Let a1, a2, a3, ... be a G.P. of increasing positive numbers. Let the sum of its 6th and 8th terms b

Q: Wait, why can't we just find $a$ and $r$ separately? It feels like we should have two equations and two variables.

That is a very natural instinct! While you *could* solve for $a$ and $r$ individually, it is often a trap in JEE Advanced problems. Solving for them separately involves taking roots and dealing with potentially messy radicals. By treating the expression $ar^3$ as a single unit, we bypass the need to find $a$ and $r$ explicitly, making the algebra much cleaner and less prone to calculation errors.

Q: Why do we reject the negative root when solving for $r^2$?

Great catch! Remember that the problem defines the G.P. as having increasing positive numbers. If $r^2 = -3$, then $r$ would be an imaginary number ($\pm i\sqrt{3}$). Since the problem specifies that the terms are real numbers, $r$ must be real. The square of any real number must be non-negative, so $r^2 = -3$ is physically impossible in this context.

Q: Is there a trick to the final expression $6(a_2 + a_4)(a_4 + a_6)$?

Yes! The trick is to express everything in terms of the 'blocks' we already know: $ar^3$ and $r^2$. Instead of expanding everything, rewrite $a_2 + a_4$ as $ar^3(\frac{1}{r^2} + 1)$ and $a_4 + a_6$ as $ar^3(1 + r^2)$. This turns a complex-looking product into a simple multiplication of known values.

Question

Let $a_1, a_2, a_3, \dots$ be a G.P. of increasing positive numbers. Let the sum of its $6^{th}$ and $8^{th}$ terms be 2 and the product of its $3^{rd}$ and $5^{th}$ terms be $\frac{1}{9}$. Then $6(a_2 + a_4)(a_4 + a_6)$ is equal to

Options:

(A) 3

(B) $3\sqrt{3}$

(C) 2

(D) $2\sqrt{2}$

Sigma Percentile · Accepted Answer

### The Elegance of Algebraic Blocks

Welcome, future engineer! Today, we are going to dismantle a beautiful problem involving a Geometric Progression (G.P.).

Many students look at a problem like this and immediately reach for the standard formulas to find the first term $a$ and the common ratio $r$. While that is a valid path, it is often the 'scenic route'—long, winding, and full of potential pitfalls.

Instead, we are going to use the 'JEE Advanced' approach: identifying the hidden structure and manipulating it with surgical precision.

### Phase 1: Decoding the G.P.

We start with a G.P. where the terms are increasing and positive. This gives us two vital constraints: $a > 0$ and $r > 1$.

The general term is defined as $a_n = ar^{n-1}$. Our goal is to find the value of $6(a_2 + a_4)(a_4 + a_6)$.

### Phase 2: The Product Trap

The problem gives us the product of the third and fifth terms: $a_3 \cdot a_5 = \frac{1}{9}$. Substituting our general term formula, we get:

$$ (ar^2) \cdot (ar^4) = \frac{1}{9} $$

Simplifying this, we arrive at $a^2 r^6 = \frac{1}{9}$.

Here is the moment of insight. Notice that $a^2 r^6$ is a perfect square: $(ar^3)^2 = \frac{1}{9}$.

Since we know $a$ and $r$ are positive, we can safely take the square root to get $ar^3 = \frac{1}{3}$. This is our first 'block'. Keep this value safe; it is the key to the entire problem.

### Phase 3: The Sum Manipulation

Next, we look at the sum condition: $a_6 + a_8 = 2$. Substituting the general terms, we get $ar^5 + ar^7 = 2$.

Now, instead of solving for $a$, we look for our block $ar^3$. We can factor it out:

$$ ar^3(r^2 + r^4) = 2 $$

See the beauty? By substituting $ar^3 = \frac{1}{3}$, the equation becomes $\frac{1}{3}(r^2 + r^4) = 2$.

Multiplying by 3, we get $r^4 + r^2 = 6$, or $r^4 + r^2 - 6 = 0$. This is just a quadratic equation in disguise!

Letting $x = r^2$, we have $x^2 + x - 6 = 0$, which factors into $(x+3)(x-2) = 0$. Since $r^2$ must be positive, we reject $-3$ and accept $r^2 = 2$.

### Phase 4: The Final Assembly

Finally, we evaluate $6(a_2 + a_4)(a_4 + a_6)$. Expanding this, we get $6(ar + ar^3)(ar^3 + ar^5)$.

We can rewrite this using our blocks:

$$ 6\left( \frac{ar^3}{r^2} + ar^3 \right) (ar^3 + ar^3 \cdot r^2) $$

Substituting $ar^3 = \frac{1}{3}$ and $r^2 = 2$, the expression becomes:

$$ 6\left( \frac{1/3}{2} + \frac{1}{3} \right) \left( \frac{1}{3} + \frac{1}{3} \cdot 2 \right) $$

This simplifies to:

$$ 6 \left( \frac{1}{6} + \frac{1}{3} \right) \left( \frac{1}{3} + \frac{2}{3} \right) = 6 \left( \frac{1}{2} \right) (1) = 3 $$

And there you have it! By focusing on the structure rather than the individual variables, we turned a daunting algebraic expression into a simple, elegant calculation.

Keep this 'block' strategy in your toolkit—it is exactly how you conquer the toughest JEE problems. The final answer is **3**.

Select Answer:

Visualized Solution

Defining the G.P. Parameters

Analyzing the Product Condition

Simplifying the Product Equation

Analyzing the Sum Condition

Factoring the Sum Equation

Substituting the Known Value

Forming a Quadratic in $r^{2}$

Solving for $r^{2}$

Validating the Value of $r^{2}$

Setting Up the Target Expression

Simplifying the Target Expression

Final Substitution

Final Calculation

The Sigma Insight: Arithmetic Progression (A.P.)

The Elegance of Algebraic Blocks

Phase 1

Phase 2

Phase 3

Phase 4

Similar Questions

Suppose $a, b, c$ are in A.P. and $a^{2}, 2 b^{2}, c^{2}$ are in G.P. If $a < b < c$ and $a + b + c = 1$ , then $9 (a^{2} + b^{2} + c^{2})$ is equal to ......... .

Suppose $a, b, c$ are in A.P. and $a^{2}, b^{2}, c^{2}$ are in G.P. if $a < b < c$ and $a + b + c = 3/2$ , then the value of $a$ is

Let $a_{1}, a_{2}, \dots, a_{n}$ be in A.P. If $a_{5} = 2 a_{7}$ and $a_{11} = 18$ , then $12 (\frac{1}{a _{10} + a _{11}} + \frac{1}{a _{11} + a _{12}} + \dots + \frac{1}{a _{17} + a _{18}})$ is equal to

Let $T_{r}$ be the $r$ th term of an A.P., for $r = 1, 2, 3, \dots$ . If for some positive integers $m, n$ we have $T_{m} = 1/ n$ and $T_{n} = 1/ m$ , then $T_{mn}$ equals

Let $a_{1} = 8, a_{2}, a_{3}, \dots, a_{n}$ be an A.P. If the sum of its first four terms is 50 and the sum of its last four terms is 170, then the product of its middle two terms is

Let $a_{1}, a_{2}, a_{3} \dots$ be terms on A.P. If $\frac{a _{1} + a _{2} + \dots + a _{p}}{a _{1} + a _{2} + \dots + a _{q}} = \frac{p ^{2}}{q ^{2}}, p \neq = q$ , then $\frac{a _{6}}{a _{21}}$ equals

If $x = \sum_{n = 0}^{\infty} a^{n}, y = \sum_{n = 0}^{\infty} b^{n}, z = \sum_{n = 0}^{\infty} c^{n}$ , where $a, b, c$ are in A.P. and $∣ a ∣ < 1, ∣ b ∣ < 1, ∣ c ∣ < 1, ab c \neq = 0$ , then

Let $a_{1}, a_{2}, a_{3}, \dots, a_{n}$ be $n$ positive consecutive terms of an arithmetic progression. If $d > 0$ is its common difference, then $lim_{n \to \infty} \frac{d}{n} (\frac{1}{a _{1} + a _{2}} + \frac{1}{a _{2} + a _{3}} + \dots + \frac{1}{a _{n - 1} + a _{n}})$ is

In an A.P., the sixth terms $a_{6} = 2$ . If the $a_{1} a_{4} a_{5}$ is the greatest, then the common difference of the A.P., is equal to

The fourth power of the common difference of an arithmetic progression with integer entries is added to the product of any four consecutive terms of it. Prove that the resulting sum is the square of an integer.