JEE Advanced 2019

LEVELJEE Main

Animated Solution for Mathematics - Sequence and Series: Let $A P (a; d)$ denote the set of all the terms of an infinite arithmetic progression with first term $a$ and common difference $d > 0$ . If $A P (1; 3) \cap A P (2; 5) \cap A P (3; 7) = A P (a; d)$ then $a + d$ equals

Q: Why do we reject the negative root or non-integer solutions for $p$ and $q$?

In the context of an Arithmetic Progression, the indices $p$ and $q$ represent the number of steps taken from the first term. These must be non-negative integers. A fractional step doesn't exist in the discrete world of APs, and a negative step would take us before the start of the sequence, which is outside our domain.

Enter Numerical Value:

Progressive Hints

Hint 1

Visualize the intersection of these three arithmetic progressions as three different clocks ticking at different rates. The moment they all align is periodic. The common difference of this new, combined sequence must be a multiple of the individual differences. What is the smallest number that is a multiple of

3

5

, and

7

Hint 2

To find the first term, you cannot simply guess. You need to treat this as a system of linear Diophantine equations. Start by finding the intersection of the first two sequences,

A P (1; 3)

and

A P (2; 5)

, by setting their general terms equal:

1 + 3 p = 2 + 5 q

Hint 3

Once you have the general term for the intersection of the first two sequences, say

x = 7 + 15 k

, you must force this to intersect with the third sequence

A P (3; 7)

. Set

7 + 15 k = 3 + 7 r

and solve for the smallest non-negative integer

k

by isolating the remainder.

Visualized Solution

The Three Arithmetic Progressions

Let $S_{1} = A P (1; 3)$
Let $S_{2} = A P (2; 5)$
Let $S_{3} = A P (3; 7)$
We need to find the intersection: $S_{1} \cap S_{2} \cap S_{3} = A P (a; d)$

Common Difference of Intersection

The intersection of APs is also an AP.
The new common difference $d$ is the Least Common Multiple (LCM) of the individual common differences.
$d = lcm (d_{1}, d_{2}, d_{3})$

Calculating $d$

$d_{1} = 3$ , $d_{2} = 5$ , $d_{3} = 7$
$d = lcm (3, 5, 7)$
Since $3, 5, 7$ are pairwise coprime:
$d = 3 \times 5 \times 7 = 105$

Finding the First Common Term

To find the first term $a$ , we first find the intersection of $S_{1}$ and $S_{2}$ .
General term of $S_{1}$ : $x = 1 + 3 p$
General term of $S_{2}$ : $x = 2 + 5 q$
where $p, q \geq 0$ are integers.

Equating $S_{1}$ and $S_{2}$

Equate the general terms:
$1 + 3 p = 2 + 5 q$
Rearrange to form a linear Diophantine equation:
$3 p - 5 q = 1$

Solving for $p$ and $q$

$3 p = 5 q + 1$
Test values for $q \geq 0$ :
If $q = 0 ⟹ 3 p = 1$ (Reject, $p$ is not an integer)
If $q = 1 ⟹ 3 p = 6 ⟹ p = 2$ (Accept)

First Common Term of $S_{1}$ and $S_{2}$

Substitute $p = 2$ into $x = 1 + 3 p$ :
$x = 1 + 3 (2) = 7$
(Check: $q = 1 ⟹ x = 2 + 5 (1) = 7$ )
The first common term is $7$ .

General Term of $S_{1} \cap S_{2}$

The intersection $S_{1} \cap S_{2}$ is an AP.
First term $= 7$
Common difference $= lcm (3, 5) = 15$
General term: $x = 7 + 15 k$ , where $k \geq 0$

Intersecting with $S_{3}$

Now, we must find terms common to this new AP and $S_{3}$ .
General term of $S_{1} \cap S_{2}$ : $x = 7 + 15 k$
General term of $S_{3}$ : $x = 3 + 7 r$
Equate them: $7 + 15 k = 3 + 7 r$

Simplifying the Equation

$7 + 15 k = 3 + 7 r$
Rearrange: $15 k + 4 = 7 r$
This means $15 k + 4$ must be a multiple of $7$ .

Solving for $k$

Rewrite $15 k$ to separate multiples of $7$ :
$15 k = 14 k + k$
Substitute back: $(14 k + k) + 4 = 7 r$
$14 k + (k + 4) = 7 r$
Since $14 k$ is divisible by $7$ , $(k + 4)$ must also be divisible by $7$ .

Finding the Smallest $k$

We need the smallest integer $k \geq 0$ such that $k + 4$ is a multiple of $7$ .
The multiples of $7$ are $0, 7, 14, 21, \dots$
The smallest multiple $\geq 4$ is $7$ .
$k + 4 = 7 ⟹ k = 3$

Calculating First Term $a$

Substitute $k = 3$ into the general term of $S_{1} \cap S_{2}$ :
$a = 7 + 15 k$
$a = 7 + 15 (3)$
$a = 7 + 45 = 52$
The first term of the final AP is $52$ .

Final Calculation

We have $a = 52$ and $d = 105$ .
The question asks for $a + d$ .
$a + d = 52 + 105 = 157$
Final Answer: $157$

00:00 / 00:00

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

Solution Diagram

Analyzing the Setup

We are given three arithmetic progressions:

S_{1} : x = 1 + 3 p

S_{2} : x = 2 + 5 q

S_{3} : x = 3 + 7 r

We seek the points where these three rhythms synchronize. This requires finding the first common term

a

and the common difference

d

of the resulting intersection sequence.

The Rhythm of the Intersection

The common difference

d

of the new sequence must be the Least Common Multiple (LCM) of the individual differences

3

5

, and

7

Since these numbers are pairwise coprime, their LCM is simply their product:

d = 3 \times 5 \times 7 = 105

This implies that once the first common term

a

is identified, every subsequent term in the intersection sequence will be of the form

x = a + 105 n

The Diophantine Dance

To find the first term

a

, we first intersect

S_{1}

and

S_{2}

by setting their general terms equal:

1 + 3 p = 2 + 5 q

Rearranging this yields the linear Diophantine equation:

3 p - 5 q = 1

Testing small values, if

q = 1

, then

3 p = 6

, which gives

p = 2

. Substituting this back into the expression for

S_{1}

x = 1 + 3 (2) = 7

Thus, the intersection of the first two sequences is defined by

x = 7 + 15 k

, where

15

is the LCM of

3

and

5

The Final Convergence

We now intersect this result with the third sequence,

S_{3}

, defined by

x = 3 + 7 r

7 + 15 k = 3 + 7 r

Simplifying this equation, we obtain:

15 k + 4 = 7 r

We observe that

15 k + 4

must be divisible by

7

. Rewriting

15 k

14 k + k

, we have:

(14 k + k) + 4 = 7 r

Since

14 k

is divisible by

7

, the term

(k + 4)

must also be divisible by

7

. The smallest non-negative integer

k

satisfying this is

k = 3

Final Calculation

Substituting

k = 3

back into the expression

x = 7 + 15 k

a = 7 + 15 (3) = 52

The first common term is

52

and the common difference is

105

. The problem asks for the sum

a + d

a + d = 52 + 105 = 157

The final result is 157.

Common Traps & FAQs

Wait, why is the common difference the LCM of the individual differences? Can't it be something else?

Think of the common difference as the 'period' of the sequence. For a term to be common to all, it must satisfy the jump requirements of every single sequence simultaneously. If sequence A jumps by

3

and sequence B by

5

, the 'beat' of their intersection must be a multiple of both

3

and

5

. The smallest such beat is the LCM, which ensures we don't miss any potential common terms.

Why do we solve for

p

and

q

using Diophantine equations? Is there a faster way?

While you might be tempted to list out terms, that is a dangerous game in an exam! Diophantine equations provide a rigorous, algebraic guarantee that you have found the *first* common term. It eliminates the risk of human error and ensures you aren't skipping over the true starting point.

Why do we reject the negative root or non-integer solutions for

p

and

q

In the context of an Arithmetic Progression, the indices

p

and

q

represent the number of steps taken from the first term. These must be non-negative integers. A fractional step doesn't exist in the discrete world of APs, and a negative step would take us before the start of the sequence, which is outside our domain.

Animated Solution for Mathematics - Sequence and Series: Let $A P (a; d)$ denote the set of all the terms of an infinite arithmetic progression with first term $a$ and common difference $d > 0$ . If $A P (1; 3) \cap A P (2; 5) \cap A P (3; 7) = A P (a; d)$ then $a + d$ equals

Enter Numerical Value:

Visualized Solution

The Three Arithmetic Progressions

Common Difference of Intersection

Calculating $d$

Finding the First Common Term

Equating $S_{1}$ and $S_{2}$

Solving for $p$ and $q$

First Common Term of $S_{1}$ and $S_{2}$

General Term of $S_{1} \cap S_{2}$

Intersecting with $S_{3}$

Simplifying the Equation

Solving for $k$

Finding the Smallest $k$

Calculating First Term $a$

Final Calculation

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

Analyzing the Setup

The Rhythm of the Intersection

The Diophantine Dance

The Final Convergence

Final Calculation

Similar Questions

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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 -

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Enter Numerical Value:

Visualized Solution

The Three Arithmetic Progressions

Common Difference of Intersection

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

Finding the First Common Term

Equating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } S1​ and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } S2​

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

First Common Term of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } S1​ and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } S2​

General Term of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } S1​∩S2​

Intersecting with .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } S3​

Simplifying the Equation

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

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

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

Final Calculation

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

Analyzing the Setup

The Rhythm of the Intersection

The Diophantine Dance

The Final Convergence

Final Calculation

Similar Questions

.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

.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. whose first term is a and common difference is d. If for some positive integers m,n,m=n,Tm​=1/n and Tn​=1/m, then a−d equals

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the sum of the first 2n terms of the A.P. 2,5,8,… is equal to the sum of the first n terms of the A.P. 57,59,61,…, then n equals

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let a1​,a2​,…,a21​ be an AP such that ∑n=120​an​an+1​1​=94​. If the sum of this AP is 189, then a6​a16​ is equal to :

Calculating $d$

Equating $S_{1}$ and $S_{2}$

Solving for $p$ and $q$

First Common Term of $S_{1}$ and $S_{2}$

General Term of $S_{1} \cap S_{2}$

Intersecting with $S_{3}$

Solving for $k$

Finding the Smallest $k$

Calculating First Term $a$

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

Let $T_{r}$ be the $r$ th term of an A.P. whose first term is $a$ and common difference is $d$ . If for some positive integers $m, n, m \neq = n, T_{m} = 1/ n$ and $T_{n} = 1/ m$ , then $a - d$ equals

Let $a_{n}$ be the $n^{t h}$ term of an A. P. If $S_{n} = a_{1} + a_{2} + a_{3} + \dots + a_{n} = 700, a_{6} = 7$ and $S_{7} = 7$ , then $a_{n}$ is equal to :

If the sum of the first $2 n$ terms of the A.P. $2, 5, 8, \dots$ is equal to the sum of the first $n$ terms of the A.P. $57, 59, 61, \dots$ , then $n$ equals

Let $a_{1}, a_{2}, \dots, a_{21}$ be an $A P$ such that $\sum_{n = 1}^{20} \frac{1}{a _{n} a _{n + 1}} = \frac{4}{9}$ . If the sum of this $A P$ is $189$ , then $a_{6} a_{16}$ is equal to :