Let a1, a2, ....., a10 be a G.P. If a3/a1 = 25, then a9/a5 equals : (A) 2(5²) (B) 4(5²) (C) 5⁴ (D) 5

Q: Wait, why does $a_1$ cancel out? Is it always safe to assume $a_1 \neq 0$?

Great observation! In a standard Geometric Progression, the first term $a_1$ is non-zero by definition. If $a_1$ were zero, every term would be zero, and the ratio $\frac{a_3}{a_1}$ would be undefined. Since we are given a valid ratio of 25, we can safely assume $a_1 \neq 0$ and cancel it out.

Q: Do I need to find the value of $r$ itself? Can $r$ be negative?

You don't need to find $r$! Since $r^2 = 25$, $r$ could indeed be $5$ or $-5$. However, notice that the target ratio is $r^4$, which is $(r^2)^2$. Whether $r$ is $5$ or $-5$, $r^4$ will always be $625$. The beauty of this problem is that the sign of $r$ doesn't affect the final result.

Q: Is there a faster way to solve this without writing out the full formula?

Absolutely. You can use the property that in a GP, the indices of the terms in a ratio determine the power of $r$. Specifically, $\frac{a_m}{a_n} = r^{m-n}$. So, $\frac{a_3}{a_1} = r^{3-1} = r^2 = 25$, and $\frac{a_9}{a_5} = r^{9-5} = r^4$. This shortcut saves time and reduces algebraic errors!

Question

Let $a_1, a_2, \dots, a_{10}$ be a G.P. If $\frac{a_3}{a_1} = 25$, then $\frac{a_9}{a_5}$ equals :

Options:

(A) $2(5^2)$

(B) $4(5^2)$

(C) $5^4$

(D) $5^3$

Sigma Percentile · Accepted Answer

### The Elegance of Geometric Progressions

Welcome, fellow traveler on the path to JEE mastery! Today, we are going to peel back the layers of a classic problem involving Geometric Progressions (G.P.).

I know that when you see indices like $a_1, a_3, a_5, a_9$, it can feel like a dense forest of notation. But I want you to take a deep breath.

Mathematics is not about memorizing formulas; it is about seeing the underlying rhythm. A G.P. is simply a sequence where each step is a multiplication by a constant factor, $r$. That is the heartbeat of this entire problem.

### Phase 1: Decoding the DNA of the Sequence

Every G.P. is defined by its starting point, $a_1$, and its growth factor, $r$. The general term is $a_n = a_1 r^{n-1}$.

Think of this as your map. If you want to reach the $n$-th term, you start at $a_1$ and take $n-1$ steps of multiplication by $r$.

When we look at the given ratio, $\frac{a_3}{a_1} = 25$, we are essentially looking at the 'growth' that occurred between the first and the third term. Using our map:

$$a_3 = a_1 r^2$$

When we divide this by $a_1$, the $a_1$ terms vanish, leaving us with $r^2 = 25$. This is a powerful realization—the first term doesn't matter for the ratio; only the common ratio $r$ dictates the relationship between terms.

### Phase 2: The Target

Now, let's look at what we are asked to find: $\frac{a_9}{a_5}$. Using our map again:

$$a_9 = a_1 r^8 \quad 	ext{and} \quad a_5 = a_1 r^4$$

When we set up the ratio $\frac{a_1 r^8}{a_1 r^4}$, the $a_1$ terms cancel out again, just like magic. We are left with $\frac{r^8}{r^4}$.

By the laws of exponents, this simplifies beautifully to $r^{8-4} = r^4$. The ratio of any two terms in a G.P. is simply the common ratio raised to the power of the difference between their indices.

### Phase 3: The Final Synthesis

We have arrived at the finish line. We know that $r^2 = 25$, and we need to find $r^4$.

Algebraically, $r^4$ is just $(r^2)^2$. Substituting our known value, we get $25^2$.

Now, look at the options provided in the question. They are expressed in powers of $5$. We know that $25 = 5^2$. Therefore:

$$25^2 = (5^2)^2$$

Using the power of a power rule, $(x^a)^b = x^{ab}$, we get $5^{2 	imes 2} = 5^4$.

### Conclusion: The Beauty of Patterns

We started with a sequence, decoded its growth factor, and used the laws of exponents to bridge the gap between the given information and the target.

The answer, **$5^4$** (or **$625$**), is not just a number; it is the result of understanding the structure of the progression. Keep looking for these patterns, keep simplifying, and most importantly, keep enjoying the process.

Animated Solution for Mathematics - Sequence and Series: Let $a_{1}, a_{2}, \dots, a_{10}$ be a G.P. If $\frac{a _{3}}{a _{1}} = 25$ , then $\frac{a _{9}}{a _{5}}$ equals :

Select Answer:

Visualized Solution

Defining the Geometric Progression

Analyzing the Given Ratio

Finding the Common Ratio Relation

Setting up the Target Ratio

Simplifying the Target Ratio

Final Computation

Final Result

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

The Elegance of Geometric Progressions

Phase 1

Phase 2

Phase 3

Conclusion

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Animated Solution for Mathematics - Sequence and Series: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let a1​,a2​,…,a10​ be a G.P. If a1​a3​​=25, then a5​a9​​ equals :