JEE Advanced 2022

LEVELJEE Main

Animated Solution for Mathematics - Sequence and Series: Let $a_{1}, a_{2}, a_{3}, \dots$ be an arithmetic progression with $a_{1} = 7$ and common difference 8. Let $T_{1}, T_{2}, T_{3}, \dots$ be such that $T_{1} = 3$ and $T_{n + 1} - T_{n} = a_{n}$ for $n \geq 1$ . Then, which of the following is/are TRUE ?

Q: Wait, why is the summation in the telescoping sum from $1$ to $n-1$ and not $n$?

That is a fantastic question! Think about the indices: $T_2 - T_1 = a_1$, $T_3 - T_2 = a_2$, and so on, up to $T_n - T_{n-1} = a_{n-1}$. When you add these equations, the left side telescopes to $T_n - T_1$. Since the last term on the left is $T_n - T_{n-1}$, the corresponding term on the right must be $a_{n-1}$. Thus, the summation must stop at $n-1$.

Q: Why did we get a quadratic expression for $T_n$?

Think of the relationship $T_{n+1} - T_n = a_n$ as a discrete version of a derivative. If the 'derivative' (the difference) is a linear function ($a_n = 8n - 1$), then the 'integral' (the sum) must be a quadratic function. Just as the integral of $x$ is $x^2/2$, the sum of $n$ is $n^2/2$. This conceptual link between calculus and discrete series is a beautiful part of mathematics!

Select Answer:

* Multiple Correct

Visualized Solution

Given Sequences $a_{n}$ and $T_{n}$

Given AP: $a_{1} = 7$ , $d = 8$
Given Sequence: $T_{1} = 3$
Relation: $T_{n + 1} - T_{n} = a_{n}$
Formula for $n$ -th term of AP: $a_{n} = a_{1} + (n - 1) d$

General Term of the AP $a_{n}$

Substitute $a_{1} = 7$ and $d = 8$ :
$a_{n} = 7 + (n - 1) 8$
Expand and simplify:
$a_{n} = 8 n - 1$

Method of Differences for $T_{n}$

Telescoping Sum: $\sum_{k = 1}^{n - 1} (T_{k + 1} - T_{k}) = \sum_{k = 1}^{n - 1} a_{k}$
This simplifies to: $T_{n} - T_{1} = \sum_{k = 1}^{n - 1} a_{k}$
Substitute $T_{1} = 3$ : $T_{n} = 3 + \sum_{k = 1}^{n - 1} a_{k}$

Setup Summation for $T_{n}$

Substitute $a_{k} = 8 k - 1$ :
$T_{n} = 3 + \sum_{k = 1}^{n - 1} (8 k - 1)$
Split the summation:
$T_{n} = 3 + 8 \sum_{k = 1}^{n - 1} k - \sum_{k = 1}^{n - 1} 1$

Calculate General Term $T_{n}$

Apply sum formulas for first $(n - 1)$ natural numbers:
$T_{n} = 3 + 8 [\frac{( n - 1 ) n}{2}] - (n - 1)$
Simplify the expression:
$T_{n} = 3 + 4 n^{2} - 4 n - n + 1$
$T_{n} = 4 n^{2} - 5 n + 4$

Check Options A and C

Check Option A ( $T_{20}$ ): $T_{20} = 4 (20)^{2} - 5 (20) + 4 = 1600 - 100 + 4 = 1504$ (False)
Check Option C ( $T_{30}$ ): $T_{30} = 4 (30)^{2} - 5 (30) + 4 = 3600 - 150 + 4 = 3454$ (True)

Setup Sum $S_{n} = \sum T_{k}$

Setup sum for $S_{n} = \sum_{k = 1}^{n} T_{k}$ :
$S_{n} = \sum_{k = 1}^{n} (4 k^{2} - 5 k + 4)$
$S_{n} = 4 \sum_{k = 1}^{n} k^{2} - 5 \sum_{k = 1}^{n} k + 4 \sum_{k = 1}^{n} 1$

Apply Summation Formulas

Apply standard summation formulas:
$S_{n} = 4 [\frac{n ( n + 1 ) ( 2 n + 1 )}{6}] - 5 [\frac{n ( n + 1 )}{2}] + 4 n$

Check Option B ( $S_{20}$ )

Substitute $n = 20$ :
$S_{20} = 4 [\frac{20 \cdot 21 \cdot 41}{6}] - 5 [\frac{20 \cdot 21}{2}] + 4 (20)$
$S_{20} = 11480 - 1050 + 80 = 10510$
Option B is True

Check Option D ( $S_{30}$ )

Substitute $n = 30$ :
$S_{30} = 4 [\frac{30 \cdot 31 \cdot 61}{6}] - 5 [\frac{30 \cdot 31}{2}] + 4 (30)$
$S_{30} = 37820 - 2325 + 120 = 35615$
Option D is False

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The Sigma Insight: Sum of Special Series

Analyzing the Setup

We are given an arithmetic progression

a_{n}

with a first term

a_{1} = 7

and a common difference

d = 8

. The general term is defined as:

a_{n} = a_{1} + (n - 1) d = 7 + (n - 1) 8 = 8 n - 1

This linear growth serves as the foundation for the sequence

T_{n}

The Telescoping Bridge

The relationship

T_{n + 1} - T_{n} = a_{n}

defines a classic telescoping structure. By summing these terms from

k = 1

n - 1

, the intermediate terms cancel out:

T_{n} - T_{1} = k = 1 \sum n - 1 a_{k}

Given

T_{1} = 3

, we express

T_{n}

as:

T_{n} = 3 + k = 1 \sum n - 1 (8 k - 1)

We split the summation and apply the standard formula

\sum_{k = 1}^{m} k = \frac{m ( m + 1 )}{2}

with

m = n - 1

T_{n} = 3 + 8 [\frac{( n - 1 ) n}{2}] - (n - 1)

Simplifying the expression, we obtain the quadratic DNA of the sequence:

T_{n} = 3 + 4 n (n - 1) - n + 1 = 4 n^{2} - 5 n + 4

Testing the Waters

Using the master key

T_{n} = 4 n^{2} - 5 n + 4

, we evaluate specific terms to verify the given options. For

n = 20

T_{20} = 4 (20)^{2} - 5 (20) + 4 = 1600 - 100 + 4 = 1504

Since Option A claims

1604

, it is false. For

n = 30

T_{30} = 4 (30)^{2} - 5 (30) + 4 = 3600 - 150 + 4 = 3454

Thus, Option C is correct.

The Grand Summation

Finally, we calculate the sum

S_{n} = \sum_{k = 1}^{n} T_{k} = \sum_{k = 1}^{n} (4 k^{2} - 5 k + 4)

. Distributing the summation yields:

S_{n} = 4 k = 1 \sum n k^{2} - 5 k = 1 \sum n k + 4 k = 1 \sum n 1

Applying the standard power sum formulas, we get:

S_{n} = 4 [\frac{n ( n + 1 ) ( 2 n + 1 )}{6}] - 5 [\frac{n ( n + 1 )}{2}] + 4 n

For

n = 20

, substituting the values results in:

S_{20} = 4 [\frac{20 \cdot 21 \cdot 41}{6}] - 5 [\frac{20 \cdot 21}{2}] + 4 (20) = 11480 - 1050 + 80 = 10510

Consequently, Option B is true. For

n = 30

, the calculation yields

35615

, meaning Option D is false.

Common Traps & FAQs

Wait, why is the summation in the telescoping sum from

1

n - 1

and not

n

That is a fantastic question! Think about the indices:

T_{2} - T_{1} = a_{1}

T_{3} - T_{2} = a_{2}

, and so on, up to

T_{n} - T_{n - 1} = a_{n - 1}

. When you add these equations, the left side telescopes to

T_{n} - T_{1}

. Since the last term on the left is

T_{n} - T_{n - 1}

, the corresponding term on the right must be

a_{n - 1}

. Thus, the summation must stop at

n - 1

Can I use mathematical induction to verify the formula for

T_{n}

Absolutely! Induction is a powerful tool. Once you hypothesize that

T_{n} = 4 n^{2} - 5 n + 4

, you can check the base case

T_{1} = 4 (1)^{2} - 5 (1) + 4 = 3

, which matches. Then, assume it holds for

n

and show that

T_{n + 1} = T_{n} + a_{n} = (4 n^{2} - 5 n + 4) + (8 n - 1) = 4 n^{2} + 3 n + 3

. If you expand

4 (n + 1)^{2} - 5 (n + 1) + 4

, you will find it equals

4 n^{2} + 8 n + 4 - 5 n - 5 + 4 = 4 n^{2} + 3 n + 3

. It works perfectly!

Why did we get a quadratic expression for

T_{n}

Think of the relationship

T_{n + 1} - T_{n} = a_{n}

as a discrete version of a derivative. If the 'derivative' (the difference) is a linear function (

a_{n} = 8 n - 1

), then the 'integral' (the sum) must be a quadratic function. Just as the integral of

x

x^{2} /2

, the sum of

n

n^{2} /2

. This conceptual link between calculus and discrete series is a beautiful part of mathematics!

Select Answer:

Visualized Solution

Given Sequences $a_{n}$ and $T_{n}$

General Term of the AP $a_{n}$

Method of Differences for $T_{n}$

Setup Summation for $T_{n}$

Calculate General Term $T_{n}$

Check Options A and C

Setup Sum $S_{n} = \sum T_{k}$

Apply Summation Formulas

Check Option B ( $S_{20}$ )

Check Option D ( $S_{30}$ )

The Sigma Insight: Sum of Special Series

Analyzing the Setup

The Telescoping Bridge

Testing the Waters

The Grand Summation

Similar Questions

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Select Answer:

Visualized Solution

Given Sequences .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } an​ and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Tn​

General Term of the AP .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } an​

Method of Differences for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Tn​

Setup Summation for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Tn​

Calculate General Term .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Tn​

Check Options A and C

Setup Sum .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Sn​=∑Tk​

Apply Summation Formulas

Check Option B ( .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } S20​)

Check Option D ( .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } S30​)

The Sigma Insight: Sum of Special Series

Analyzing the Setup

The Telescoping Bridge

Testing the Waters

The Grand Summation

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Method of Differences for $T_{n}$

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Check Option D ( $S_{30}$ )

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Let ${a_{n}}_{n = 1}^{\infty}$ be a sequence such that $a_{1} = 1, a_{2} = 1$ and $a_{n + 2} = 2 a_{n + 1} + a_{n}$ for all $n \geq 1$ . Then the value of $47 \sum_{n = 1}^{\infty} \frac{a _{n}}{2 ^{3 n}}$ is equal to

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Let $a_{n}$ be $n^{t h}$ term of the series $5 + 8 + 14 + 23 + 35 + 50 + \dots$ and $S_{n} = \sum_{k = 1}^{n} a_{k}$ . Then $S_{30} - a_{40}$ is equal to

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