JEE Main 2022 (25 June Shift 1)

LEVELBoard

Animated Solution for Mathematics - Sequence and Series: For a natural number $n$ , let $a_{n} = 1 9^{n} - 1 2^{n}$ . Then, the value of $\frac{31 a _{9} - a _{10}}{57 a _{8}}$ is

Q: Where did the $31$ come from, and why is it important?

Great observation! The number $31$ is not random; it is $19 + 12$. When you expand the numerator, you will see terms like $31 \cdot 19^9 - 19^{10}$. By writing $31$ as $(19 + 12)$, you get $(19 + 12) \cdot 19^9 - 19^{10}$, which simplifies beautifully to $19^{10} + 12 \cdot 19^9 - 19^{10} = 12 \cdot 19^9$. This is the 'aha!' moment that collapses the entire expression.

Q: Is there a general recurrence relation for this sequence?

Yes, absolutely! For any sequence of the form $a_n = p^n - q^n$, it satisfies the linear recurrence $a_{n+1} = (p+q)a_n - (pq)a_{n-1}$. Here, $p=19$ and $q=12$, so $p+q=31$ and $pq=228$. Thus, $a_{10} = 31a_9 - 228a_8$. Rearranging this gives $31a_9 - a_{10} = 228a_8$. This confirms our algebraic result instantly!

Enter Numerical Value:

Progressive Hints

Hint 1

The Spark: Do not attempt to calculate the massive powers

1 9^{9}

1 2^{9}

. Instead, look at the structure of the expression. Notice that the coefficient

31

is actually the sum of the two bases,

19 + 12

. This is the hidden key to unlocking the problem.

Hint 2

The Arsenal: Use the distributive property of multiplication over subtraction. Expand the numerator

31 a_{9} - a_{10}

by substituting

a_{n} = 1 9^{n} - 1 2^{n}

. Once expanded, group the terms with base

19

together and the terms with base

12

together.

Hint 3

The Execution: After grouping, you will find that you can factor out

1 9^{9}

from the base-19 terms and

- 1 2^{9}

from the base-12 terms. This will reveal a common factor of

(1 9^{8} - 1 2^{8})

, which is exactly

a_{8}

. Start your algebraic journey with the expression:

31 (1 9^{9} - 1 2^{9}) - (1 9^{10} - 1 2^{10})

Visualized Solution

$a_{n} = 1 9^{n} - 1 2^{n}$

Given sequence: $a_{n} = 1 9^{n} - 1 2^{n}$
Target expression: $\frac{31 a _{9} - a _{10}}{57 a _{8}}$

$Focus on Numerator$

Let's isolate the numerator: $31 a_{9} - a_{10}$

$Substitute a_{9} and a_{10}$

Substitute $n = 9$ : $a_{9} = 1 9^{9} - 1 2^{9}$
Substitute $n = 10$ : $a_{10} = 1 9^{10} - 1 2^{10}$
Raw substitution: $31 (1 9^{9} - 1 2^{9}) - (1 9^{10} - 1 2^{10})$

$Expand the Brackets$

Expand the terms: $31 \cdot 1 9^{9} - 31 \cdot 1 2^{9} - 1 9^{10} + 1 2^{10}$

$Group Powers of 19$

Group terms with base $19$ : $31 \cdot 1 9^{9} - 1 9^{10}$
Factor out $1 9^{9}$ : $1 9^{9} (31 - 19)$

$Group Powers of 12$

Group terms with base $12$ : $- 31 \cdot 1 2^{9} + 1 2^{10}$
Factor out $- 1 2^{9}$ : $- 1 2^{9} (31 - 12)$

$Simplify the Coefficients$

Calculate $31 - 19 = 12$
Calculate $31 - 12 = 19$
Substitute back: $1 9^{9} (12) - 1 2^{9} (19)$

$Factor Out Common Terms$

Notice the common factor: $19 \cdot 12$
Factor it out: $(19 \cdot 12) (1 9^{8} - 1 2^{8})$

$Recognize a_{8}$

Recall the definition: $a_{n} = 1 9^{n} - 1 2^{n}$
Therefore, $a_{8} = 1 9^{8} - 1 2^{8}$
Numerator becomes: $(19 \cdot 12) \cdot a_{8} = 228 a_{8}$

$Final Calculation$

Original fraction: $\frac{31 a _{9} - a _{10}}{57 a _{8}}$
Substitute simplified numerator: $\frac{228 a _{8}}{57 a _{8}}$
Cancel $a_{8}$ : $\frac{228}{57}$
Final Answer: $4$

00:00 / 00:00

The Sigma Insight: Sum of Special Series

Analyzing the Setup

The problem asks us to evaluate the expression:

\frac{31 a _{9} - a _{10}}{57 a _{8}}

where the sequence is defined by

a_{n} = 1 9^{n} - 1 2^{n}

. The sheer size of these powers is designed to intimidate, but we must look past the numbers to find the underlying algebraic structure.

The Algebraic Soul

We begin by focusing solely on the numerator:

31 a_{9} - a_{10}

. Substituting the definitions of

a_{9}

and

a_{10}

, we obtain:

31 (1 9^{9} - 1 2^{9}) - (1 9^{10} - 1 2^{10})

Expanding this expression carefully, we get:

31 \cdot 1 9^{9} - 31 \cdot 1 2^{9} - 1 9^{10} + 1 2^{10}

The Beauty of Symmetry

To simplify, we group the terms by their respective bases. For the base

19

terms, we have

31 \cdot 1 9^{9} - 1 9^{10}

, which factors as:

1 9^{9} (31 - 19) = 1 9^{9} (12)

For the base

12

terms, we have

- 31 \cdot 1 2^{9} + 1 2^{10}

, which factors as:

- 1 2^{9} (31 - 12) = - 1 2^{9} (19)

Combining these results, our numerator becomes:

12 \cdot 1 9^{9} - 19 \cdot 1 2^{9}

The Final Cancellation

We observe that both terms share a common factor of

19 \cdot 12

. Factoring this out, we get:

(19 \cdot 12) (1 9^{8} - 1 2^{8})

Since

1 9^{8} - 1 2^{8} = a_{8}

, the numerator simplifies to

228 \cdot a_{8}

. Returning to our original fraction:

\frac{228 \cdot a _{8}}{57 \cdot a _{8}}

The

a_{8}

terms cancel out, leaving us with the final calculation:

\frac{228}{57} = 4

The final answer is 4.

Common Traps & FAQs

Wait, why can't I just calculate

1 9^{9}

and

1 2^{9}

directly?

That is the classic trap! In JEE Advanced, you are rarely expected to perform brute-force calculations with such large numbers. The problem is designed to test your ability to spot algebraic patterns. If you try to calculate these values, you will waste precious time and likely make a calculation error. Always look for the underlying structure first.

Where did the

31

come from, and why is it important?

Great observation! The number

31

is not random; it is

19 + 12

. When you expand the numerator, you will see terms like

31 \cdot 1 9^{9} - 1 9^{10}

. By writing

31

(19 + 12)

, you get

(19 + 12) \cdot 1 9^{9} - 1 9^{10}

, which simplifies beautifully to

1 9^{10} + 12 \cdot 1 9^{9} - 1 9^{10} = 12 \cdot 1 9^{9}

. This is the 'aha!' moment that collapses the entire expression.

Is there a general recurrence relation for this sequence?

Yes, absolutely! For any sequence of the form

a_{n} = p^{n} - q^{n}

, it satisfies the linear recurrence

a_{n + 1} = (p + q) a_{n} - (pq) a_{n - 1}

. Here,

p = 19

and

q = 12

, so

p + q = 31

and

pq = 228

. Thus,

a_{10} = 31 a_{9} - 228 a_{8}

. Rearranging this gives

31 a_{9} - a_{10} = 228 a_{8}

. This confirms our algebraic result instantly!

Animated Solution for Mathematics - Sequence and Series: For a natural number $n$ , let $a_{n} = 1 9^{n} - 1 2^{n}$ . Then, the value of $\frac{31 a _{9} - a _{10}}{57 a _{8}}$ is

Enter Numerical Value:

Visualized Solution

$a_{n} = 1 9^{n} - 1 2^{n}$

$Focus on Numerator$

$Substitute a_{9} and a_{10}$

$Expand the Brackets$

$Group Powers of 19$

$Group Powers of 12$

$Simplify the Coefficients$

$Factor Out Common Terms$

$Recognize a_{8}$

$Final Calculation$

The Sigma Insight: Sum of Special Series

Analyzing the Setup

The Algebraic Soul

The Beauty of Symmetry

The Final Cancellation

Similar Questions

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

Let ${a_{n}}_{n = 0}^{\infty}$ be a sequence such that $a_{0} = a_{1} = 0$ and $a_{n + 2} = 2 a_{n + 1} - a_{n} + 1$ for all $n \geq 0$ . Then, $\sum_{n = 2}^{\infty} \frac{a _{n}}{7 ^{n}}$ is equal to

For positive integers n, if $4 a_{n} = (n^{2} + 5 n + 6)$ and $S_{n} = \sum_{k = 1}^{n} (\frac{1}{a _{k}})$ , then the value of $507 S_{2025}$ is:

If $a_{n} = \frac{- 2}{4 n ^{2} - 16 n + 15}$ , then $a_{1} + a_{2} + \dots + a_{25}$ is equal to :

Let $S_{n} = \sum_{k = 1}^{4 n} (- 1)^{\frac{k ( k + 1 )}{2}} k^{2}$ . Then $S_{n}$ can take value(s)

For $k \in N$ , if the sum of the series $1 + \frac{4}{k} + \frac{8}{k ^{2}} + \frac{13}{k ^{3}} + \frac{19}{k ^{4}} + \dots$ is 10, then the value of $k$ is

If $\frac{1 ^{3} + 2 ^{3} + 3 ^{3} + \dots upto n terms}{1 \cdot 3 + 2 \cdot 5 + 3 \cdot 7 + \dots upto n terms} = \frac{9}{5}$ , then the value of $n$ is

Let $⟨ a_{n} ⟩$ be a sequence such that $a_{0} = 0$ , $a_{1} = \frac{1}{2}$ and $2 a_{n + 2} = 5 a_{n + 1} - 3 a_{n}$ , $n = 0, 1, 2, 3, ...$ Then $\sum_{k = 1}^{100} a_{k}$ is

Let ${a_{n}}_{n = 0}^{\infty}$ be a sequence such that $a_{0} = a_{1} = 0$ and $a_{n + 2} = 3 a_{n + 1} - 2 a_{n} + 1, \forall n \geq 0$ . Then $a_{25} a_{23} - 2 a_{25} a_{22} - 2 a_{23} a_{24} + 4 a_{22} a_{24}$ is equal to:

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

Animated Solution for Mathematics - Sequence and Series: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } For a natural number n, let an​=19n−12n. Then, the value of 57a8​31a9​−a10​​ is

Enter Numerical Value:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } an​=19n−12n

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Focus on Numerator

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Substitute a9​ and a10​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Expand the Brackets

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Group Powers of 19

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Group Powers of 12

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Simplify the Coefficients

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Factor Out Common Terms

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Recognize a8​

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

The Sigma Insight: Sum of Special Series

Analyzing the Setup

The Algebraic Soul

The Beauty of Symmetry

The Final Cancellation

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let {an​}n=1∞​ be a sequence such that a1​=1,a2​=1 and an+2​=2an+1​+an​ for all n≥1. Then the value of 47∑n=1∞​23nan​​ is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let {an​}n=0∞​ be a sequence such that a0​=a1​=0 and an+2​=2an+1​−an​+1 for all n≥0. Then, ∑n=2∞​7nan​​ is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } For positive integers n, if 4an​=(n2+5n+6) and Sn​=∑k=1n​(ak​1​), then the value of 507S2025​ is:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If an​=4n2−16n+15−2​, then a1​+a2​+⋯+a25​ is equal to :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let Sn​=∑k=14n​(−1)2k(k+1)​k2. Then Sn​ can take value(s)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } For k∈N, if the sum of the series 1+k4​+k28​+k313​+k419​+… is 10, then the value of k is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If 1⋅3+2⋅5+3⋅7+…upto n terms13+23+33+…upto n terms​=59​, then the value of n is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let ⟨an​⟩ be a sequence such that a0​=0, a1​=21​ and 2an+2​=5an+1​−3an​, n=0,1,2,3,... Then ∑k=1100​ak​ is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let an​ be nth term of the series 5+8+14+23+35+50+… and Sn​=∑k=1n​ak​. Then S30​−a40​ is equal to

Animated Solution for Mathematics - Sequence and Series: For a natural number $n$ , let $a_{n} = 1 9^{n} - 1 2^{n}$ . Then, the value of $\frac{31 a _{9} - a _{10}}{57 a _{8}}$ is

$a_{n} = 1 9^{n} - 1 2^{n}$

$Focus on Numerator$

$Substitute a_{9} and a_{10}$

$Expand the Brackets$

$Group Powers of 19$

$Group Powers of 12$

$Simplify the Coefficients$

$Factor Out Common Terms$

$Recognize a_{8}$

$Final Calculation$

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

Let ${a_{n}}_{n = 0}^{\infty}$ be a sequence such that $a_{0} = a_{1} = 0$ and $a_{n + 2} = 2 a_{n + 1} - a_{n} + 1$ for all $n \geq 0$ . Then, $\sum_{n = 2}^{\infty} \frac{a _{n}}{7 ^{n}}$ is equal to

For positive integers n, if $4 a_{n} = (n^{2} + 5 n + 6)$ and $S_{n} = \sum_{k = 1}^{n} (\frac{1}{a _{k}})$ , then the value of $507 S_{2025}$ is:

If $a_{n} = \frac{- 2}{4 n ^{2} - 16 n + 15}$ , then $a_{1} + a_{2} + \dots + a_{25}$ is equal to :

Let $S_{n} = \sum_{k = 1}^{4 n} (- 1)^{\frac{k ( k + 1 )}{2}} k^{2}$ . Then $S_{n}$ can take value(s)

For $k \in N$ , if the sum of the series $1 + \frac{4}{k} + \frac{8}{k ^{2}} + \frac{13}{k ^{3}} + \frac{19}{k ^{4}} + \dots$ is 10, then the value of $k$ is

If $\frac{1 ^{3} + 2 ^{3} + 3 ^{3} + \dots upto n terms}{1 \cdot 3 + 2 \cdot 5 + 3 \cdot 7 + \dots upto n terms} = \frac{9}{5}$ , then the value of $n$ is

Let $⟨ a_{n} ⟩$ be a sequence such that $a_{0} = 0$ , $a_{1} = \frac{1}{2}$ and $2 a_{n + 2} = 5 a_{n + 1} - 3 a_{n}$ , $n = 0, 1, 2, 3, ...$ Then $\sum_{k = 1}^{100} a_{k}$ is

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