JEE Main 2022 (29 June Shift 2)

LEVELJEE Main

Animated Solution for Mathematics - Sequence and Series: Let 3, 6, 9, 12, ... upto 78 terms and 5, 9, 13, 17, ... upto 59 terms be two series. Then, the sum of the terms common to both the series is equal to ______.

Q: What if the first common term wasn't obvious? Is there a general method?

Absolutely! You can write the general terms $T_1 = 3 + (n_1-1)3 = 3n_1$ and $T_2 = 5 + (n_2-1)4 = 4n_2 + 1$. Set them equal: $3n_1 = 4n_2 + 1$. You can solve this Diophantine equation for the smallest positive integers $n_1$ and $n_2$ that satisfy it. For small numbers, listing is faster, but for larger ones, this algebraic approach is your best friend.

Enter Numerical Value:

Progressive Hints

Hint 1

Think of these two arithmetic progressions as two trains moving along the same track at different speeds. The common terms are the stations where both trains stop. To find the first station, simply list the first few terms of each series until you find the first number that appears in both.

Hint 2

The common terms themselves form a new arithmetic progression. The first term is the one you found in the first hint. The common difference of this new sequence is not arbitrary; it is the Least Common Multiple (LCM) of the common differences of the two original series. Why? Because a term must be a multiple of both step-sizes relative to the starting point to land on the same value.

Hint 3

The common sequence is bounded by the shorter of the two original series. Since the common terms must exist in both series, they cannot exceed the smaller of the two maximum limits. Set up the inequality

a + (n - 1) d \leq min (l_{1}, l_{2})

to find the maximum number of terms

n

Visualized Solution

Analyze Series 1

Series 1: $S_{1} : 3, 6, 9, 12, \dots$ (78 terms)
First term ( $a_{1}$ ) = $3$
Common difference ( $d_{1}$ ) = $3$

Last Term of $S_{1}$

$l_{1} = a_{1} + (n_{1} - 1) d_{1}$
$l_{1} = 3 + (78 - 1) 3$
$l_{1} = 3 + 231 = 234$

Analyze Series 2

Series 2: $S_{2} : 5, 9, 13, 17, \dots$ (59 terms)
First term ( $a_{2}$ ) = $5$
Common difference ( $d_{2}$ ) = $4$

Last Term of $S_{2}$

$l_{2} = a_{2} + (n_{2} - 1) d_{2}$
$l_{2} = 5 + (59 - 1) 4$
$l_{2} = 5 + 232 = 237$

First Common Term ( $a$ )

Compare $S_{1}$ and $S_{2}$
First common term ( $a$ ) = $9$

Common Difference ( $d$ )

Common terms form a new AP
$d = LCM (d_{1}, d_{2})$
$d = LCM (3, 4) = 12$

The Common AP

Common AP: $9, 21, 33, \dots$

Upper Bound for Common Terms

Common terms must be $\leq min (l_{1}, l_{2})$
Upper bound = $min (234, 237) = 234$

Number of Common Terms ( $n$ )

Let $n$ be the number of terms
$T_{n} = a + (n - 1) d \leq 234$

Substitute Values

Substitute $a = 9$ and $d = 12$
$9 + (n - 1) 12 \leq 234$

Solve for $n$

$12 (n - 1) \leq 234 - 9$
$12 (n - 1) \leq 225$

Calculate $n$

$n - 1 \leq \frac{225}{12} = 18.75$
$n \leq 19.75$
Since $n$ is an integer, $n = 19$

Sum of Common Terms ( $S_{n}$ )

$S_{n} = \frac{n}{2} [2 a + (n - 1) d]$
Substitute $n = 19, a = 9, d = 12$

Compute $S_{19}$

$S_{19} = \frac{19}{2} [2 (9) + (19 - 1) 12]$
$S_{19} = \frac{19}{2} [18 + 216]$

Final Calculation

$S_{19} = \frac{19}{2} [234]$
$S_{19} = 19 \times 117 = 2223$

00:00 / 00:00

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

Solution Diagram

Analyzing the Setup

The first sequence,

S_{1}

, is an arithmetic progression starting at

a_{1} = 3

with a common difference

d_{1} = 3

. It contains

n_{1} = 78

terms.

The second sequence,

S_{2}

, starts at

a_{2} = 5

with a common difference

d_{2} = 4

. It contains

n_{2} = 59

terms.

Defining the Boundaries

To find the range of intersection, we calculate the last term of each sequence using the formula

l = a + (n - 1) d

For

S_{1}

l_{1} = 3 + (78 - 1) 3 = 3 + 231 = 234

For

S_{2}

l_{2} = 5 + (59 - 1) 4 = 5 + 232 = 237

Any common term must satisfy the condition that it is less than or equal to the smaller of these two boundaries. Therefore, the maximum value for a common term is

min (234, 237) = 234

The Rhythm of Intersection

The first common term observed in both sequences is

9

. This serves as the first term,

a

, of our new intersection sequence.

The common difference

d

of this new sequence is the Least Common Multiple of the individual differences

d_{1} = 3

and

d_{2} = 4

. Thus,

d = lcm (3, 4) = 12

The intersection sequence is defined by

T_{n} = 9 + (n - 1) 12

Determining the Number of Terms

We must find the number of terms

n

such that

T_{n} \leq 234

. Substituting the values into the inequality:

9 + (n - 1) 12 \leq 234

12 (n - 1) \leq 225

n - 1 \leq 18.75 \Rightarrow n \leq 19.75

Since

n

must be an integer, there are exactly

n = 19

common terms.

The Grand Summation

To find the sum of these 19 terms, we use the arithmetic series sum formula:

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

Substituting

n = 19

a = 9

, and

d = 12

S_{19} = \frac{19}{2} [2 (9) + (19 - 1) 12]

S_{19} = \frac{19}{2} [18 + 216]

S_{19} = \frac{19}{2} [234] = 19 \times 117

The final sum of the common terms is 2223.

Common Traps & FAQs

Wait, why is the common difference of the new series the LCM of the original differences?

Think of it this way: if you are jumping by 3 and your friend is jumping by 4, you only land on the same spot when you have both jumped a distance that is a multiple of both 3 and 4. The smallest such distance is the LCM, which is 12. Any common term must be 12 units away from the previous common term to maintain the pattern in both sequences.

Why do we use the minimum of the two last terms as the upper bound?

A common term must exist in *both* series. If Series 1 stops at 234 and Series 2 stops at 237, any number greater than 234 cannot be in Series 1. Therefore, it cannot be a common term. We are constrained by the 'shortest' series.

What if the first common term wasn't obvious? Is there a general method?

Absolutely! You can write the general terms

T_{1} = 3 + (n_{1} - 1) 3 = 3 n_{1}

and

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

. Set them equal:

3 n_{1} = 4 n_{2} + 1

. You can solve this Diophantine equation for the smallest positive integers

n_{1}

and

n_{2}

that satisfy it. For small numbers, listing is faster, but for larger ones, this algebraic approach is your best friend.

Animated Solution for Mathematics - Sequence and Series: Let 3, 6, 9, 12, ... upto 78 terms and 5, 9, 13, 17, ... upto 59 terms be two series. Then, the sum of the terms common to both the series is equal to ______.

Enter Numerical Value:

Visualized Solution

Analyze Series 1

Last Term of $S_{1}$

Analyze Series 2

Last Term of $S_{2}$

First Common Term ( $a$ )

Common Difference ( $d$ )

The Common AP

Upper Bound for Common Terms

Number of Common Terms ( $n$ )

Substitute Values

Solve for $n$

Calculate $n$

Sum of Common Terms ( $S_{n}$ )

Compute $S_{19}$

Final Calculation

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

Analyzing the Setup

Defining the Boundaries

The Rhythm of Intersection

Determining the Number of Terms

The Grand Summation

Similar Questions

The $8^{t h}$ common term of the series $S_{1} = 3 + 7 + 11 + 15 + 19 + \dots$ , $S_{2} = 1 + 6 + 11 + 16 + 21 + \dots$ is

The series of positive multiples of 3 is divided into sets : ${3}, {6, 9, 12}, {15, 18, 21, 24, 27}, \dots$ Then the sum of the elements in the $1 1^{t h}$ set is equal to ________.

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

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

Different A.P.'s are constructed with the first term 100, the last term 199, and integral common differences. The sum of the common differences of all such A.P.'s having at least 3 terms and at most 33 terms is ________.

The number of common terms in the progressions $4, 9, 14, 19, \dots$ , up to $2 5^{t h}$ term and $3, 6, 9, 12, \dots$ , up to $3 7^{t h}$ term is :

The number of terms common to the A.P.’s 3, 7, 11, … 407 and 2, 9, 16, … 709 is _______.

If the sum of the first 40 terms of the series, $3 + 4 + 8 + 9 + 13 + 14 + 18 + 19 + .....$ is $(102) m$ , then $m$ is equal to:

Let $s_{1}, s_{2}, s_{3}, \dots, s_{10}$ respectively be the sum of 12 terms of 10 A.Ps whose first terms are $1, 2, 3, \dots, 10$ and the common differences are $1, 3, 5, \dots, 19$ respectively. Then $\sum_{i = 1}^{10} s_{i}$ is equal to

If the sum of the first 40 terms of the series, $3 + 4 + 8 + 9 + 13 + 14 + 18 + 19 + \dots$ is $(102) m$ , then m is equal to :

Animated Solution for Mathematics - Sequence and Series: Let 3, 6, 9, 12, ... upto 78 terms and 5, 9, 13, 17, ... upto 59 terms be two series. Then, the sum of the terms common to both the series is equal to ______.

Enter Numerical Value:

Visualized Solution

Analyze Series 1

Last Term of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } S1​

Analyze Series 2

Last Term of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } S2​

First Common Term ( .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a)

Common Difference ( .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } d)

The Common AP

Upper Bound for Common Terms

Number of Common Terms ( .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } n)

Substitute Values

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

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

Sum of Common Terms ( .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Sn​)

Compute .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } S19​

Final Calculation

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

Analyzing the Setup

Defining the Boundaries

The Rhythm of Intersection

Determining the Number of Terms

The Grand Summation

Similar Questions

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.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

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Last Term of $S_{1}$

Last Term of $S_{2}$

First Common Term ( $a$ )

Common Difference ( $d$ )

Number of Common Terms ( $n$ )

Solve for $n$

Calculate $n$

Sum of Common Terms ( $S_{n}$ )

Compute $S_{19}$

The $8^{t h}$ common term of the series $S_{1} = 3 + 7 + 11 + 15 + 19 + \dots$ , $S_{2} = 1 + 6 + 11 + 16 + 21 + \dots$ is

The series of positive multiples of 3 is divided into sets : ${3}, {6, 9, 12}, {15, 18, 21, 24, 27}, \dots$ Then the sum of the elements in the $1 1^{t h}$ set is equal to ________.

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

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

The number of common terms in the progressions $4, 9, 14, 19, \dots$ , up to $2 5^{t h}$ term and $3, 6, 9, 12, \dots$ , up to $3 7^{t h}$ term is :

If the sum of the first 40 terms of the series, $3 + 4 + 8 + 9 + 13 + 14 + 18 + 19 + .....$ is $(102) m$ , then $m$ is equal to:

If the sum of the first 40 terms of the series, $3 + 4 + 8 + 9 + 13 + 14 + 18 + 19 + \dots$ is $(102) m$ , then m is equal to :