Let a1, a2, a3 ... be terms on A.P. If (a1 + a2 + ... + ap) / (a1 + a2 + ... + aq) = p² / q², p ≠ q,

Q: Wait, why can't we just solve for $a$ and $d$ individually? We have an equation, right?

That is a classic trap! You have one equation with two unknowns ($a$ and $d$). You cannot find the specific values of $a$ and $d$ because they are not unique—only their ratio matters. In JEE, we often look for relationships, not just raw values. We are solving for the ratio, which is an invariant property of this specific AP.

Q: Why did we divide by 2 in the middle of the problem?

Great observation. We did that to align our algebraic form with the target. The sum formula naturally gives us $2a$, but the term formula $a_n = a + (n-1)d$ starts with a single $a$. By dividing by $2$, we force the equation to speak the language of the term formula. It is a bridge between the 'sum world' and the 'term world'.

Q: Is there a shortcut for this? I feel like this takes too long.

Yes, there is a beautiful property! If $S_n$ is proportional to $n^2$, then the $n$-th term $a_n$ is proportional to $(2n-1)$. So, $\frac{a_6}{a_{21}} = \frac{2(6)-1}{2(21)-1} = \frac{11}{41}$. While this shortcut is powerful, understanding the algebraic derivation we did is crucial for when the problem gets more complex!

Question

Let $a_1, a_2, a_3 \dots$ be terms on A.P. If $\frac{a_1 + a_2 + \dots + a_p}{a_1 + a_2 + \dots + a_q} = \frac{p^2}{q^2}, p 
eq q$, then $\frac{a_6}{a_{21}}$ equals

Options:

(A) 41/11

(B) 7/2

(C) 2/7

(D) 11/41

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Every journey begins with a single step, and in Arithmetic Progressions (AP), that step is the sum formula. We know that the sum of the first $n$ terms is given by:

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

This formula is our bedrock. It tells us how the first term $a$ and the common difference $d$ conspire to create the sum.

When we look at the given condition $\frac{S_p}{S_q} = \frac{p^2}{q^2}$, we are essentially looking at a relationship between the sum of $p$ terms and the sum of $q$ terms. Let us write this out explicitly:

$$\frac{\frac{p}{2}[2a + (p-1)d]}{\frac{q}{2}[2a + (q-1)d]} = \frac{p^2}{q^2}$$

Take a deep breath. The $\frac{1}{2}$ cancels out immediately, leaving us with:

$$\frac{p[2a + (p-1)d]}{q[2a + (q-1)d]} = \frac{p^2}{q^2}$$

### The Algebraic Bridge

Now, we must be clever. Since $p$ and $q$ are indices of terms, they are non-zero integers. We can safely divide both sides by $\frac{p}{q}$ to simplify the expression:

$$\frac{2a + (p-1)d}{2a + (q-1)d} = \frac{p}{q}$$

This is the core relationship connecting the sum ratio to the term structure. However, we want to find the ratio of the sixth term to the twenty-first term, $\frac{a_6}{a_{21}}$.

We know that the $n$-th term is $a_n = a + (n-1)d$. Therefore, our target ratio is:

$$\frac{a_6}{a_{21}} = \frac{a + 5d}{a + 20d}$$

### The Transformation

Notice the discrepancy: our current equation has a $2a$ in the numerator and denominator, but our target has a single $a$. To align these, we divide the numerator and the denominator of our equation by $2$:

$$\frac{a + \frac{p-1}{2}d}{a + \frac{q-1}{2}d} = \frac{p}{q}$$

The structure is now identical to our target ratio $\frac{a + 5d}{a + 20d}$. For these two expressions to be equal, the coefficients of $d$ must match perfectly.

We set the coefficients as follows:

$$\frac{p-1}{2} = 5 \Rightarrow p-1 = 10 \Rightarrow p = 11$$

$$\frac{q-1}{2} = 20 \Rightarrow q-1 = 40 \Rightarrow q = 41$$

### Final Calculation

We have found the values of $p$ and $q$ that map our general sum ratio to our specific term ratio. Substituting these back into our equation, we find:

$$\frac{a_6}{a_{21}} = \frac{p}{q} = \frac{11}{41}$$

There it is! The final answer is **$\frac{11}{41}$**. It is the realization that the sum of an AP and the terms of an AP are deeply intertwined through these algebraic symmetries.

Animated Solution for Mathematics - Sequence and Series: 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

Select Answer:

Visualized Solution

Given Condition

Sum of $n$ terms

Ratio of Sums

Simplifying the Ratio

Further Simplification

Target Ratio: $\frac{a _{6}}{a _{21}}$

Comparing Expressions

The Substitution Trick

Finding the value of $p$

Finding the value of $q$

Final Calculation

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

Analyzing the Setup

The Algebraic Bridge

The Transformation

Final Calculation

Similar Questions

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Select Answer:

Visualized Solution

Given Condition

Sum of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } n terms

Ratio of Sums

Simplifying the Ratio

Further Simplification

Target Ratio: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a21​a6​​

Comparing Expressions

The Substitution Trick

Finding the value of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } p

Finding the value of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } q

Final Calculation

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

Analyzing the Setup

The Algebraic Bridge

The Transformation

Final Calculation

Similar Questions

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