Let a1, a2, a3, ... be a G. P. of increasing positive numbers. If a3*a5 = 729 and a2 + a4 = 111/4, t

Q: Why do we reject $r = -6$?

The problem explicitly states the G.P. consists of 'increasing positive numbers'. If $r$ were negative, the terms would alternate between positive and negative, violating the condition. If $r$ were $6$, the sequence grows. If $r$ were $-6$, it oscillates. Thus, $r=6$ is the only valid solution.

Q: Can I use the sum of $n$ terms formula here?

You could, but it is overkill. The question asks for the sum of only the first three terms: $a_1 + a_2 + a_3$. Using the general formula $S_n = \frac{a(r^n - 1)}{r - 1}$ is valid, but simply writing out $a + ar + ar^2$ is faster and less prone to calculation errors.

Q: Is there a faster way to find $a$ and $r$?

You could use logarithms, but that is unnecessary complexity. The system of equations approach we used is the most robust. Always look for the 'product-sum' relationship in G.P. problems—it is a recurring pattern in JEE Advanced.

Question

Let $a_1, a_2, a_3, \dots$ be a G. P. of increasing positive numbers. If $a_3 a_5 = 729$ and $a_2 + a_4 = \frac{111}{4}$, then $24 (a_1 + a_2 + a_3)$ is equal to

Options:

(A) 131

(B) 130

(C) 129

(D) 128

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, future engineers! Today, we are going to dissect a beautiful problem involving a Geometric Progression (G.P.). When you see a G.P. problem in the JEE Advanced, do not just start crunching numbers. First, pause and look at the 'DNA' of the sequence.

We define our terms as $a_1 = a$, $a_2 = ar$, $a_3 = ar^2$, $a_4 = ar^3$, and $a_5 = ar^4$. The problem gives us two critical constraints: the numbers are positive, and the sequence is increasing.

This immediately tells us that $a > 0$ and $r > 1$. If $r$ were less than $1$, the terms would shrink; if $r$ were negative, the terms would flip-flop between positive and negative. Keep these constraints in your pocket—they are the key to rejecting extraneous solutions later.

### The Product Trap

The problem provides the product of the third and fifth terms: $a_3 a_5 = 729$. Let us translate this into our algebraic language:

$$(ar^2)(ar^4) = 729$$

When you multiply these, remember the laws of exponents: $a \cdot a = a^2$ and $r^2 \cdot r^4 = r^6$. So, we have $a^2 r^6 = 729$.

Now, look at this expression. It is a perfect square! We can rewrite it as $(ar^3)^2 = 729$. Taking the square root of both sides, we get:

$$ar^3 = 27 \quad 	ext{--- (i)}$$

This is the 'Aha!' moment. We have reduced a complex product into a clean, linear-looking relationship.

### The Summation Bridge

Next, we are given the sum of the second and fourth terms: $a_2 + a_4 = \frac{111}{4}$. Translating this, we get:

$$ar + ar^3 = \frac{111}{4} \quad 	ext{--- (ii)}$$

Now, look at the beauty of the substitution method. We already know from Equation (i) that $ar^3 = 27$. Why solve for $a$ and $r$ separately when we can just plug this value in?

Substituting $27$ into our sum equation, we get $ar + 27 = \frac{111}{4}$. Subtracting $27$ from both sides, we find:

$$ar = \frac{111}{4} - 27$$

Calculating this, $27 	imes 4 = 108$, so $ar = \frac{111 - 108}{4} = \frac{3}{4}$. We label this Equation (iii):

$$ar = \frac{3}{4} \quad 	ext{--- (iii)}$$

### The Final Assembly

Now we have two simple equations: $ar^3 = 27$ and $ar = \frac{3}{4}$. To isolate $r$, we divide the first by the second:

$$\frac{ar^3}{ar} = \frac{27}{3/4}$$

The $a$ terms cancel out, leaving $r^2 = 27 	imes \frac{4}{3} = 9 	imes 4 = 36$. Since $r > 1$, we take the positive root:

$$r = 6$$

With $r$ in hand, finding $a$ is trivial. Using $ar = \frac{3}{4}$, we get $a(6) = \frac{3}{4}$, which simplifies to:

$$a = \frac{3}{24} = \frac{1}{8}$$

We have successfully decoded the sequence! The final task is to calculate $24(a_1 + a_2 + a_3)$, which is $24(a + ar + ar^2)$. Factoring out $a$, we get:

$$24a(1 + r + r^2)$$

Substituting our values:

$$24 	imes \frac{1}{8} 	imes (1 + 6 + 6^2) = 3 	imes (1 + 6 + 36) = 3 	imes 43 = 129$$

There it is! The logic holds, the math is clean, and the **final answer is 129**. Remember, in JEE, it is never just about the final number; it is about the elegance of the path you take to get there.

Animated Solution for Mathematics - Sequence and Series: Let $a_{1}, a_{2}, a_{3}, \dots$ be a G. P. of increasing positive numbers. If $a_{3} a_{5} = 729$ and $a_{2} + a_{4} = \frac{111}{4}$ , then $24 (a_{1} + a_{2} + a_{3})$ is equal to

Select Answer:

Visualized Solution

Defining the G.P. Terms

Analyzing the Product $a_{3} a_{5}$

Simplifying the Product Equation

Analyzing the Sum $a_{2} + a_{4}$

Substituting $a r^{3}$ into the Sum

Solving for $a r$

Finding the Common Ratio $r$

Finding the First Term $a$

Setting up the Final Expression

Final Calculation

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

Analyzing the Setup

The Product Trap

The Summation Bridge

The Final Assembly

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