JEE Advanced 2011

LEVELJEE Main

Animated Solution for Mathematics - Sequence and Series: Let $a_{1}, a_{2}, a_{3}, \dots, a_{100}$ be an arithmetic progression with $a_{1} = 3$ and $S_{p} = \sum_{i = 1}^{p} a_{i}, 1 \leq p \leq 100$ . For any integer $n$ with $1 \leq n \leq 20$ , let $m = 5 n$ . If $S_{m} / S_{n}$ does not depend on $n$ , then $a_{2}$ is \dots.

Q: Why can't we just pick a value for $n$ and solve for $d$?

That is a clever shortcut! If you pick $n=1$, you get $S_5/S_1 = \text{constant}$. This gives you an equation in $d$. However, in JEE Advanced, you must ensure that this $d$ works for *all* $n$. Picking one value is a necessary condition, but proving it works for all $n$ is the sufficient condition. By solving the general algebraic form, you guarantee it works for every $n$.

Q: Why did we set $6-d=0$? What if the ratio was something else?

Great question! We set $6-d=0$ because the ratio of the coefficients of $n$ was $25/1 = 25$, but the ratio of the constant terms was $5(6-d)/(6-d) = 5$. Since $25 \neq 5$, the only way to avoid a dependency on $n$ is to make the constant terms zero. If the ratio of coefficients had matched the ratio of constants, we wouldn't have needed to set them to zero!

Q: Is there a general property for $S_{kn}/S_n$?

Yes! For an AP, $S_{kn}/S_n$ is independent of $n$ if and only if the constant term in the sum formula is zero. This happens when $2a_1 - d = 0$, or $d = 2a_1$. In our case, $a_1=3$, so $d=2(3)=6$. It's a beautiful property that links the first term directly to the common difference!

Enter Numerical Value:

Visualized Solution

Given Conditions of the AP

Arithmetic Progression: $a_{1}, a_{2}, \dots, a_{100}$
First term: $a_{1} = 3$
Sum of first $p$ terms: $S_{p} = \sum_{i = 1}^{p} a_{i}$
Given condition: $\frac{S _{m}}{S _{n}}$ is independent of $n$ , where $m = 5 n$

Sum Formula for an AP

The sum of the first $p$ terms of an AP is given by:
$S_{p} = \frac{p}{2} [2 a_{1} + (p - 1) d]$
Here, $d$ is the common difference.

Substituting the First Term

We know the first term $a_{1} = 3$ .
Substituting this into our sum formula:
$S_{p} = \frac{p}{2} [2 (3) + (p - 1) d]$
$S_{p} = \frac{p}{2} [6 + (p - 1) d]$

Setting up the Ratio $\frac{S _{m}}{S _{n}}$

We need to analyze the ratio $\frac{S _{m}}{S _{n}}$ .
We are given that $m = 5 n$ .
So, the ratio becomes $\frac{S _{5 n}}{S _{n}}$ .

Expanding the Ratio

Substitute $p = 5 n$ for the numerator and $p = n$ for the denominator.
Numerator: $S_{5 n} = \frac{5 n}{2} [6 + (5 n - 1) d]$
Denominator: $S_{n} = \frac{n}{2} [6 + (n - 1) d]$
Ratio: $\frac{S _{m}}{S _{n}} = \frac{\frac{5 n}{2} [ 6 + ( 5 n - 1 ) d ]}{\frac{n}{2} [ 6 + ( n - 1 ) d ]}$

Simplifying the Ratio

Notice the common factor of $\frac{n}{2}$ in both the numerator and denominator.
Canceling $\frac{n}{2}$ gives:
$\frac{S _{m}}{S _{n}} = \frac{5 [ 6 + ( 5 n - 1 ) d ]}{6 + ( n - 1 ) d}$

Rearranging Terms

Expand the inner brackets and group the terms with $n$ and the constant terms.
Numerator: $5 [6 + 5 n d - d] = 5 [(6 - d) + 5 n d]$
Denominator: $6 + n d - d = (6 - d) + n d$
$\frac{S _{m}}{S _{n}} = 5 \frac{( 6 - d ) + 5 n d}{( 6 - d ) + n d}$

Condition for Independence from $n$

For this ratio to be independent of $n$ , it must be a constant.
A rational function $\frac{A + B n}{C + D n}$ is constant if $\frac{A}{C} = \frac{B}{D}$ .
Here, $\frac{5 ( 6 - d )}{6 - d} = \frac{25 d}{d} ⟹ 5 = 25$ , which is impossible!
The only exception is if the constant terms are zero, making it $\frac{25 n d}{n d} = 25$ .

Equating Constant Term to Zero

Therefore, the constant term $(6 - d)$ must be equal to zero.
$6 - d = 0$

Solving for $d$

Solving for $d$ :
$d = 6$
If $d = 6$ , the ratio becomes $5 \frac{5 n d}{n d} = 25$ , which is indeed independent of $n$ .

Finding the Second Term $a_{2}$

We need to find the second term $a_{2}$ .
$a_{2} = a_{1} + d$
$a_{2} = 3 + 6 = 9$

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

Solution Diagram

Analyzing the Setup

We are given an arithmetic progression (AP) where the first term is

a_{1} = 3

and the common difference is

d

. We are told that the ratio of the sum of the first

5 n

terms to the sum of the first

n

terms is independent of

n

This implies that for any positive integer

n

, the ratio

\frac{S _{5 n}}{S _{n}}

remains a constant value. Our goal is to determine the value of

d

that satisfies this condition.

The Tool of the Trade

To solve this, we utilize the standard formula for the sum of the first

p

terms of an AP:

S_{p} = \frac{p}{2} [2 a_{1} + (p - 1) d]

Substituting the known value

a_{1} = 3

into this formula, we obtain:

S_{p} = \frac{p}{2} [6 + (p - 1) d]

The Algebraic Challenge

We now construct the ratio

\frac{S _{5 n}}{S _{n}}

using our formula:

\frac{S _{5 n}}{S _{n}} = \frac{\frac{5 n}{2} [ 6 + ( 5 n - 1 ) d ]}{\frac{n}{2} [ 6 + ( n - 1 ) d ]}

By canceling the common factor

\frac{n}{2}

from the numerator and the denominator, the expression simplifies to:

\frac{S _{5 n}}{S _{n}} = 5 \cdot \frac{6 + ( 5 n - 1 ) d}{6 + ( n - 1 ) d}

To analyze the dependency on

n

, we expand the terms inside the brackets:

\frac{S _{5 n}}{S _{n}} = 5 \cdot \frac{( 6 - d ) + 5 n d}{( 6 - d ) + n d}

The 'Aha!' Moment

For the expression to be independent of

n

, the variable

n

must be eliminated from the ratio. This occurs if the constant terms in the numerator and denominator are zero, specifically when

(6 - d) = 0

(6 - d) = 0

, the expression simplifies beautifully:

\frac{S _{5 n}}{S _{n}} = 5 \cdot \frac{5 n d}{n d} = 5 \cdot 5 = 25

Since

25

is a constant, the condition of independence is satisfied.

Final Calculation

Setting the constant term to zero, we find:

6 - d = 0 \Rightarrow d = 6

The problem asks for the second term of the progression,

a_{2}

. Using the definition

a_{2} = a_{1} + d

a_{2} = 3 + 6 = 9

The common difference is

6

, and the second term of the sequence is 9.

Common Traps & FAQs

Why can't we just pick a value for

n

and solve for

d

That is a clever shortcut! If you pick

n = 1

, you get

S_{5} / S_{1} = constant

. This gives you an equation in

d

. However, in JEE Advanced, you must ensure that this

d

works for *all*

n

. Picking one value is a necessary condition, but proving it works for all

n

is the sufficient condition. By solving the general algebraic form, you guarantee it works for every

n

Why did we set

6 - d = 0

? What if the ratio was something else?

Great question! We set

6 - d = 0

because the ratio of the coefficients of

n

was

25/1 = 25

, but the ratio of the constant terms was

5 (6 - d) / (6 - d) = 5

. Since

25 \neq = 5

, the only way to avoid a dependency on

n

is to make the constant terms zero. If the ratio of coefficients had matched the ratio of constants, we wouldn't have needed to set them to zero!

Is there a general property for

S_{k n} / S_{n}

Yes! For an AP,

S_{k n} / S_{n}

is independent of

n

if and only if the constant term in the sum formula is zero. This happens when

2 a_{1} - d = 0

, or

d = 2 a_{1}

. In our case,

a_{1} = 3

, so

d = 2 (3) = 6

. It's a beautiful property that links the first term directly to the common difference!

Enter Numerical Value:

Visualized Solution

Given Conditions of the AP

Sum Formula for an AP

Substituting the First Term

Setting up the Ratio $\frac{S _{m}}{S _{n}}$

Expanding the Ratio

Simplifying the Ratio

Rearranging Terms

Condition for Independence from $n$

Equating Constant Term to Zero

Solving for $d$

Finding the Second Term $a_{2}$

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

Analyzing the Setup

The Tool of the Trade

The Algebraic Challenge

The 'Aha!' Moment

Final Calculation

Similar Questions

Let $a_{1}, a_{2}, a_{3}, \dots, a_{n}$ be $n$ positive consecutive terms of an arithmetic progression. If $d > 0$ is its common difference, then $lim_{n \to \infty} \frac{d}{n} (\frac{1}{a _{1} + a _{2}} + \frac{1}{a _{2} + a _{3}} + \dots + \frac{1}{a _{n - 1} + a _{n}})$ is

Suppose $a_{1}, a_{2}, \dots, a_{n}, \dots$ be an arithmetic progression of natural numbers. If the ratio of the sum of the first five terms to the sum of first nine terms of the progression is $5 : 17$ and $110 < a_{15} < 120$ , then the sum of the first ten terms of the progression is equal to -

Let $S_{n}$ be the sum of the first $n$ terms of an arithmetic progression. If $S_{3 n} = 3 S_{2 n}$ , then the value of $\frac{S _{4 n}}{S _{2 n}}$ is :

Let $S_{n}$ denote the sum of the first $n$ terms of an arithmetic progression. If $S_{10} = 390$ and the ratio of the tenth and the fifth terms is $15 : 7$ , then $S_{15} - S_{5}$ is equal to:

If $lo g_{3} 2, lo g_{3} (2^{x} - 5)$ , and $lo g_{3} (2^{x} - \frac{7}{2})$ are in arithmetic progression, determine the value of $x$ .

Let $T_{r}$ be the rth term of an A.P. If for some m, $T_{m} = \frac{1}{25}$ and $T_{25} = \frac{1}{20}$ , then $5 m \sum_{r = m}^{2 m} T_{r}$ is equal to Note: $20 \sum_{r = 1}^{25} T_{r} = 13$ is given condition in Hindi version.

Let $3, 7, 11, 15, \dots, 403$ and $2, 5, 8, 11, \dots, 404$ be two arithmetic progressions. Then the sum, of the common terms in them, is equal to

Let $S_{n}$ denote the sum of first $n$ terms an arithmetic progression. If $S_{20} = 790$ and $S_{10} = 145$ , then $S_{15} - S_{5}$ is :

Let $T_{r}$ be the $r$ th term of an A.P., for $r = 1, 2, 3, \dots$ . If for some positive integers $m, n$ we have $T_{m} = 1/ n$ and $T_{n} = 1/ m$ , then $T_{mn}$ equals

The sum of the common terms of the following three arithmetic progressions. $3, 7, 11, 15, \dots, 399$ , $2, 5, 8, 11, \dots, 359$ and $2, 7, 12, 17, \dots, 197$ , is equal to

Enter Numerical Value:

Visualized Solution

Given Conditions of the AP

Sum Formula for an AP

Substituting the First Term

Setting up the Ratio .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Sn​Sm​​

Expanding the Ratio

Simplifying the Ratio

Rearranging Terms

Condition for Independence from .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } n

Equating Constant Term to Zero

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } d

Finding the Second Term .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a2​

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

Analyzing the Setup

The Tool of the Trade

The Algebraic Challenge

The 'Aha!' Moment

Final Calculation

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let Sn​ be the sum of the first n terms of an arithmetic progression. If S3n​=3S2n​, then the value of S2n​S4n​​ is :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If log3​2,log3​(2x−5), and log3​(2x−27​) are in arithmetic progression, determine the value of x.

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let 3,7,11,15,…,403 and 2,5,8,11,…,404 be two arithmetic progressions. Then the sum, of the common terms in them, is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let Sn​ denote the sum of first n terms an arithmetic progression. If S20​=790 and S10​=145, then S15​−S5​ is :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let Tr​ be the rth term of an A.P., for r=1,2,3,…. If for some positive integers m,n we have Tm​=1/n and Tn​=1/m, then Tmn​ equals

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The sum of the common terms of the following three arithmetic progressions. 3,7,11,15,…,399, 2,5,8,11,…,359 and 2,7,12,17,…,197, is equal to

Setting up the Ratio $\frac{S _{m}}{S _{n}}$

Condition for Independence from $n$

Solving for $d$

Finding the Second Term $a_{2}$

Let $S_{n}$ be the sum of the first $n$ terms of an arithmetic progression. If $S_{3 n} = 3 S_{2 n}$ , then the value of $\frac{S _{4 n}}{S _{2 n}}$ is :

If $lo g_{3} 2, lo g_{3} (2^{x} - 5)$ , and $lo g_{3} (2^{x} - \frac{7}{2})$ are in arithmetic progression, determine the value of $x$ .

Let $3, 7, 11, 15, \dots, 403$ and $2, 5, 8, 11, \dots, 404$ be two arithmetic progressions. Then the sum, of the common terms in them, is equal to

Let $S_{n}$ denote the sum of first $n$ terms an arithmetic progression. If $S_{20} = 790$ and $S_{10} = 145$ , then $S_{15} - S_{5}$ is :

Let $T_{r}$ be the $r$ th term of an A.P., for $r = 1, 2, 3, \dots$ . If for some positive integers $m, n$ we have $T_{m} = 1/ n$ and $T_{n} = 1/ m$ , then $T_{mn}$ equals

The sum of the common terms of the following three arithmetic progressions. $3, 7, 11, 15, \dots, 399$ , $2, 5, 8, 11, \dots, 359$ and $2, 7, 12, 17, \dots, 197$ , is equal to