Let a1, a2, a3, ... be an A.P. If (a1+a2+...+a10)/(a1+a2+...+ap) = 100/p², p ≠ 10, then a11/a10 is e

Q: Wait, why can we just cancel $(p-10)$ from both sides? Is that always allowed?

That is a fantastic observation! In algebra, we can only divide by an expression if it is non-zero. The problem explicitly states $p \neq 10$, which is our 'green light' to divide by $(p-10)$. If $p$ were allowed to be 10, the ratio would be $100/100 = 1$, which is a trivial identity and wouldn't help us find the relationship between $a_1$ and $d$.

Q: Why did the sum formula simplify so nicely? Is it always this easy?

It feels like magic, doesn't it? The reason it simplifies is that the sum of an A.P. is a quadratic in $n$ of the form $S_n = An^2 + Bn$. When you are given a ratio like $\frac{S_n}{n^2} = \text{constant}$, it forces the linear term $B$ to be zero. This is a deep property of sequences that you'll start to recognize as you solve more problems!

Q: I got $d = 2a_1$. Does this mean the A.P. is fixed?

Not exactly! It means the A.P. belongs to a specific family of sequences where the common difference is twice the first term. For example, if $a_1 = 1$, then $d = 2$, and the sequence is $1, 3, 5, 7, \dots$. If $a_1 = 5$, then $d = 10$, and the sequence is $5, 15, 25, 35, \dots$. Notice that in both cases, the ratio $\frac{a_{11}}{a_{10}}$ remains constant because the $a_1$ terms cancel out!

Q: Can I use the property $S_n = n \times \text{average of first and last term}$ instead?

Absolutely! That is a brilliant way to think about it. Using $S_n = \frac{n}{2}(a_1 + a_n)$, you can write the ratio as $\frac{10(a_1 + a_{10})}{p(a_1 + a_p)} = \frac{100}{p^2}$. This is often faster and less prone to algebraic errors. Always look for the most elegant path!

Q: What if I made a mistake in the expansion? How do I check my work?

The best way to check is to pick a simple value for $p$ (other than 10). Let $p=1$. Then $\frac{S_{10}}{S_1} = \frac{100}{1^2} = 100$. Since $S_1 = a_1$, this means $S_{10} = 100a_1$. Using $S_{10} = 5(2a_1 + 9d)$, we get $10a_1 + 45d = 100a_1$, which simplifies to $45d = 90a_1$, or $d = 2a_1$. It matches perfectly!

Question

Let $a_1, a_2, a_3, \dots$ be an $A.P.$ If $\frac{a_1+a_2+\dots+a_{10}}{a_1+a_2+\dots+a_p} = \frac{100}{p^2}, p 
e 10$, then $\frac{a_{11}}{a_{10}}$ is equal to:

Options:

(A) $\frac{19}{21}$

(B) $\frac{100}{121}$

(C) $\frac{21}{19}$

(D) $\frac{121}{100}$

Sigma Percentile · Accepted Answer

### The Symphony of Sequences: Unlocking the A.P. Mystery

Welcome, fellow traveler on the road to JEE excellence. Today, we are not just solving a problem; we are peeling back the layers of an Arithmetic Progression (A.P.) to reveal the elegant structure hidden beneath.

When you look at a problem like this, it is easy to feel overwhelmed by the variables. But remember, mathematics is the art of finding patterns in chaos. Let us embark on this journey together.

### The Setup: Visualizing the Ratio

We are given an A.P. with terms $a_1, a_2, a_3, \dots$ and a curious ratio of their sums:

$$
\frac{S_{10}}{S_p} = \frac{100}{p^2}
$$

Imagine you are standing on a staircase where each step is higher than the last by a constant amount $d$. The sum $S_n$ represents the total height you have climbed after $n$ steps.

The problem tells us that the ratio of the total height at the 10th step to the height at the $p$-th step is exactly the ratio of the squares of the step numbers. This is a profound hint! It suggests that the sum is growing quadratically, which is the hallmark of an A.P. where the first term and the common difference are locked in a specific, beautiful dance.

### The Toolkit: Summoning the Sum Formula

To break this open, we reach for our most reliable tool: the sum formula for an A.P. $S_n = \frac{n}{2}[2a_1 + (n-1)d]$. We apply this to both $S_{10}$ and $S_p$.

When we substitute these into our ratio, we get:

$$
\frac{\frac{10}{2}[2a_1 + (10-1)d]}{\frac{p}{2}[2a_1 + (p-1)d]} = \frac{100}{p^2}
$$

Take a deep breath. I know it looks like a mess of variables, but look closely. The $\frac{1}{2}$ terms cancel out immediately. We are left with:

$$
\frac{10(2a_1 + 9d)}{p(2a_1 + (p-1)d)} = \frac{100}{p^2}
$$

### The Algebraic Dance: Simplifying the Chaos

Now, let us simplify. We can divide both sides by 10 and multiply by $p$. This leaves us with:

$$
\frac{2a_1 + 9d}{2a_1 + (p-1)d} = \frac{10}{p}
$$

This is the moment of truth. We cross-multiply to clear the denominators:

$$
p(2a_1 + 9d) = 10(2a_1 + (p-1)d)
$$

Expanding this, we get:

$$
2a_1p + 9dp = 20a_1 + 10dp - 10d
$$

Now, let us group the terms. Bring all the $a_1$ terms to one side and the $d$ terms to the other:

$$
2a_1p - 20a_1 = 10dp - 9dp - 10d
$$

Factorizing both sides, we find:

$$
2a_1(p - 10) = d(p - 10)
$$

Because the problem guarantees $p 
eq 10$, we can safely divide by $(p-10)$. The result is a stunningly simple relationship:

$$
d = 2a_1
$$

### The Grand Finale: Finding the Target

We have discovered the secret DNA of this sequence! Every term in this A.P. is governed by the rule $d = 2a_1$.

Now, we turn our attention to the goal: finding the ratio $\frac{a_{11}}{a_{10}}$. Using the general term formula $a_n = a_1 + (n-1)d$, we write:

$$
\frac{a_{11}}{a_{10}} = \frac{a_1 + 10d}{a_1 + 9d}
$$

Substitute our discovery, $d = 2a_1$, into this expression:

$$
\frac{a_{11}}{a_{10}} = \frac{a_1 + 10(2a_1)}{a_1 + 9(2a_1)} = \frac{a_1 + 20a_1}{a_1 + 18a_1}
$$

Finally, the $a_1$ terms cancel out, leaving us with the elegant result:

$$
\frac{a_{11}}{a_{10}} = \frac{21}{19}
$$

### Reflection

Look at what we have achieved. We started with a complex ratio and, through systematic simplification, uncovered a fundamental property of the sequence.

This is the essence of JEE Advanced mathematics—not just calculating, but understanding the underlying structure. You have mastered the logic, and now, you are ready for the next challenge. Keep pushing, keep questioning, and most importantly, keep falling in love with the process.

Animated Solution for Mathematics - Sequence and Series: Let $a_{1}, a_{2}, a_{3}, \dots$ be an $A . P .$ If $\frac{a _{1} + a _{2} + \dots + a _{10}}{a _{1} + a _{2} + \dots + a _{p}} = \frac{100}{p ^{2}}, p \neq = 10$ , then $\frac{a _{11}}{a _{10}}$ is equal to:

Select Answer:

Visualized Solution

Understand the Given Problem

Recall the Sum Formula for an A.P.

Substitute $n = 10$ and $n = p$

Simplify the Ratio

Cross-Multiply to Solve for $a_{1}$ and $d$

Expand the Expressions

Group and Factorize

Establish the Relation between $a_{1}$ and $d$

Express the Target Ratio

Substitute $d = 2 a_{1}$

Final Calculation

Conclusion and Key Takeaway

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

The Symphony of Sequences

The Setup

The Toolkit

The Algebraic Dance

The Grand Finale

Reflection

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