Let an be the nth term of a G.P. of positive terms. If Σ(from n=1 to 100) a(2n+1) = 200 and Σ(from n

Q: Wait, why does the sum of odd terms start from $a_3$ and not $a_1$?

Great observation! Look at the index $2n+1$. When $n=1$, the term is $a_{2(1)+1} = a_3$. The summation notation is a strict instruction manual. It tells us exactly where to start and stop. Don't assume it starts at $a_1$ just because it's an 'odd' sum; always plug in the lower limit of the summation to see the first term.

Q: Why can't we just use the standard sum formula $S_n = \frac{a(r^n - 1)}{r-1}$?

You absolutely could, but it would be a trap! You have two unknowns, $a$ and $r$, and two equations. Solving that system algebraically is a nightmare of exponents. By using the property of the common ratio, we bypass the need to find the first term $a$ entirely. Always look for the 'ratio relationship' before diving into the standard formulas.

Q: How do I know if $r$ is positive or negative?

The problem statement explicitly mentions that the G.P. consists of 'positive terms'. If $r$ were negative, the terms would oscillate between positive and negative values. Since all terms are positive, $r$ must be positive. This is a subtle but crucial piece of information that validates our calculation.

Question

Let $a_n$ be the $n^{th}$ term of a G.P. of positive terms. If $\sum_{n=1}^{100} a_{2n+1} = 200$ and $\sum_{n=1}^{100} a_{2n} = 100$, then $\sum_{n=1}^{200} a_n$ is equal to:

Options:

(A) 175

(B) 150

(C) 300

(D) 225

Sigma Percentile · Accepted Answer

### The Hidden Symmetry of Geometric Progressions

Welcome, fellow traveler of the mathematical landscape. Today, we are not just solving a problem; we are uncovering the hidden symmetry within a Geometric Progression (G.P.).

When you first look at a problem involving sums of indices like $\sum_{n=1}^{100} a_{2n+1}$ and $\sum_{n=1}^{100} a_{2n}$, it is easy to feel overwhelmed. It looks like a mountain of algebra.

But I want you to take a breath. In the world of JEE Advanced, the most complex-looking problems often hide the most elegant, simple truths. Let us peel back the layers together.

### Phase 1: Decoding the Even Terms

We are given two pieces of information. First, the sum of the even-indexed terms is $100$. Let us write this out explicitly to see the pattern:

$$a_2 + a_4 + a_6 + \dots + a_{200} = 100$$

This is our anchor. It is a concrete value. Now, look at the second piece of information: the sum of the odd-indexed terms is $200$.

But wait—look at the index! It is $2n+1$. When $n=1$, we get $a_3$. When $n=2$, we get $a_5$. So the sum is:

$$a_3 + a_5 + a_7 + \dots + a_{201} = 200$$

### Phase 2: The Bridge of the Common Ratio

Here is where the magic happens. In any G.P., the relationship between consecutive terms is defined by the common ratio, $r$. Specifically, $a_{k+1} = a_k \cdot r$.

Look at our two sums side-by-side. If we take the even sum ($a_2 + a_4 + \dots + a_{200}$) and multiply every single term by $r$, what happens?

$$a_2 \cdot r = a_3$$
$$a_4 \cdot r = a_5$$
$$a_6 \cdot r = a_7$$

Do you see it? By multiplying the even sum by $r$, we generate the odd sum! Mathematically, we can express this as:

$$\sum_{n=1}^{100} a_{2n+1} = r \cdot \sum_{n=1}^{100} a_{2n}$$

Substituting our known values, we get $200 = r \cdot 100$. Instantly, the fog clears, and we find that **$r = 2$**. We have cracked the code of the progression.

### Phase 3: The Final Assembly

Our goal is to find the total sum of the first $200$ terms, which is $\sum_{n=1}^{200} a_n$. We can partition this total sum into two distinct groups: the sum of all odd-indexed terms and the sum of all even-indexed terms:

$$S_{total} = (a_1 + a_3 + \dots + a_{199}) + (a_2 + a_4 + \dots + a_{200})$$

We already know the second part—the even sum is $100$. We just need the 'true' odd sum, starting from $a_1$. Let us use the same ratio logic.

We know that:

$$(a_2 + a_4 + \dots + a_{200}) = r \cdot (a_1 + a_3 + \dots + a_{199})$$

Substituting $r=2$ and the even sum of $100$:

$$100 = 2 \cdot (	ext{Odd Sum})$$

Therefore, the sum of the odd terms is $50$.

### The Grand Finale

Now, we simply add our two components together. The total sum is the sum of the odd terms plus the sum of the even terms:

$$S_{total} = 50 + 100 = 150$$

And there it is. We didn't need to find the first term $a_1$, nor did we need to use the complex sum formula. We simply used the inherent symmetry of the progression.

Remember, in physics and math, always look for the relationship between the variables before you start calculating. You have the tools; now go forth and conquer the next problem with this same clarity. The final answer is **$150$**.

Animated Solution for Mathematics - Sequence and Series: Let $a_{n}$ be the $n^{t h}$ term of a G.P. of positive terms. If $\sum_{n = 1}^{100} a_{2 n + 1} = 200$ and $\sum_{n = 1}^{100} a_{2 n} = 100$ , then $\sum_{n = 1}^{200} a_{n}$ is equal to:

Select Answer:

Visualized Solution

Visualizing the Geometric Progression

The Sum of Even Terms

The Sum of Odd Terms (Shifted)

Relating the Two Sums

Extracting the Common Ratio $r$

Target: Sum of First $200$ Terms

Finding the First Odd Sum

Calculating the First Odd Sum

Final Calculation

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

The Hidden Symmetry of Geometric Progressions

Phase 1

Phase 2

Phase 3

The Grand Finale

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