NCERT Solutions for class 11 Maths Chapter 9: Sequences and Series

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NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series are added in the article. A sequence is a list of elements that can be repeated in any order, whereas a series is the total of all elements. An arithmetic progression is one of the most common examples of sequence and series.

Key concepts covered in NCERT Solutions Class 11 Maths Chapter 9 Sequence and Series are:

Download: NCERT Solutions for Class 11 Mathematics Chapter 9 pdf

Class 11 Maths NCERT Solutions Chapter 9 Sequence and Series

Class 11 Maths NCERT Solutions Chapter 9 Sequence and Series are provided below:

Also check: Sequence and Series

Important Topics for Class 11 Maths NCERT Solutions Chapter 9 Sequence and Series

Important Topics for Class 11 Maths NCERT Solutions Chapter 9 Sequence and Series are elaborated below:

Arithmetic Progression (A.P.)

An Arithmetic Progression (AP) is a sequence where differences between every two consecutive terms are the same. There is a possibility to derive a formula for the n^th term.

Example: Find the general term of the arithmetic progression -3, -(1/2), 2…

Solution: Given sequence is -3, -(1/2),2…

Here, first term is a=-3, and common difference is:

d = -(1/2) -(-3) = -(1/2)+3 = 5/2

By AP formulas, the general term of an AP is calculated by the formula:

an = a+(n-1)d

an = -3 +(n-1) 5/2

= -3+ (5/2)n - 5/2
= 5n/2 - 11/2

Thus, general term of the given AP is: an = 5n/2 - 11/2

Geometric Progression (G. P.)

In a geometric progression, each successive term is obtained by multiplying the common ratio to its preceding term. The formula for the nth term of a geometric progression whose first term is a and common ratio is r is:

a_n=ar_n-1

Geometric Mean (G.M.)

Geometric Mean (GM) is average value or mean which signifies central tendency of the set of numbers by finding product of their values.

Example: What is the geometric mean of 4,8.3,9 and 17?

Solution: Multiply the numbers together and then take the 5th root (because there are 5 numbers) = (4 * 8 * 3 * 9 * 17)(1/5) = 6.81

NCERT Solutions For Class 11 Maths Chapter 9 Exercises:

The detailed solutions for all the NCERT Solutions for Chapter 9 Sequence and Series under different exercises are as follows:

Also check:

Important Chapter Related Links
Harmonic Mean	Sum of Squares	Difference between Sequence and Series
Summation Formula	Geometric Series Formula	Sum of Arithmetic Sequence Formula
Sum Of Cubes Formula	Arithmetic Mean Formula	Arithmetic Sequence Explicit Formula
Arithmetic Sequence Recursive Formula	Difference of Cubes Formula	Difference of Squares Formula
Fourier Series Formula	Infinite Geometric Series Formula	Infinite Series Formula
Taylor Series Formula	Prime Number Formula	Arithmetic Geometric Sequence

Also check:

Math Study Guides
Maths Important Formulas	Maths MCQs	Comparison Articles in Maths
Difference Between Topics in Maths	Math Study Material	Arithmetic

CBSE CLASS XII Related Questions

1.
Find the foot of the perpendicular from the point \( (0, 2, 3) \) on the line \( \frac{-x-3}{-5} = \frac{1-y}{-2} = \frac{3z+12}{9} \) and hence find the length of the perpendicular.
View Solution
2.
Find the value of \( p \) if the shortest distance between the lines \( \vec{r} = (\hat{i} + 2\hat{j} + \hat{k}) + \lambda(\hat{i} - \hat{j} + \hat{k}) \) and \( \vec{r} = (p\hat{i} - \hat{j} - \hat{k}) + \mu(2\hat{i} + \hat{j} + 2\hat{k}) \) is \( \frac{3}{\sqrt{2}} \) units.
View Solution
3.
If \( y\sqrt{x^2+1} = \log (\sqrt{x^2+1}-x) \), show that \( (x^2+1) \frac{dy{dx} + xy + 1 = 0 \).}
View Solution
4.
Solve the differential equation \( (x - \sin y) \, dy + \tan y \, dx = 0 \).
View Solution
5.
The volume of a wooden block in the shape of a cube increases at a constant rate as the air becomes moist during the rainy season. Show that the rate of change of its surface area varies inversely as the length of edge of the cube.
View Solution
6.
A box contains 6 cards numbered 1 to 6. A student is asked to pick up two cards, one by one after replacement and note down the numbers on the cards. Let A be the event of getting sum of the numbers on two cards as 10, and B, the event of a number other than 4 on the first card selected. Find \( P(A \text{ and } B) \) and find whether the events A and B are independent events or not.
View Solution

View All

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NCERT Solutions for class 11 Maths Chapter 9: Sequences and Series

Class 11 Maths NCERT Solutions Chapter 9 Sequence and Series

Important Topics for Class 11 Maths NCERT Solutions Chapter 9 Sequence and Series

CBSE CLASS XII Related Questions

1.
Find the foot of the perpendicular from the point \( (0, 2, 3) \) on the line \( \frac{-x-3}{-5} = \frac{1-y}{-2} = \frac{3z+12}{9} \) and hence find the length of the perpendicular.

2.
Find the value of \( p \) if the shortest distance between the lines \( \vec{r} = (\hat{i} + 2\hat{j} + \hat{k}) + \lambda(\hat{i} - \hat{j} + \hat{k}) \) and \( \vec{r} = (p\hat{i} - \hat{j} - \hat{k}) + \mu(2\hat{i} + \hat{j} + 2\hat{k}) \) is \( \frac{3}{\sqrt{2}} \) units.

3.
If \( y\sqrt{x^2+1} = \log (\sqrt{x^2+1}-x) \), show that \( (x^2+1) \frac{dy{dx} + xy + 1 = 0 \).}

4.
Solve the differential equation \( (x - \sin y) \, dy + \tan y \, dx = 0 \).

5.
The volume of a wooden block in the shape of a cube increases at a constant rate as the air becomes moist during the rainy season. Show that the rate of change of its surface area varies inversely as the length of edge of the cube.

Similar Mathematics Concepts

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NCERT Solutions for class 11 Maths Chapter 9: Sequences and Series

Class 11 Maths NCERT Solutions Chapter 9 Sequence and Series

Important Topics for Class 11 Maths NCERT Solutions Chapter 9 Sequence and Series

CBSE CLASS XII Related Questions

1.Find the foot of the perpendicular from the point \( (0, 2, 3) \) on the line \( \frac{-x-3}{-5} = \frac{1-y}{-2} = \frac{3z+12}{9} \) and hence find the length of the perpendicular.

2.Find the value of \( p \) if the shortest distance between the lines \( \vec{r} = (\hat{i} + 2\hat{j} + \hat{k}) + \lambda(\hat{i} - \hat{j} + \hat{k}) \) and \( \vec{r} = (p\hat{i} - \hat{j} - \hat{k}) + \mu(2\hat{i} + \hat{j} + 2\hat{k}) \) is \( \frac{3}{\sqrt{2}} \) units.

3.If \( y\sqrt{x^2+1} = \log (\sqrt{x^2+1}-x) \), show that \( (x^2+1) \frac{dy{dx} + xy + 1 = 0 \).}

4.Solve the differential equation \( (x - \sin y) \, dy + \tan y \, dx = 0 \).

5.The volume of a wooden block in the shape of a cube increases at a constant rate as the air becomes moist during the rainy season. Show that the rate of change of its surface area varies inversely as the length of edge of the cube.

Similar Mathematics Concepts

Comments

SUBSCRIBE TO OUR NEWS LETTER

1.
Find the foot of the perpendicular from the point \( (0, 2, 3) \) on the line \( \frac{-x-3}{-5} = \frac{1-y}{-2} = \frac{3z+12}{9} \) and hence find the length of the perpendicular.

2.
Find the value of \( p \) if the shortest distance between the lines \( \vec{r} = (\hat{i} + 2\hat{j} + \hat{k}) + \lambda(\hat{i} - \hat{j} + \hat{k}) \) and \( \vec{r} = (p\hat{i} - \hat{j} - \hat{k}) + \mu(2\hat{i} + \hat{j} + 2\hat{k}) \) is \( \frac{3}{\sqrt{2}} \) units.

3.
If \( y\sqrt{x^2+1} = \log (\sqrt{x^2+1}-x) \), show that \( (x^2+1) \frac{dy{dx} + xy + 1 = 0 \).}

4.
Solve the differential equation \( (x - \sin y) \, dy + \tan y \, dx = 0 \).

5.
The volume of a wooden block in the shape of a cube increases at a constant rate as the air becomes moist during the rainy season. Show that the rate of change of its surface area varies inversely as the length of edge of the cube.