Let a1, a2, a3, ... be G.P. of increasing positive terms such that a2 · a3 · a4 = 64 and a1 + a3 + a

Q: Wait, why do we reject the negative root or the fraction for $r^2$?

Great observation! The problem explicitly states that the G.P. consists of 'increasing positive terms'. This implies the common ratio $r$ must be greater than 1. If $r^2 = \frac{1}{28}$, then $r = \frac{1}{\sqrt{28}}$, which is less than 1, violating our constraint. Thus, we must have $r^2 = 28$.

Q: Why can't we just solve for $a$ and $r$ using the two equations?

You absolutely could, but it would be algebraically tedious. By using the symmetry of the G.P. (expressing terms relative to $a_3$), we avoid solving for $a$ directly and simplify the system into a single variable equation in $r^2$. In JEE, time is your most valuable asset—always look for symmetry!

Q: Is there a faster way to calculate $a_3 + a_5 + a_7$?

Yes! Once you have $a_3 = 4$ and $r^2 = 28$, notice that $a_3 + a_5 + a_7 = a_3(1 + r^2 + r^4)$. Instead of calculating $r^4$ separately, you can just plug in $r^2 = 28$ and compute $4(1 + 28 + 28^2)$. It's much cleaner.

Question

Let $a_1, a_2, a_3, \dots$ be G.P. of increasing positive terms such that $a_2 \cdot a_3 \cdot a_4 = 64$ and $a_1 + a_3 + a_5 = \frac{813}{7}$. Then $a_3 + a_5 + a_7$ is equal to :

Options:

(A) 3244

(B) 3248

(C) 3252

(D) 3256

Sigma Percentile · Accepted Answer

### The Elegance of Geometric Symmetry

Welcome, future engineer! Today, we are going to dive into a problem that, at first glance, might look like a standard algebra grind. But, as we peel back the layers, you will see that it is actually a beautiful dance of symmetry.

When you encounter a Geometric Progression (G.P.) in JEE Advanced, your first instinct should not be to blindly write out the general terms $a, ar, ar^2, \dots$. Instead, pause. Look for the structure. Look for the middle ground.

### Phase 1: The Product Trap

We are given a sequence of increasing positive terms where $a_2 \cdot a_3 \cdot a_4 = 64$. If you write this as $(ar) \cdot (ar^2) \cdot (ar^3) = 64$, you get $a^3 r^6 = 64$.

Now, look at that expression again. $a^3 r^6$ is exactly $(ar^2)^3$. And what is $ar^2$? It is the third term, $a_3$!

So, we have $(a_3)^3 = 64$. Taking the cube root, we immediately find $a_3 = 4$.

This is the 'spark'—the moment where the complexity collapses into simplicity. In any G.P., the product of three consecutive terms is always the cube of the middle term. Remember this; it is a powerful tool in your arsenal.

### Phase 2: The Summation Challenge

Now, we move to the second condition: $a_1 + a_3 + a_5 = \frac{813}{7}$. We know $a_3 = 4$.

Let's express $a_1$ and $a_5$ in terms of $a_3$ and the common ratio $r$. Since $a_3 = a_1 r^2$, we have $a_1 = \frac{a_3}{r^2}$. Similarly, $a_5 = a_3 r^2$.

Our equation becomes:

$$ \frac{a_3}{r^2} + a_3 + a_3 r^2 = \frac{813}{7} $$

Substituting $a_3 = 4$, we get:

$$ 4 \left( \frac{1}{r^2} + 1 + r^2 ight) = \frac{813}{7} $$

### Phase 3: The Algebraic Dance

Divide both sides by 4, and we are left with:

$$ \frac{1}{r^2} + 1 + r^2 = \frac{813}{28} $$

Subtracting 1 from both sides gives us:

$$ r^2 + \frac{1}{r^2} = \frac{813}{28} - 1 = \frac{785}{28} $$

Now, here is where the magic happens. Don't rush to solve a quadratic equation. Look at the fraction $\frac{785}{28}$.

Notice that $28^2 = 784$. So, $\frac{785}{28} = \frac{784 + 1}{28} = 28 + \frac{1}{28}$.

By simple observation, we see that $r^2$ must be 28 (or $\frac{1}{28}$). Given the sequence is increasing, $r > 1$, so $r^2 = 28$ is our only valid solution.

### Phase 4: The Final Victory

We need to find $a_3 + a_5 + a_7$. Using our symmetry trick again, this is $a_3 + a_3 r^2 + a_3 r^4 = a_3(1 + r^2 + r^4)$.

Substituting $a_3 = 4$ and $r^2 = 28$, we get:

$$ 4(1 + 28 + 28^2) $$

Since $28^2 = 784$, the expression inside the bracket is $1 + 28 + 784 = 813$.

Finally, $4 	imes 813 = 3252$. We have arrived at the destination! The final answer is **3252**.

### Conclusion

This problem wasn't about brute-force calculation; it was about recognizing the inherent symmetry of the G.P. Whenever you see products of consecutive terms, think of the middle term.

Whenever you see sums, think of expressing them relative to a known term. Keep practicing these patterns, and you will find that even the most intimidating JEE problems start to feel like old friends. Keep pushing, keep learning, and stay curious!

Animated Solution for Mathematics - Sequence and Series: Let $a_{1}, a_{2}, a_{3}, \dots$ be G.P. of increasing positive terms such that $a_{2} \cdot a_{3} \cdot a_{4} = 64$ and $a_{1} + a_{3} + a_{5} = \frac{813}{7}$ . Then $a_{3} + a_{5} + a_{7}$ is equal to :

Select Answer:

Visualized Solution

Define the G.P. Terms

Analyze the Product Condition

Find the Third Term $a_{3}$

Set up the Sum Equation

Simplify the Equation

Solve for $r^{2} + \frac{1}{r ^{2}}$

Determine the value of $r^{2}$

Set up the Final Expression

Final Calculation

Conclusion & Key Takeaway

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

The Elegance of Geometric Symmetry

Phase 1

Phase 2

Phase 3

Phase 4

Conclusion

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