Find the 5th term of the sequence defined by $a_n = 2n + 3$.

Correct Option: C. Solution: Substitute $n = 5$ into the formula: $a_5 = 2(5) + 3 = 10 + 3 = 15$.

Find the geometric mean of 16 and 64.

Correct Option: A. Solution: The geometric mean of two numbers $a$ and $b$ is $\sqrt{ab}$. Here, $a = 16$ and $b = 64$, so the geometric mean is $\sqrt{16 \times 64} = \sqrt{1024} = 32$.

If a sequence is defined by $a_n = 2n + 5$, what is the 3rd term?

Correct Option: B. Solution: Substitute $n = 3$ into $a_n = 2n + 5$: $a_3 = 2 \times 3 + 5 = 11$.

If a sequence is defined by $a_n = 3n + 2$, what is the 10th term?

Correct Option: C. Solution: Substitute $n = 10$ into the formula: $a_{10} = 3(10) + 2 = 30 + 2 = 32$.

A sequence is defined by the rule $a_n = 3n + 1$. What is the 5th term of this sequence?

Correct Option: A. Solution: Substitute $n = 5$ into the formula: $a_5 = 3(5) + 1 = 16$.

Sequences and Series

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Summary

Chapter 8: Sequences and Series

Summary

A sequence is an ordered collection of objects with an identified first member, second member, etc.
Sequences can represent various phenomena, such as population growth or financial deposits.
Sequences that follow specific patterns are called progressions, including arithmetic progressions (A.P.) and geometric progressions (G.P.).
Important concepts include:
- Arithmetic Mean (A.M.)
- Geometric Mean (G.M.)
- Relationships between A.M. and G.M.
- Special series for sums of natural numbers, squares, and cubes.
A series is the sum of the terms of a sequence, represented in sigma notation (Σ).
Finite sequences have a limited number of terms, while infinite sequences do not.
The Fibonacci sequence is a notable example of a sequence generated by a recurrence relation.

Learning Objectives

Understand the concept of geometric progression (G.P.) and its properties.
Identify the first term and common ratio in a G.P.
Calculate specific terms in a G.P. using the formula for the n-th term.
Derive the sum of the first n terms of a G.P.
Solve problems involving the relationship between terms in a G.P.
Apply G.P. concepts to real-world scenarios, such as population growth and financial calculations.

Detailed Notes

Sequences and Series Notes

Introduction to Sequences

A sequence is an ordered collection of objects with identified members (first, second, third, etc.).
Examples include population growth over time, bank deposits, and depreciated values of commodities.

Geometric Progression (G.P.)

A sequence is called a geometric progression if each term after the first is obtained by multiplying the previous term by a constant called the common ratio (r).
General form: a, ar, ar², ar³, ... where a is the first term.

Examples of G.P.

Example 1: 2, 4, 8, 16 (common ratio = 2)
Example 2: 9, 27, 81 (common ratio = 3)
Example 3: 0.1, 0.01, 0.001 (common ratio = 0.1)

Important Properties of G.P.

If the first term is a and the common ratio is r, the nth term can be expressed as:

T_n = ar^(n-1)
The sum of the first n terms (S_n) of a G.P. can be calculated using the formula:

S_n = a(1 - r^n) / (1 - r) for r ≠ 1

Miscellaneous Examples

Example: If the first term of a G.P. is 5 and the common ratio is 2, find the sum of the first n terms.
Example: The sum of three numbers in G.P. is 56. If we subtract 1, 7, 21 from these numbers, they form an arithmetic progression.

Exercises

Find the 20th term of the G.P. 2, 4, 8, ...
Show that the 5th, 8th, and 11th terms of a G.P. satisfy the relation q² = ps.
If a, b, c, d are in G.P., prove that (a² + b² + c²)(b² + c² + d²) = (ab + bc + cd)².

Conclusion

Sequences and series, particularly geometric progressions, play a crucial role in various mathematical applications and real-world scenarios.

Exam Tips & Common Mistakes

Common Mistakes and Exam Tips

Common Pitfalls

Misunderstanding Sequences and Series: Students often confuse sequences with series. Remember, a sequence is a list of numbers, while a series is the sum of the terms of a sequence.
Incorrectly Identifying G.P.: Ensure that the ratio of consecutive terms is constant when identifying a geometric progression (G.P.).
Ignoring Conditions: When solving problems involving A.M. and G.M., do not overlook the conditions given in the problem, such as the relationship between the two means.
Misapplying Formulas: Be careful when applying formulas for the sum of G.P. or A.P. Ensure you are using the correct formula based on the context of the problem.

Tips for Success

Practice with Examples: Work through various examples of sequences and series to become familiar with different types of problems.
Understand the Concepts: Focus on understanding the underlying concepts of A.M. and G.M. rather than just memorizing formulas.
Check Your Work: Always double-check your calculations, especially when dealing with series sums and terms in G.P.
Use Visual Aids: Drawing diagrams or charts can help visualize sequences and series, making it easier to understand their relationships.

Practice & Assessment

Multiple Choice Questions

Correct Answer: C

Solution:

Let the terms be

a/r

a

ar

. Then,

a/r + a + ar = 13

and

(a/r) \times a \times ar = 27

. Solving these gives

r = 3

x, y, z

are in arithmetic progression

x, y, z

are in geometric progression

x, y, z

are in harmonic progression

None of the above

Correct Answer: B

Solution:

Since the terms are from the same G.P., they must also form a G.P.

Correct Answer: B

Solution:

The sum of the first

n

terms of a G.P. is given by

S_n = a \frac{r^n - 1}{r - 1}

. Substituting the given values, we have

605 = 5 \frac{3^n - 1}{3 - 1}

. Simplifying,

605 = 5 \frac{3^n - 1}{2}

. This gives

1210 = 5(3^n - 1)

, or

3^n = 243

. Since

3^5 = 243

n = 5

-1

-2

Correct Answer: A

Solution:

Let the terms be

a/r, a, ar

. Then

a/r + a + ar = 12

and

(a/r) \times a \times ar = -1

. Solving gives

a = -1

a, b, c are in arithmetic progression

a, b, c are in geometric progression

a, b, c are in harmonic progression

None of the above

Correct Answer: B

Solution:

In a geometric progression, if the pᵗʰ, qᵗʰ, and rᵗʰ terms are a, b, and c, then

b^2 = ac

. This confirms that a, b, c are in geometric progression.

Rs 15000

Rs 20000

Rs 25000

Rs 30000

Correct Answer: B

Solution:

Simple interest is calculated as

SI = \frac{P \times R \times T}{100}

. Here,

P = 10000

R = 5

T = 20

. So,

SI = \frac{10000 \times 5 \times 20}{100} = 10000

. Total amount = Principal + Interest = Rs 10000 + Rs 10000 = Rs 20000.

768

384

1536

1024

Correct Answer: C

Solution:

The formula for the nᵗʰ term of a G.P. is

a_n = ar^{n-1}

. Given

a_8 = 192

and

r = 2

, we find

a = \frac{192}{2^7} = 1.5

. Now,

a_{12} = 1.5 \times 2^{11} = 1536

Correct Answer: B

Solution:

The sequence is arithmetic with a common difference of 2. Starting from

a_1 = 1

, we have

a_2 = 3

a_3 = 5

a_4 = 7

, and

a_5 = 9

Correct Answer: B

Solution:

Let the terms be

a/r

a

ar

. Then

a/r + a + ar = 13

and

(a/r) \times a \times ar = 27

. Solving these gives

a = 3

Correct Answer: C

Solution:

Substitute

n = 5

into the formula:

a_5 = 2(5) + 3 = 10 + 3 = 15

Correct Answer: A

Solution:

The geometric mean of two numbers

a

and

b

\sqrt{ab}

. Here,

a = 16

and

b = 64

, so the geometric mean is

\sqrt{16 \times 64} = \sqrt{1024} = 32

Rs. 8192

Rs. 10240

Rs. 9830

Rs. 10000

Correct Answer: A

Solution:

The value of the machine after 5 years is given by

15625 \times (0.8)^5 = 8192

102

Correct Answer: B

Solution:

The sum of the first

n

terms of a G.P. is given by the formula

S_n = a \frac{r^n - 1}{r - 1}

. Substituting

a = 3

r = 2

, and

n = 5

, we get

S_5 = 3 \frac{2^5 - 1}{2 - 1} = 3 \times 31 = 93

. Therefore, the correct option is 96.

Rs 1157.63

Rs 1100.00

Rs 1050.00

Rs 1150.00

Correct Answer: A

Solution:

The amount

A

after

n

years with compound interest is given by

A = P(1 + \frac{r}{100})^n

. Here,

P = 1000

r = 5

n = 3

. So,

A = 1000(1 + 0.05)^3 = 1000 \times 1.157625 = 1157.63

Correct Answer: B

Solution:

The sequence is defined recursively. Calculating step-by-step:

a_2 = 3(2) + 1 = 7

a_3 = 3(7) + 1 = 22

, and

a_4 = 3(22) + 1 = 67

. Therefore, the correct answer is option b.

Correct Answer: B

Solution:

Substitute

n = 3

into

a_n = 2n + 5

a_3 = 2 \times 3 + 5 = 11

Correct Answer: A

Solution:

The common ratio

r

can be found using the formula for the

n^{th}

term of a G.P.,

a_n = ar^{n-1}

. Given

a_8 = 192

and

a_{12} = 3072

, we have

ar^7 = 192

and

ar^{11} = 3072

. Dividing the second equation by the first gives

r^4 = \frac{3072}{192} = 16

, so

r = 2

Rs 1295.50

Rs 1300.00

Rs 1200.00

Rs 1100.00

Correct Answer: A

Solution:

The formula for compound interest is

A = P(1 + r)^n

. Here,

P = 500

r = 0.10

, and

n = 10

. Thus,

A = 500(1.10)^{10} = 1295.50

Correct Answer: A

Solution:

The terms are

ar^3 = x

ar^9 = y

ar^{15} = z

. The ratio

\frac{y}{x} = r^6

and

\frac{z}{y} = r^6

, hence

x

y

z

are in G.P. with common ratio

r^6

. If

r = 2

, then

r^6 = 64

, which is consistent.

Correct Answer: A

Solution:

Given

a_3 = ar^2 = 24

and

a_6 = ar^5 = 192

. Dividing the second equation by the first gives

r^3 = 192/24 = 8

, so

r = 2

Correct Answer: C

Solution:

Substitute

n = 10

into the formula:

a_{10} = 3(10) + 2 = 30 + 2 = 32

Correct Answer: B

Solution:

The sum of the first

n

terms of a G.P. is given by

S_n = a \frac{r^n - 1}{r - 1}

. Substituting the given values, we have

93 = 3 \frac{2^n - 1}{2 - 1}

. Simplifying,

2^n - 1 = 31

, which gives

2^n = 32

, and hence

n = 5

. Therefore, the correct answer is option b.

Correct Answer: A

Solution:

The arithmetic mean of two numbers

a

and

b

is given by

\frac{a+b}{2} = 8

, thus

a+b = 16

. The geometric mean is given by

\sqrt{ab} = 5

, thus

ab = 25

. Therefore, the correct answer is option a.

Correct Answer: A

Solution:

Substitute

n = 5

into the formula:

a_5 = 3(5) + 1 = 16

Rs. 8192

Rs. 10240

Rs. 9830

Rs. 10000

Correct Answer: A

Solution:

The value of the machine after

n

years can be calculated using the formula for depreciation:

V_n = V_0 (1 - r)^n

where

V_0 = 15625

r = 0.20

, and

n = 5

. Substituting these values, we get:

V_5 = 15625 \times (1 - 0.20)^5

V_5 = 15625 \times 0.8^5

V_5 = 15625 \times 0.32768

V_5 = 8192

120

240

480

960

Correct Answer: C

Solution:

The number of bacteria after 4 hours is

30 \times 2^4 = 480

x^2 = yz

y^2 = xz

z^2 = xy

x + y + z = 0

Correct Answer: B

Solution:

In a G.P., the

n

th term is given by

a_n = ar^{n-1}

. For the 4th, 10th, and 16th terms, we have

x = ar^3

y = ar^9

, and

z = ar^{15}

. The relationship

y^2 = xz

can be shown as follows:

y^2 = (ar^9)^2 = a^2r^{18}

xz = (ar^3)(ar^{15}) = a^2r^{18}

Therefore,

y^2 = xz

162

486

Correct Answer: B

Solution:

The 4th term of a G.P. is given by

a_4 = ar^{3}

. Here,

a = 2

and

r = 3

. Therefore,

a_4 = 2 \times 3^3 = 54

Correct Answer: C

Solution:

Let the first term be

a

and the common ratio be

r

. The 4th term is

ar^3 = 16

and the 10th term is

ar^9 = 1024

. Dividing the second equation by the first, we get

r^6 = \frac{1024}{16} = 64

, so

r = 64^{1/6} = 4

. Therefore, the correct answer is option c.

Rs 1295.50

Rs 1300.00

Rs 1296.87

Rs 1320.00

Correct Answer: C

Solution:

The formula for compound interest is

A = P(1 + r/n)^{nt}

, where

P = 500

r = 0.10

n = 1

, and

t = 10

. Substituting these values, we get

A = 500(1 + 0.10)^10 = 500(2.59374) = 1296.87

. Therefore, the correct answer is option c.

Rs 1157.63

Rs 1157.50

Rs 1150.00

Rs 1160.00

Correct Answer: A

Solution:

The compound interest formula is

A = P(1 + \frac{r}{100})^n

, where

P = 1000

r = 5

, and

n = 3

. Calculating,

A = 1000(1 + 0.05)^3 = 1000(1.157625) = 1157.63

x^2 - 16x + 25 = 0

x^2 - 10x + 25 = 0

x^2 - 12x + 25 = 0

x^2 - 14x + 25 = 0

Correct Answer: A

Solution:

Let the numbers be

a

and

b

. The A.M. is

\frac{a+b}{2} = 8

, giving

a+b = 16

. The G.M. is

\sqrt{ab} = 5

, giving

ab = 25

. The quadratic equation is given by

x^2 - (a+b)x + ab = 0

, which becomes

x^2 - 16x + 25 = 0

768

1536

3072

6144

Correct Answer: C

Solution:

Given

a_3 = ar^2 = 24

and

a_6 = ar^5 = 192

. Dividing the second equation by the first gives

r^3 = 8

, so

r = 2

. Substituting

r = 2

into

ar^2 = 24

, we find

a = 6

. Thus, the 10th term is

a_{10} = ar^9 = 6 \times 2^9 = 3072

480

960

Correct Answer: B

Solution:

The number of bacteria after

n

hours can be calculated using the formula

N = N_0 \times 2^n

, where

N_0

is the initial number of bacteria. Substituting

N_0 = 30

and

n = 5

, we get

N = 30 \times 2^5 = 30 \times 32 = 960

x^2 - 16x + 25 = 0

x^2 - 10x + 16 = 0

x^2 - 8x + 25 = 0

x^2 - 8x + 15 = 0

Correct Answer: A

Solution:

Let the roots be

\alpha

and

\beta

. The A.M. is given by

\frac{\alpha + \beta}{2} = 8

, so

\alpha + \beta = 16

. The G.M. is given by

\sqrt{\alpha \beta} = 5

, so

\alpha \beta = 25

. The quadratic equation is

x^2 - (\alpha + \beta)x + \alpha \beta = 0

, which is

x^2 - 16x + 25 = 0

Correct Answer: A

Solution:

The sum of the first

n

terms of a G.P. is

S_n = a \frac{r^n - 1}{r - 1}

. Here,

a = 5

r = 3

n = 4

. So,

S_4 = 5 \frac{3^4 - 1}{3 - 1} = 5 \frac{81 - 1}{2} = 5 \times 40 = 200

Correct Answer: B

Solution:

Let the terms be

a

ar

, and

ar^2

. The sum is

a(1 + r + r^2) = 21

and the product is

a^3r^3 = 216

. Solving

a^3r^3 = 216

, we have

ar = 6

. Substituting

ar = 6

in the sum equation,

6(1 + r + r^2) = 21

. Solving

1 + r + r^2 = 3.5

, we find

r = 3

Correct Answer: C

Solution:

Let the terms of the G.P. be

a

ar

, and

ar^2

. The conditions give:

a + ar + ar^2 = 56

Subtracting 1, 7, and 21 gives an A.P.:

a - 1, ar - 7, ar^2 - 21

For an A.P., the difference between consecutive terms is constant:

(ar - 7) - (a - 1) = (ar^2 - 21) - (ar - 7)

Simplifying, we get:

ar - a + 6 = ar^2 - ar - 14

ar - a + 6 = ar^2 - ar - 14

Solving these equations:

ar - a + 6 = ar^2 - ar - 14

2ar = ar^2 + 20

r = 2

Substituting

r = 2

into the sum equation:

a + 2a + 4a = 56

7a = 56

a = 8

The first term is

16

768

512

1024

1536

Correct Answer: A

Solution:

The 10th term of a G.P. is given by

a_{10} = ar^{9}

. Here,

a = 3

and

r = 2

. So,

a_{10} = 3 \times 2^{9} = 3 \times 512 = 1536

256

1024

Correct Answer: B

Solution:

In the 3rd round, each of the 16 people from the 2nd round writes to 4 more people:

4^3 = 64

letters.

1/3

1/2

Correct Answer: B

Solution:

Let the terms be

a/r

a

, and

ar

. Then

a/r + a + ar = 39

and

a^3 = 1

. Hence,

a = 1

. Substituting

a = 1

into the sum equation gives

1/r + 1 + r = 39

. Solving

r + 1/r = 38

, we find

r = 3

r = 1/3

. Since the product is 1,

r = 3

is valid.

-1

-3

Correct Answer: C

Solution:

Let the terms be

a/r

a

ar

. Given

a/r + a + ar = 12

and

a^3 = -1

. Solving these gives

r = -3

Correct Answer: A

Solution:

Let the first term be

a

and the common ratio be

r

. Then

ar^3 = 16

and

ar^6 = 128

. Dividing the second equation by the first gives

r^3 = 8

, so

r = 2

1/3

1/2

Correct Answer: C

Solution:

Let the terms be

a/r

a

, and

ar

. The sum is

a/r + a + ar = 39

and the product is

a^3 = 1

, so

a = 1

. Substituting

a = 1

into the sum equation gives

1/r + 1 + r = 39

. Solving for

r

, we find

r = 1/3

Correct Answer: C

Solution:

Starting with

a_1 = 3

, we calculate

a_2 = 2(3) + 1 = 7

a_3 = 2(7) + 1 = 15

, and

a_4 = 2(15) + 1 = 31

Correct Answer: B

Solution:

Substituting the values into the formula,

100 = \frac{n}{2}(2(4) + (n-1)3)

. Simplifying,

200 = n(8 + 3n - 3)

, which gives

200 = n(5 + 3n)

. Solving the quadratic

3n^2 + 5n - 200 = 0

, we find

n = 10

Rs. 163.84

Rs. 327.68

Rs. 655.36

Rs. 819.20

Correct Answer: B

Solution:

The number of letters in the

n

th set is

4^n

. For the 8th set, the number of letters is

4^8

. The cost for mailing one letter is 0.50 Rs (50 paise). Therefore, the total cost is:

\text{Cost} = 4^8 \times 0.50

\text{Cost} = 65536 \times 0.50

\text{Cost} = 32768 \times 0.01

\text{Cost} = Rs. 327.68

Correct Answer: D

Solution:

Let the first term be

a

. Then, the sum of the first three terms is

a + ar + ar^2 = 13.5

. Substituting

r = 0.5

, we get

a + 0.5a + 0.25a = 13.5

. Solving,

1.75a = 13.5

, hence

a = 6

y^2 = xz

x^2 = yz

z^2 = xy

x + y + z = 0

Correct Answer: A

Solution:

In a G.P., the

n

-th term is given by

a_n = ar^{n-1}

. Therefore,

x = ar^3

y = ar^9

, and

z = ar^{15}

. Thus,

y^2 = (ar^9)^2 = a^2r^{18}

and

xz = (ar^3)(ar^{15}) = a^2r^{18}

. Hence,

y^2 = xz

128

Correct Answer: B

Solution:

The number of ancestors doubles each generation. In the 5th generation, there are

2^5 = 32

ancestors.

3072

3840

768

1536

Correct Answer: A

Solution:

The 8th term is

ar^7 = 192

. Solving gives

a = 3

. The 12th term is

ar^{11} = 3 \times 2^{11} = 3072

Correct Answer: B

Solution:

The first three terms are 5, 10, and 20. Their sum is 5 + 10 + 20 = 35.

Correct Answer: B

Solution:

The sum of the first

n

terms of a G.P. is given by

S_n = a \frac{r^n - 1}{r - 1}

where

a = 5

r = 3

, and

S_n = 605

. Substituting these values, we get:

605 = 5 \frac{3^n - 1}{3 - 1}

605 = \frac{5(3^n - 1)}{2}

1210 = 5(3^n - 1)

242 = 3^n - 1

243 = 3^n

Since

243 = 3^5

, we have

n = 5

True

False

Cannot be determined

None of the above

Correct Answer: A

Solution:

The terms are

ar^3 = x

ar^9 = y

ar^{15} = z

. The ratio

\frac{y}{x} = r^6

and

\frac{z}{y} = r^6

, hence

x

y

z

are in G.P.

-2

-1

Correct Answer: B

Solution:

Let the first term be

a

and the common ratio be

r

. The first two terms are

a

and

ar

. Given

a + ar = -4

and

ar^4 = 4(ar^2)

. Solving these gives

r = -2

Correct Answer: A

Solution:

The 5th term of a G.P. is given by

a_5 = ar^{n-1} = 3 \times 2^{5-1} = 3 \times 16 = 48

1/3

1/2

Correct Answer: C

Solution:

Let the terms be

a/r, a, ar

. Then

a/r + a + ar = 39

and

(a/r) \times a \times ar = 1

. Solving gives

r = 1/3

Correct Answer: B

Solution:

The sequence is an arithmetic progression with first term

a_1 = 1

and common difference

d = 2

. The

n

-th term is given by

a_n = a_1 + (n-1)d

. Substituting

n = 20

a_{20} = 1 + 19 \times 2 = 41

512

1024

2046

4096

Correct Answer: B

Solution:

The number of ancestors is a G.P. with first term 2 and common ratio 2. The 10th term is

2^{10} = 1024

True or False

Correct Answer: True

Solution:

The Fibonacci sequence was indeed discovered by the Italian mathematician Leonardo Fibonacci around the 12th century.

Correct Answer: False

Solution:

If the number of bacteria doubles every hour starting with 30, after 4 hours, the number will be

30 \times 2^4 = 480

Correct Answer: True

Solution:

The geometric mean (G.M.) of two positive numbers

a

and

b

is indeed given by

\sqrt{ab}

Correct Answer: False

Solution:

The sum of the first

n

terms of a geometric progression can be finite or infinite, depending on the common ratio.

Correct Answer: True

Solution:

This formula is used to calculate the sum of the first

n

terms of a G.P. when the common ratio

r

is not equal to 1.

Correct Answer: True

Solution:

The Fibonacci sequence is defined by the recurrence relation

a_1 = a_2 = 1

, and

a_n = a_{n-1} + a_{n-2}

Solution:

The Fibonacci sequence is generated by the recurrence relation

a_n = a_{n-1} + a_{n-2}

(n+1)

th to the

Solution:

In a geometric progression, each term is obtained by multiplying the previous term by a constant ratio.

Correct Answer: False

Solution:

Solution:

A sequence is called a geometric progression if the ratio of any term to its preceding term is the same throughout.

Correct Answer: True

Solution:

This is the correct formula for the sum of the first

n

terms of a geometric progression where

r

is the common ratio and

a

is the first term.

Correct Answer: False

Solution:

The common ratio in a geometric progression can be any non-zero number, including values less than 1.

Correct Answer: False

Solution:

The sequence of prime numbers does not follow a specific algebraic formula; it is described by a verbal description.

Correct Answer: True

Solution:

The sequence of even natural numbers is 2, 4, 6, ..., which can be expressed as

a_n = 2n

for natural numbers

n

Correct Answer: False

Solution:

The geometric mean of two numbers

a

and

b

is given by

\sqrt{ab}

Solution:

The geometric mean of two positive numbers

a

and

b

is indeed

\sqrt{ab}

Sequences and Series

AI Learning Assistant

Summary

Chapter 8: Sequences and Series

Summary

Learning Objectives

Detailed Notes

Sequences and Series Notes

Introduction to Sequences

Geometric Progression (G.P.)

Examples of G.P.

Important Properties of G.P.

Miscellaneous Examples

Exercises

Conclusion

Exam Tips & Common Mistakes

Common Mistakes and Exam Tips

Common Pitfalls

Tips for Success

Practice & Assessment

Multiple Choice Questions

Q1.The sum of the first three terms of a G.P. is 13 and their product is 27. What is the common ratio?

Correct Answer: C

Q2.If the 4th, 10th, and 16th terms of a G.P. are xxx, yyy, and zzz, respectively, what is the relationship between xxx, yyy, and zzz?

Correct Answer: B

Q3.Consider a geometric progression where the first term a=5a = 5a=5 and the common ratio r=3r = 3r=3. If the sum of the first nnn terms is 605, find the value of nnn.

Correct Answer: B

Q4.The sum of the first three terms of a G.P. is 12, and their product is -1. What is the first term?

Correct Answer: A

Q5.If the pᵗʰ, qᵗʰ, and rᵗʰ terms of a G.P. are a, b, and c respectively, which of the following is true?

Correct Answer: B

Q6.A person deposited Rs 10000 in a bank at 5% simple interest annually. What will be the total amount after 20 years?

Correct Answer: B

Q7.What is the 12th term of a geometric progression (G.P.) if the 8th term is 192 and the common ratio is 2?

Correct Answer: C

Q8.A sequence is defined by a1=1a_1 = 1a1​=1 and an=an−1+2a_n = a_{n-1} + 2an​=an−1​+2 for n≥2n \geq 2n≥2. What is the 5th term?

Correct Answer: B

Q9.If the sum of the first three terms of a G.P. is 13 and their product is 27, what is the first term?

Correct Answer: B

Q10.Find the 5th term of the sequence defined by an=2n+3a_n = 2n + 3an​=2n+3.

Correct Answer: C

Q11.Find the geometric mean of 16 and 64.

Correct Answer: A

Q12.A manufacturer reckons that the value of a machine, which costs Rs. 15625, will depreciate each year by 20%. What will be its value at the end of 5 years?

Correct Answer: A

Q13.Consider a geometric progression with the first term a=3a = 3a=3 and common ratio r=2r = 2r=2. What is the sum of the first 5 terms of this G.P.?

Correct Answer: B

Q14.A person deposits Rs 1000 in a bank at an annual interest rate of 5% compounded annually. What will be the amount after 3 years?

Correct Answer: A

Q15.A sequence is defined by a1=2a_1 = 2a1​=2 and an=3an−1+1a_n = 3a_{n-1} + 1an​=3an−1​+1 for n≥2n \geq 2n≥2. What is the 4th term of this sequence?

Correct Answer: B

Q16.If a sequence is defined by an=2n+5a_n = 2n + 5an​=2n+5, what is the 3rd term?

Correct Answer: B

Q17.What is the common ratio of the geometric progression where the 8th term is 192 and the 12th term is 3072?

Correct Answer: A

Q18.A person deposits Rs 500 in a bank with an annual interest rate of 10% compounded annually. What will be the amount after 10 years?

Correct Answer: A

Q19.If the 4th, 10th, and 16th terms of a G.P. are xxx, yyy, and zzz, respectively, prove that xxx, yyy, zzz are in G.P. What is the common ratio?

Correct Answer: A

Q20.In a G.P., the 3rd term is 24 and the 6th term is 192. What is the common ratio?

Correct Answer: A

Q21.If a sequence is defined by an=3n+2a_n = 3n + 2an​=3n+2, what is the 10th term?

Correct Answer: C

Q22.Consider a geometric progression where the first term is a=3a = 3a=3 and the common ratio is r=2r = 2r=2. If the sum of the first nnn terms is 93, find the value of nnn.

Correct Answer: B

Q23.If the arithmetic mean (A.M.) and geometric mean (G.M.) of two numbers are 8 and 5 respectively, what is the product of these two numbers?

Correct Answer: A

Q24.A sequence is defined by the rule an=3n+1a_n = 3n + 1an​=3n+1. What is the 5th term of this sequence?

Correct Answer: A

Q25.A manufacturer estimates that a machine's value depreciates by 20% each year. If the initial value is Rs. 15625, what will be its estimated value at the end of 5 years?

Correct Answer: A

Q26.How many bacteria will be present at the end of the 4th hour if the number of bacteria doubles every hour starting with 30 bacteria?

Correct Answer: C

Q27.If the 4th, 10th, and 16th terms of a geometric progression are xxx, yyy, and zzz, respectively, which of the following is true?

Correct Answer: B

Q28.If the first term of a geometric progression (G.P.) is 2 and the common ratio is 3, what is the 4th term?

Correct Answer: B

Q29.In a geometric progression, the 4th term is 16 and the 10th term is 1024. What is the common ratio?

Correct Answer: C

Q30.A person deposits Rs 500 in a bank that offers an annual interest rate of 10% compounded annually. What will be the amount after 10 years?