Let a_1, a_2, a_3, ... be a G.P. of increasing positive terms. If a_1 a_5 = 28 and a_2 + a_4 = 29 th

Q: Wait, why do we reject the negative root for $r$ or the value less than 1?

Great question! The problem explicitly states that the geometric progression consists of 'increasing positive terms'. If $r$ were less than 1, the terms would decrease (e.g., $10, 5, 2.5...$). If $r$ were negative, the terms would oscillate between positive and negative. Since the problem demands an increasing sequence of positive terms, we must have $r > 1$. Always check your constraints against the physical reality of the problem!

Q: Why can't we just use logarithms to solve $a^2 r^4 = 28$?

You could, but it would be like using a sledgehammer to crack a nut! Logarithms are powerful, but here they would introduce unnecessary complexity. We have a beautiful algebraic structure where $a^2 r^4$ is a perfect square of $ar^2$. Recognizing that $a^2 r^4 = (ar^2)^2$ allows us to simplify the problem into a standard quadratic equation, which is much cleaner and less prone to calculation errors.

Q: Is it possible that $a_1$ is negative?

The problem states the terms are 'positive'. This is a vital constraint. If $a_1$ were negative, the entire sequence would be negative (since $r$ must be positive to keep the terms increasing and positive). Thus, $a_1$ must be positive. Always look for these small, seemingly insignificant words—they are often the keys to the entire solution.

Question

Let $a_{1},a_{2},a_{3},...$ be a G.P. of increasing positive terms. If $a_{1}a_{5}=28$ and $a_{2}+a_{4}=29$ then $a_{6}$ is equal to:

Options:

(A) 628

(B) 812

(C) 526

(D) 784

Sigma Percentile · Accepted Answer

### The Elegance of Geometric Progressions

Imagine you are standing at the start of a path, looking at a sequence of numbers that grow with a rhythmic, predictable intensity. This is the world of a Geometric Progression (G.P.).

In a G.P., each step forward is not just an addition, but a multiplication by a constant factor, the common ratio $r$. Today, we are going to unravel a mystery involving such a sequence.

We are given two clues: the product of the first and fifth terms is $28$, and the sum of the second and fourth terms is $29$. Our goal is to find the sixth term, $a_6$.

### Decoding the Constraints

First, we define our G.P. as $a, ar, ar^2, ar^3, ar^4, ar^5, \dots$. The problem states that the terms are positive and the sequence is strictly increasing.

This tells us immediately that $a > 0$ and $r > 1$. This is not just flavor text; it is the boundary condition that will guide us to the correct answer later.

### The Algebraic Dance

Let us translate our clues into the language of algebra. The first condition is $a_1 \cdot a_5 = 28$.

Substituting our G.P. terms, we get $a \cdot (ar^4) = 28$, which simplifies to $a^2 r^4 = 28$. Taking the square root of both sides, and knowing $a$ and $r$ are positive, we find:

$$ar^2 = \sqrt{28} = 2\sqrt{7}$$

Now, look at the second condition: $a_2 + a_4 = 29$. This translates to $ar + ar^3 = 29$.

If we factor out an $r$, we get $r(a + ar^2) = 29$. Notice the symmetry; we have an $ar^2$ term sitting right there, waiting to be replaced by our previous result.

### The Quadratic Crossroads

Substituting $ar^2 = 2\sqrt{7}$ into our sum equation, we get $r(a + 2\sqrt{7}) = 29$. We still have two variables, $a$ and $r$.

Let us isolate $a$ from our earlier finding: $a = \frac{2\sqrt{7}}{r^2}$. Now, substitute this into our sum equation:

$$r\left(\frac{2\sqrt{7}}{r^2} + 2\sqrt{7}\right) = 29$$

Distributing the $r$, we get $\frac{2\sqrt{7}}{r} + 2\sqrt{7}r = 29$. To clear the fraction, multiply the entire equation by $r$ and rearrange:

$$2\sqrt{7}r^2 - 29r + 2\sqrt{7} = 0$$

This is a classic quadratic equation in $r$. Using the quadratic formula $r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, we calculate the discriminant:

$$D = (-29)^2 - 4(2\sqrt{7})(2\sqrt{7}) = 841 - 112 = 729$$

The square root of $729$ is $27$. Thus, $r = \frac{29 \pm 27}{4\sqrt{7}}$. This gives us two potential values: $r = \frac{56}{4\sqrt{7}} = 2\sqrt{7}$ or $r = \frac{2}{4\sqrt{7}} = \frac{1}{2\sqrt{7}}$.

### The Final Ascent

Recall our constraint: the sequence is increasing, so $r > 1$. Since $2\sqrt{7} > 1$ and $\frac{1}{2\sqrt{7}} < 1$, we must reject the second value.

Our common ratio is $r = 2\sqrt{7}$. Now, finding $a$ is trivial:

$$a = \frac{2\sqrt{7}}{(2\sqrt{7})^2} = \frac{2\sqrt{7}}{28} = \frac{1}{2\sqrt{7}}$$

Finally, we calculate $a_6 = ar^5$. Substituting our values:

$$a_6 = \left(\frac{1}{2\sqrt{7}}\right) \cdot (2\sqrt{7})^5 = (2\sqrt{7})^4$$

Squaring $2\sqrt{7}$ gives $28$, and squaring $28$ gives $784$. The journey is complete, and the answer is **$784$**.

Animated Solution for Mathematics - Sequence and Series: Let $a_{1}, a_{2}, a_{3}, ...$ be a G.P. of increasing positive terms. If $a_{1} a_{5} = 28$ and $a_{2} + a_{4} = 29$ then $a_{6}$ is equal to:

Select Answer:

Visualized Solution

Defining the Geometric Progression

Analyzing the Product $a_{1} a_{5} = 28$

Setting up the Sum $a_{2} + a_{4} = 29$

Logic Bridge: Connecting the Equations

Expressing $a$ in terms of $r$

Simplifying to a Quadratic Equation

Solving the Quadratic Equation (Raw Setup)

Atomic Compute: The Discriminant

Finding the Values of $r$

Applying Constraints (The Trap)

Finding the First Term $a$

Setting up the Target $a_{6}$

Final Computation

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

The Elegance of Geometric Progressions

Decoding the Constraints

The Algebraic Dance

The Quadratic Crossroads

The Final Ascent

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