JEE Main 2021 (March)

LEVELJEE Main

Animated Solution for Mathematics - Sequence and Series: Consider an arithmetic series and a geometric series having four initial terms from the set $11, 8, 21, 16, 26, 32, 4$ . If the last terms of these series are the maximum possible four digit numbers, then the number of common terms in these two series is equal to

Enter Numerical Value:

Visualized Solution

Analyzing the Given Set

Given set: ${11, 8, 21, 16, 26, 32, 4}$
We need to find four initial terms for an Arithmetic Progression (AP) and a Geometric Progression (GP).

Identifying the AP

Look for numbers with a constant difference.
AP candidate: $11, 16, 21, 26$
Common difference $d = 5$

Identifying the GP

Look for numbers with a constant ratio.
GP candidate: $4, 8, 16, 32$
Common ratio $r = 2$

General Terms of AP and GP

AP General Term: $T_{m} = 11 + (m - 1) 5 = 5 m + 6$
GP General Term: $T_{n} = 4 \cdot 2^{n - 1} = 2^{n + 1}$

Equating for Common Terms

Condition for a common term: $T_{n} = T_{m}$
$2^{n + 1} = 5 m + 6$

Applying Modular Arithmetic

Notice the right side: $5 m + 6$
When divided by $5$ , the remainder is $1$ .
Therefore, $2^{n + 1} \equiv 1 (mod 5)$

Analyzing Powers of $2$ Modulo $5$

Let's check powers of $2$ :
$2^{1} \equiv 2 (mod 5)$
$2^{2} \equiv 4 (mod 5)$
$2^{3} \equiv 3 (mod 5)$
$2^{4} \equiv 1 (mod 5)$
The cycle repeats every $4$ powers.

Finding the First Common Term

First valid exponent is $4$ .
$2^{4} = 16$
Check: $16 = 5 (2) + 6$ (Valid AP term)

Finding the Second Common Term

Next valid exponent is $8$ (multiple of $4$ ).
$2^{8} = 256$
Check: $256 = 5 (50) + 6$ (Valid AP term)

Finding the Third Common Term

Next valid exponent is $12$ .
$2^{12} = 4096$
Check: $4096 = 5 (818) + 6$ (Valid AP term)

Checking the Fourth Term

Next valid exponent is $16$ .
$2^{16} = 65536$
Constraint Check: The problem states the maximum possible terms are four-digit numbers.
$65536$ is a five-digit number. (Rejected)

Final Conclusion

The valid common terms are $16, 256,$ and $4096$ .
Total number of common terms = 3

00:00 / 00:00

The Sigma Insight: Geometric Progression (G.P.)

Solution Diagram

Analyzing the Setup

The given set of numbers is

{11, 8, 21, 16, 26, 32, 4}

. Our objective is to identify two distinct sequences—an Arithmetic Progression (AP) and a Geometric Progression (GP)—each containing four terms.

For the AP, we observe the sequence

11, 16, 21, 26

. This sequence has a first term

a = 11

and a common difference

d = 5

. The general term is given by:

T_{m} = 11 + (m - 1) 5 = 5 m + 6

For the GP, we identify the sequence

4, 8, 16, 32

. This sequence has a first term

a = 4

and a common ratio

r = 2

. The general term is given by:

T_{n} = 4 \cdot 2^{n - 1} = 2^{n + 1}

The Master Equation

To find the common terms between these two sequences, we equate their general terms:

2^{n + 1} = 5 m + 6

We seek integer values for

n

and

m

that satisfy this equality. By applying modular arithmetic, we observe the right side modulo

5

5 m + 6 \equiv 1 (mod 5)

This implies that

2^{n + 1}

must also satisfy the condition

2^{n + 1} \equiv 1 (mod 5)

The Cycle of Powers

We examine the powers of

2

modulo

5

2^{1} \equiv 2 (mod 5)

2^{2} \equiv 4 (mod 5)

2^{3} \equiv 8 \equiv 3 (mod 5)

2^{4} \equiv 16 \equiv 1 (mod 5)

The cycle repeats every

4

powers. Therefore,

n + 1

must be a multiple of

4

. Let

n + 1 = 4 k

, where

k

is a positive integer.

Final Calculation

We evaluate the potential terms generated by

2^{4 k}

: For

k = 1

2^{4} = 16

. For

k = 2

2^{8} = 256

. For

k = 3

2^{12} = 4096

. For

k = 4

2^{16} = 65536

Given the constraint that the terms must be four-digit numbers, we accept

16, 256,

and

4096

. The value

65536

is rejected as it exceeds the four-digit limit.

Thus, there are exactly 3 common terms that satisfy the conditions.

Common Traps & FAQs

Wait, why do we reject

65536

as a common term?

Great catch! The problem explicitly states that the last terms of these series are the maximum possible four-digit numbers. Since

65536

is a five-digit number, it exceeds the constraint given in the problem, so we must exclude it from our count.

How did you know the AP was

11, 16, 21, 26

and not some other combination?

That is the detective work! We look for a constant difference

d

. If you test combinations, you will find that

11, 16, 21, 26

is the only set of four numbers from the given list that maintains a constant difference of

5

. The same logic applies to the GP with a common ratio of

2

Is there a faster way to solve

2^{n + 1} = 5 m + 6

without checking every power?

Yes, modular arithmetic is your best friend here! By looking at the equation modulo

5

, we see

2^{n + 1} \equiv 1 (mod 5)

. We know the powers of

2

modulo

5

follow a cycle of

4

(

2, 4, 3, 1

). This immediately tells us that

n + 1

must be a multiple of

4

, saving us from testing every single integer.

Enter Numerical Value:

Visualized Solution

Analyzing the Given Set

Identifying the AP

Identifying the GP

General Terms of AP and GP

Equating for Common Terms

Applying Modular Arithmetic

Analyzing Powers of $2$ Modulo $5$

Finding the First Common Term

Finding the Second Common Term

Finding the Third Common Term

Checking the Fourth Term

Final Conclusion

The Sigma Insight: Geometric Progression (G.P.)

Analyzing the Setup

The Master Equation

The Cycle of Powers

Final Calculation

Similar Questions

If the sum and product of four positive consecutive terms of a G.P., are 126 and 1296, respectively, then the sum of common ratios of all such GPs is

For the two positive numbers $a, b$ , if $a, b$ and $\frac{1}{18}$ are in a geometric progression, while $\frac{1}{a}, 10$ and $\frac{1}{b}$ are in an arithmetic progression, then, $16 a + 12 b$ is equal to

If in a G.P. of 64 terms, the sum of all the terms is 7 times the sum of the odd terms of the G.P, then the common ratio of the G.P. is equal to

Let $a$ and $b$ be two distinct positive real numbers. Let $1 1^{t h}$ term of a GP, whose first term is $a$ and third term is $b$ , is equal to $p^{t h}$ term of another GP, whose first term is $a$ and fifth term is $b$ . Then $p$ is equal to

In an increasing, geometric series, the sum of the second and the sixth term is $\frac{25}{2}$ and the product of the third and fifth term is 25. Then, the sum of $4^{t h}, 6^{t h}$ and $8^{t h}$ terms is equal to :

The $4^{t h}$ term of GP is 500 and its common ratio is $\frac{1}{m}, m \in N$ . Let $S_{n}$ denote the sum of the first $n$ terms of this GP. If $S_{6} > S_{5} + 1$ and $S_{7} < S_{6} + \frac{1}{2}$ , then the number of possible values of $m$ is

Consider an infinite geometric series with first term $a$ and common ratio $r$ . If its sum is $4$ and the second term is $3/4$ , then

Enter Numerical Value:

Visualized Solution

Analyzing the Given Set

Identifying the AP

Identifying the GP

General Terms of AP and GP

Equating for Common Terms

Applying Modular Arithmetic

Analyzing Powers of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 2 Modulo .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 5

Finding the First Common Term

Finding the Second Common Term

Finding the Third Common Term

Checking the Fourth Term

Final Conclusion

The Sigma Insight: Geometric Progression (G.P.)

Analyzing the Setup

The Master Equation

The Cycle of Powers

Final Calculation

Similar Questions

If the sum and product of four positive consecutive terms of a G.P., are 126 and 1296, respectively, then the sum of common ratios of all such GPs is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } For the two positive numbers a,b, if a,b and 181​ are in a geometric progression, while a1​,10 and b1​ are in an arithmetic progression, then, 16a+12b is equal to

If in a G.P. of 64 terms, the sum of all the terms is 7 times the sum of the odd terms of the G.P, then the common ratio of the G.P. is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Consider an infinite geometric series with first term a and common ratio r. If its sum is 4 and the second term is 3/4, then

Analyzing Powers of $2$ Modulo $5$

For the two positive numbers $a, b$ , if $a, b$ and $\frac{1}{18}$ are in a geometric progression, while $\frac{1}{a}, 10$ and $\frac{1}{b}$ are in an arithmetic progression, then, $16 a + 12 b$ is equal to

Consider an infinite geometric series with first term $a$ and common ratio $r$ . If its sum is $4$ and the second term is $3/4$ , then