Class 11 Mathematics Chapter 8 Sequences And Series

Key Concept: Sum of n Terms (r≠1

d) 80

[Solution Description] The sum of the first $n$ terms of a G.P. when $r \neq 1$ is given by the formula $S_n = \frac{a(1-r^n)}{1-r}$. Here, $a = 2$, $r = 3$, and $n = 4$.

Substitute the given values into the formula:

$$S_4 = \frac{2(1-3^4)}{1-3}$$

Calculate $3^4$:

$$3^4 = 81$$

Now plug in the value and simplify:

$$S_4 = \frac{2(1-81)}{-2} = \frac{2(-80)}{-2} = \frac{-160}{-2} = 80$$

Thus, the sum of the first 4 terms is $80$.

Key Concept: Definition Recall

b) A sum of terms from a sequence

[Solution Description] A series is defined as the sum of the terms of a sequence. It involves adding all the terms together to get the total.

Key Concept: Term Identification

b) 96

[Solution Description]

To find the sixth term of the given G.P., we use the formula for the general term:

$$a_n = ar^{n-1}$$

Given that $a = 3$, $r = 2$, and $n = 6$, substitute these values into the formula:

$$a_6 = 3 \times 2^{6-1} = 3 \times 2^5$$

Calculate $2^5$:

$$2^5 = 32$$

Therefore,

$$a_6 = 3 \times 32 = 96$$

Key Concept: Formula Application

d) 1/9

[Solution Description]

To find the 5th term of the H.P., calculate the 5th term of the A.P. using:

$$a_n = a + (n-1)d$$

Given $a = 5$, $n = 5$, and $d = 1$:

$$a_5 = 5 + (5-1)\times1 = 5 + 4 = 9$$

Thus, the 5th term of the A.P. is 9. Therefore, the 5th term of the H.P. is its reciprocal:

$$h_5 = \frac{1}{9}$$

Key Concept: Basic G.M. Calculation

c) 8

[Solution Description] The geometric mean (G.M.) of two numbers $a$ and $b$ is given by the formula $\sqrt{ab}$. Here, $a = 4$ and $b = 16$.

Calculate:

$$G.M. = \sqrt{4 \times 16} = \sqrt{64} = 8$$

Therefore, the answer is 8.

Key Concept: Common Ratio

c) 3

[Solution Description]

To find the common ratio $r$, we divide any term by its preceding term. Taking the second term divided by the first term:

$$r = \frac{9}{3} = 3$$

Therefore, the common ratio is 3.

Key Concept: Geometric Mean

c) 12

[Solution Description]

The formula for the geometric mean (G.M.) of two numbers $a$ and $b$ is:

$$G.M. = \sqrt{ab}$$

Substituting $a = 16$ and $b = 9$:

$$G.M. = \sqrt{16 \times 9} = \sqrt{144} = 12$$

Hence, the geometric mean is 12.

Key Concept: Sum of First n Terms

b) 10

[Solution Description] Since the common ratio $r = 1$, the formula for the sum of the first $n$ terms is $S_n = na$. Here, $a = 2$ and $n = 5$.

Calculating the sum:

$$S_5 = 5 \times 2$$

$$= 10$$

Thus, the sum of the first 5 terms is 10.

Key Concept: Sequence Definition

b) An ordered list of numbers following a specific rule

[Solution Description] A sequence is an ordered list of numbers that follows a specific rule or pattern. Each number in the sequence is called a term, and the position of each term is fixed.

Key Concept: Sum of n Terms

b) 45

[Solution Description]

The sum of the first $n$ terms of a G.P. can be calculated using $S_n = a \frac{1-r^n}{1-r}$ for $r \neq 1$. Here, $a = 3$, $r = \frac{6}{3} = 2$, and $n = 4$.

Substitute these into the formula:

$$S_4 = 3 \cdot \frac{1 – 2^4}{1 – 2} = 3 \cdot \frac{1 – 16}{-1}$$

$$= 3 \cdot \frac{-15}{-1} = 3 \times 15 = 45$$

Therefore, the sum of the first 4 terms is 45.

Key Concept: Sequence Definition

d) All of the above

[Solution Description] A sequence is an arrangement of numbers in a definite order according to some rule.

Key Concept: Basic Formula

b) $\sqrt{ab}$

[Solution Description] The formula for the Geometric Mean (G.M.) of two numbers $a$ and $b$ is calculated using the square root of their product. This is represented by:

$$G = \sqrt{ab}$$

Hence, the correct answer is $\sqrt{ab}$.

Key Concept: Common Ratio Calculation

a) 3

[Solution Description] Let’s calculate the common ratio using the formula:

$$r = \left(\frac{b}{a}\right)^{\frac{1}{n+1}}$$

where $a = 2$, $b = 162$, and $n = 4$.

$$r = \left(\frac{162}{2}\right)^{\frac{1}{4+1}} = \left(81\right)^{\frac{1}{5}} = 3$$

Therefore, the common ratio is 3.

Key Concept: Calculate nth Term

c) 26

[Solution Description] To find the 8th term ($a_8$), we use the formula for the nth term:

$$a_n = a + (n-1)d$$

Here, $a = 5$, $d = 3$, and $n = 8$. Plug these values into the formula to get:

$$a_8 = 5 + (8-1) \cdot 3 = 5 + 21 = 26$$

Therefore, the 8th term is 26.

Key Concept: Identify Pattern

c) $\left(\frac{n(n+1)}{2}\right)^2$

[Solution Description]

The sum of cubes of the first $n$ natural numbers is given by the formula: $\left( \frac{n(n+1)}{2} \right)^2$. This indicates that the sum of cubes follows the pattern of being equivalent to the square of the sum of the first $n$ natural numbers.

Key Concept: A.M. and G.M. Relationship

c) 26.58 and 5.42

[Solution Description] Let the two numbers be $a$ and $b$. The formulas for A.M. and G.M are:

$$A.M. = \frac{a+b}{2} = 16$$

$$G.M. = \sqrt{ab} = 12$$

From the A.M. equation:

$$a + b = 32$$

From the G.M. equation:

$$ab = 144$$

Solving these equations simultaneously, we express one number in terms of another from the A.M. equation:

$$a = 32 – b$$

Substitute into the G.M. equation:

$$(32-b)b = 144$$

Solve this quadratic equation for $$b$$:

$$b^2 – 32b + 144 = 0$$

Using the quadratic formula $$b = \frac{-B \pm \sqrt{B^2-4AC}}{2A}$$:

$$b = \frac{32 \pm \sqrt{1024-576}}{2}$$

$$b = \frac{32 \pm \sqrt{448}}{2}$$

$$b = \frac{32 \pm 21.166}{2}$$

Approximating $\sqrt{448} = 21.166$ gives us solutions:

$$b \approx 26.583 \quad \text{or} \quad b \approx 5.417$$

Correspondingly, $a \approx 5.417$ or $a \approx 26.583$. So the numbers are approximations of 26.58 and 5.42.

Key Concept: Fibonacci Sequence

d) Leonardo Fibonacci

[Solution Description] Leonardo Fibonacci is known for developing the Fibonacci sequence which appears in his book “Liber Abaci”.

Key Concept: Sum Formula Identification

a) $S_n = \frac{n(n+1)}{2}$

[Solution Description] The formula for the sum of the first $n$ natural numbers is given by $S_n = \frac{n(n+1)}{2}$.

Key Concept: Definition Recall

d) The ratio of a term to its previous term.

[Solution Description] In a geometric progression, the common ratio is obtained by dividing any term by its preceding term. It remains constant throughout the sequence.

Key Concept: Sequence Definition

b) An ordered collection of numbers

[Solution Description] A sequence is an ordered collection of numbers where each number has a specific position, such as first, second, third, etc. Therefore, the definition that accurately describes this concept is “An ordered collection of numbers.”

Key Concept: Common Ratio Calculation

b) 3

[Solution Description] To find the common ratio of a geometric progression, divide any term by its preceding term. Here, $r = \frac{9}{3} = 3$ or $\frac{27}{9} = 3$. Therefore, the common ratio is 3.

Key Concept: Basic G.P. Calculation

b) 3

[Solution Description] In a geometric progression, the nth term is given by $a_n = ar^{n-1}$, where $a$ is the first term and $r$ is the common ratio. Given $a = 9$, $a_3 = 81$, we have:

$$ar^2 = 81$$

Substituting for $a = 9$:

$$9r^2 = 81$$

Divide both sides by 9:

$$r^2 = 9$$

Taking the square root on both sides gives $r = 3$ or $r = -3$. Therefore, the positive common ratio is 3.

Key Concept: Geometric Progression Basics

c) 2

[Solution Description]

To find the common ratio $r$ of a geometric progression (G.P.), we divide any term by its preceding term. Here, dividing the second term by the first term:

$$r = \frac{10}{5} = 2.$$

Thus, the common ratio is 2.

Key Concept: Geometric Progression Identification

b) 3, 6, 12, 24

[Solution Description] A geometric progression has a constant ratio between consecutive terms. For option B, the sequence 3, 6, 12, 24 has a common ratio of 2.

Key Concept: Finite vs Infinite

b) Infinite

[Solution Description] This series represents a geometric progression where the first term is 3 and the common ratio is also 3. Since there is no last term specified, it continues indefinitely, making it an infinite series.

Key Concept: Definition Recall

c) 1 + 3 + 5 + 7

[Solution Description]

A series is the sum of the terms of a sequence. The correct option is an expression involving addition.

Key Concept: Geometric Progression Identification

a) 3, 9, 27, 81

[Solution Description] A geometric progression has a constant ratio between consecutive terms. In option a), each term is obtained by multiplying the previous term by 3, so it forms a G.P.

Key Concept: Relationship Understanding

c) 144

[Solution Description]

The relationship between A.M. and G.M. for two numbers $a$ and $b$ states that the A.M. is $\frac{a+b}{2} = 13$ and the G.M. is $\sqrt{ab} = 12$. We need to find the product of the two numbers, which is $ab$.

Start by squaring the equation for G.M:

$$ab = 12^2 = 144$$

Therefore, the product of the two numbers is 144.

Note: You don’t need to solve for $a$ and $b$ separately since the question only asks for their product.

Key Concept: Geometric Mean

c) 36

[Solution Description] The geometric mean (G.M.) between two numbers $a$ and $b$ is calculated as:

$$\text{G.M.} = \sqrt{ab}$$

For this problem, $a = 9$ and $b = 144$. Substituting these values gives:

$$\text{G.M.} = \sqrt{9 \times 144} = \sqrt{1296} = 36$$

Therefore, the geometric mean is 36.

Key Concept: Common Ratio

b) 3

[Solution Description] In a geometric progression, the common ratio $r$ can be found by dividing any term by its preceding term. Here, dividing the second term by the first: $\frac{12}{4} = 3$. Thus, the common ratio is 3.

Key Concept: Term Position Identification

b) 7th

[Solution Description] The nth term of a geometric progression is given by $a_n = ar^{n-1}$. Here, the first term $a = 1$ and the common ratio $r = 3$. We need $a_n = 729$, so:

$$1 \cdot 3^{n-1} = 729$$

So,

$$3^{n-1} = 729$$

Expressing 729 as a power of 3 gives:

$$729 = 3^6$$

So,

$$n-1 = 6$$

Solving for $n$ gives:

$$n = 7$$

Hence, the term is the 7th term.

Key Concept: Definition of Terms

c) A list of numbers arranged in a definite order according to some rule.

[Solution Description] A sequence is an arrangement of numbers in a definite order according to some rule.

Key Concept: Sum of Terms with r=1

c) 12

[Solution Description] When the common ratio $r = 1$, the sum of the first $n$ terms is given by:

$$S_n = na.$$

Here, $n = 6$ and $a = 2$. So,

$$S_6 = 6 \times 2 = 12.$$

Therefore, the sum of the first 6 terms is 12.

Key Concept: Find nth Term

c) 729

[Solution Description] The first term $a$ is 3, and the common ratio $r$ is $\frac{9}{3} = 3$. The formula for the nth term of a G.P. is given by $a_n = ar^{n-1}$. For the 6th term:

$$a_6 = 3 \times 3^{6-1} = 3 \times 3^5 = 3 \times 243 = 729.$$

Key Concept: Series Definition

c) The sum of the terms of a sequence

[Solution Description] A series is defined as the sum of the terms in a sequence. It involves adding all the terms of the sequence together.

Key Concept: Common Difference Calculation

c) 5

The common difference $d$ in an arithmetic progression is found by subtracting any term from the subsequent term. Here, we can calculate $d = 12 – 7 = 5$.

Key Concept: Term Calculation

a), 1/5

[Solution Description]

In a Harmonic Progression, the nth term is the reciprocal of the nth term of the corresponding A.P. Here, the terms are 1, 1/2, 1/3,… The corresponding A.P. is 1, 2, 3,… The 5th term in this A.P. is 5. Therefore, the 5th term of the H.P. is the reciprocal of 5, which is 1/5.

Key Concept: Advanced Sequence Analysis, Geometric Mean Calculation

b) It is equal to $\frac{1}{9}$

[Solution Description] The first three terms are $b_1 = \frac{1}{3}$, $b_2 = \frac{1}{9}$, and $b_3 = \frac{1}{27}$. The geometric mean of these terms is calculated as:

$$\text{Geometric Mean} = \sqrt[3]{b_1 \cdot b_2 \cdot b_3} = \sqrt[3]{\frac{1}{3} \cdot \frac{1}{9} \cdot \frac{1}{27}} = \sqrt[3]{\frac{1}{729}} = \frac{1}{9}.$$

Therefore, the geometric mean of the first three terms is $\frac{1}{9}$.

Key Concept: Sequence Rule

b) $a_n = 2n + 1$

[Solution Description] The formula for the nth term of an arithmetic sequence is given by $a_n = a_1 + (n-1) \cdot d$. Substituting $a_1 = 3$ and $d = 2$, we get:

$$a_n = 3 + (n-1) \cdot 2 = 3 + 2n – 2 = 2n + 1$$

Key Concept: Equation Solving

a) (27, 3)

[Solution Description]

Let the numbers be $a$ and $b$. From the given data, $A = \frac{a + b}{2} = 15$ and $G = \sqrt{ab} = 9$.

So, we have:

$$a + b = 30$$

and

$$ab = 81$$

These form a pair of equations that can be solved simultaneously:

From $a+b = 30$, we get $b = 30 – a$.

Substitute in $ab = 81$:

$$a(30-a) = 81$$

$$30a – a^2 = 81$$

$$a^2 – 30a + 81 = 0$$

Solving this quadratic equation using the quadratic formula:

$$a = \frac{-(-30) \pm \sqrt{(-30)^2 -4 \times 1 \times 81}}{2 \times 1}$$

$$a = \frac{30 \pm \sqrt{900 – 324}}{2}$$

$$a = \frac{30 \pm \sqrt{576}}{2}$$

$$a = \frac{30 \pm 24}{2}$$

The solutions are $a = \frac{54}{2} = 27$ or $a = \frac{6}{2} = 3$.

Thus, the pairs $(a, b)$ are $(27, 3)$ and $(3, 27)$.

Key Concept: Finite vs Infinite Series

a) Finite

[Solution Description] The upper limit of the summation is 10, which means it stops at $k = 10$. Therefore, the series is finite since it has a limited number of terms.

Key Concept: Mathematician’s Contribution

d) James Gregory

[Solution Description] James Gregory is known for his work in classifying mathematical series.

Key Concept: Complex Sum Calculation, Real World Application

b) 40500

[Solution Description] Initially, the bacteria count is 500. Since it triples every two hours, after 2 hours, the count becomes $500 \times 3 = 1500$. The sequence for every two hours will then be: 500, 1500, 4500, 13500, etc.

Now, let’s find the total count after 8 hours, which corresponds to 4 intervals of doubling:

– After first interval (2 hours): $1500$

– After second interval (4 hours): $1500 \times 3 = 4500$

– After third interval (6 hours): $4500 \times 3 = 13500$

– After fourth interval (8 hours): $13500 \times 3 = 40500$

Thus, the total count at the end of 8 hours will be 40500.