and $ar^3$. Their sum is given as:

$$ar + ar^2 + ar^3 = 3$$

Factoring out $ar$, we obtain a symmetric expression:

$$ar(1 + r + r^2) = 3 \quad \dots (1)$$

The second snapshot involves the sixth, seventh, and eighth terms, which are $ar^5$, $ar^6$, and $ar^7$. Their sum is $243$. Factoring out $ar^5$, we get:

$$ar^5(1 + r + r^2) = 243 \quad \dots (2)$$

### The Power of Division

We can simplify these equations by observing the common term $(1 + r + r^2)$. By dividing equation (2) by equation (1), we perform a mathematical operation that isolates the common ratio:

$$\frac{ar^5(1 + r + r^2)}{ar(1 + r + r^2)} = \frac{243}{3}$$

This simplifies significantly to:

$$r^4 = 81$$

Since we know $r$ must be positive, we take the fourth root of $81$, which yields **$r = 3$**. We have successfully unlocked the growth factor of our sequence.

### Finding the Starting Point

With $r = 3$ determined, finding the first term $a$ is a matter of substitution. Plugging $r = 3$ back into our first equation:

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

$$3a(1 + 3 + 9) = 3$$

$$39a = 3 \implies a = \frac{1}{13}$$

We have now fully defined our sequence. We know where it starts at **$a = \frac{1}{13}$** and how it grows at **$r = 3$**.

### The Final Ascent

We are asked for the sum of the first 50 terms. The formula for the sum of the first $n$ terms of a G.P. is:

$$S_n = \frac{a(r^n - 1)}{r - 1}$$

Substituting our values $n = 50$, $a = \frac{1}{13}$, and $r = 3$:

$$S_{50} = \frac{\frac{1}{13}(3^{50} - 1)}{3 - 1}$$

This simplifies to the final result:

**$$S_{50} = \frac{1}{26}(3^{50} - 1)$$**

The complexity of the problem dissolves into this clean, elegant expression. Remember, the goal is not just to reach the answer, but to see the underlying symmetry; when you see a G.P., always look for the ratio and the common factor.

Animated Solution for Mathematics - Sequence and Series: If the sum of the second, third and fourth terms of a positive term G.P. is 3 and the sum of its sixth, seventh and.eighth terms is 243, then the sum of the first 50 terms of this G.P. is :

Select Answer:

Visualized Solution

Define the G.P. Parameters

Equation for Terms 2, 3, and 4

Equation for Terms 6, 7, and 8

Divide Equations to Eliminate $a$

Solve for Common Ratio $r$

Solve for First Term $a$

Sum of First 50 Terms Formula

Final Calculation

Conclusion & Key Takeaway

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

Analyzing the Setup

Defining the DNA of the Sequence

The Power of Division

Finding the Starting Point

The Final Ascent

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