Let the sum of an infinite G.P., whose first term is a and the common ratio is r, be 5. Let the sum

Q: Wait, why didn't we need to find the exact values of $a$ and $r$?

That is the beauty of this problem! Often in JEE Advanced, the question is designed to test your ability to recognize structural relationships rather than just brute-force calculation. By noticing that $S_{21}$ is a multiple of the middle term $a_{11}$, we bypass the need for the specific values of $a$ and $r$ entirely. It saves time and reduces the chance of calculation errors.

Q: Is there a shortcut for the sum of an A.P. with an odd number of terms?

Yes! For any A.P. with an odd number of terms $n$, the sum is always $n$ times the middle term. Since $21$ is odd, the middle term is the $\frac{21+1}{2} = 11$th term. Thus, $S_{21} = 21 \cdot a_{11}$. This is a powerful tool to keep in your arsenal.

Q: The problem gives $S_{\infty}$ and $S_5$. Do I always need to solve for $a$ and $r$ first?

Not always. While solving for $a$ and $r$ is a safe, standard approach, always pause for a moment to look at the final expression you need to find. If the final expression can be simplified or related to the given information without finding the variables, you can save precious minutes during the exam.

Question

Let the sum of an infinite G.P., whose first term is $a$ and the common ratio is $r$, be 5. Let the sum of its first five terms be $\frac{98}{25}$. Then the sum of the first 21 terms of an AP, whose first term is $10ar$, $n^{th}$ term is $a_n$ and the common difference is $10ar^2$, is equal to :

Options:

(A) $21 a_{11}$

(B) $22 a_{11}$

(C) $15 a_{16}$

(D) $14 a_{16}$

Sigma Percentile · Accepted Answer

### The Hidden Symmetry of Sequences

Welcome, fellow traveler on the road to JEE mastery. Today, we are going to dissect a problem that, at first glance, looks like a standard algebra grind.

You see an infinite geometric progression, a finite sum, and then a shift into an arithmetic progression. It is easy to feel the urge to immediately start solving for $a$ and $r$.

But I want you to pause. Take a deep breath. In the world of competitive physics and mathematics, the most elegant solutions often come from observing the structure before diving into the algebra.

### Phase 1: The Infinite G.P. Anchor

We start with an infinite geometric progression. We know the sum of an infinite G.P. is given by the elegant formula:

$$S_{\infty} = \frac{a}{1-r}, \quad |r| < 1$$

The problem gives us $S_{\infty} = 5$. This is our anchor.

We immediately have the relationship:

$$\frac{a}{1-r} = 5 \quad \Rightarrow \quad a = 5(1-r)$$

This is a solid foundation, but let's hold off on solving for $a$ and $r$ just yet.

### Phase 2: The Structural Shortcut

Next, we are given the sum of the first five terms:

$$S_5 = \frac{a(1-r^5)}{1-r} = \frac{98}{25}$$

Now, here is where the 'Aha!' moment happens. Instead of substituting $a = 5(1-r)$ and creating a mess, look at the expression for $S_5$.

It contains the term $\frac{a}{1-r}$! We already know this is equal to $5$. So, the equation becomes:

$$5(1-r^5) = \frac{98}{25}$$

This is a massive simplification. We are not just solving for variables; we are manipulating the structure of the problem to reveal its secrets.

### Phase 3: The A.P. Connection

Now, the problem shifts gears. We are introduced to an arithmetic progression with first term $A = 10ar$ and common difference $D = 10ar^2$.

We need the sum of the first $21$ terms, $S_{21}$. The formula for the sum of an A.P. is:

$$S_n = \frac{n}{2} [2A + (n-1)D]$$

Substituting $n=21$, we get:

$$S_{21} = \frac{21}{2} [2A + 20D]$$

Look at that expression inside the bracket: $2A + 20D$. If we factor out a $2$, we get $2(A + 10D)$.

The $2$ in the numerator and the $2$ in the denominator cancel out perfectly, leaving us with:

$$S_{21} = 21(A + 10D)$$

### The Grand Finale

Now, consider the general term of an A.P.: $a_n = A + (n-1)D$. If we calculate the $11$th term, $a_{11}$, we get:

$$a_{11} = A + (11-1)D = A + 10D$$

Do you see it? The expression inside our sum formula is exactly the $11$th term!

Therefore, **$S_{21} = 21 \cdot a_{11}$**.

We have arrived at the answer without ever needing to calculate the specific values of $a$ and $r$. The problem was designed to reward those who look for the underlying symmetry.

Remember, in JEE Advanced, your greatest tool is not just your calculator or your ability to solve equations—it is your ability to see the beauty in the structure. Keep practicing this, and you will find that these problems stop being obstacles and start being puzzles waiting to be solved.

Select Answer:

Visualized Solution

Infinite G.P. Parameters

Equation from Infinite Sum

Sum of First Five Terms

Structural Substitution

Solving for $r^{5}$

Defining the A.P.

Sum Formula for A.P.

Simplifying $S_{21}$

Relating to $a_{11}$

Final Answer

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

The Hidden Symmetry of Sequences

Phase 1

Phase 2

Phase 3

The Grand Finale

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