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NCERT solutions for class 11 maths chapter 9 sequences and series are discussed here. Sequence means the progression of numbers in a definite order and series means the sum of the objects of the sequence. In the previous classes, you have studied about arithmetic progression(A.P). In this NCERT Book chapter, we will discuss more arithmetic progression(A.P) and geometric progression(G.P). In this article, you will get sequences and series class 11 NCERT solutions. these NCERT solutions are prepared by experts in keeping in mind of latest syllabus of CBSE 2023.
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The chapter 9 class 11 maths includes topics such as arithmetic progression(A.P), geometric progression(G.P), arithmetic means(A.M), geometric mean(G.M), the relationship between A.M. and G.M, sum to n terms of special series, sum to n terms of squares and cubes of natural numbers are covered in this NCERT Book chapter. Students can find all NCERT Solutions for Class 11 here and practice them. You will get questions related to these topics in the class 11 maths chapter 9 NCERT solutions. In this ch 9 maths class 11, there are two types of sequence.
Also read:
Progression:
A sequence whose terms follow certain patterns is known as a progression.
Arithmetic Progression (AP):
An arithmetic progression (A.P.) is a sequence where the terms either increase or decrease regularly by the same constant called the common difference (d).
The first term is denoted by a, and the last term of an AP is denoted by l.
The nth term of an AP: a_{n} = a + (n – 1)d.
nth term of an AP from the last term: a’_{n} = a_{n} – (n – 1)d.
Common difference of an AP: d = a_{n} – a_{n-1}, for all n > 1.
Sum of n Terms of an AP: Sn = n/2 [2a + (n – 1)d] = n/2 (a_{1}+ a_{n}).
A sequence is an AP if the sum of n terms is of the form An_{2} + B_{n}, where A and B are constants, and A = half of common difference, i.e., 2A = d.
Arithmetic Mean:
If a, A, and b are in an AP, then A = (a + b)/2 is called the arithmetic mean of a and b.
If a_{1}, a_{2}, a_{3},…, a_{n} are n numbers, then their arithmetic mean is given by: A = (a_{1} + a_{2} + a_{3} + ... + a_{n})/n.
The common difference d is given as: d = (b – a)/(n + 1).
The Sum of n arithmetic means between a and b is n (a + b)/2.
Geometric Progression (GP):
A sequence in which the ratio of two consecutive terms is constant is called a geometric progression (GP).
The constant ratio is called the common ratio (r), i.e., r = a_{n+1}/a_{n}, for all n > 1.
The general term or nth term of GP: an = ar^{(n-1)}.
nth term of a GP from the end: a’n = 1/r^{n-1}, where l is the last term.
If a, b, and c are three consecutive terms of a GP, then b^{2} = ac.
Geometric Mean (GM):
If a, G, and b are in a GP, then G is called the geometric mean of a and b and is given by G = √(ab).
If a, G_{1}, G_{2}, G_{3}, …, G_{n}, b are in GP, then G_{1}, G_{2}, G_{3}, …, G_{n} are in GMs between a and b.
The common ratio r is given as: r = (b/a)^{(1/n+1)}.
The GM of a_{1}, a_{2}, a_{3},…, an is given by: GM = (a_{1} . a_{2} . a_{3} …a_{n})^{(1/n)}.
Product of n GMs is G_{1} × G_{2} × G_{3} × … × G_{n} = G_{n} = (ab)^{(n/2)}.
Sum of First n Natural Numbers:
Sum of the first n natural numbers is:
Σn = 1 + 2 + 3 + … + n = n(n+1)/2.
Sum of Squares of First n Natural Numbers:
Sum of squares of the first n natural numbers is:
Σn^{2} = 1^{2} + 2^{2} + 3^{2} + … + n^{2} = n(n+1)(2n+1)/6.
Sum of Cubes of First n Natural Numbers:
Sum of cubes of the first n natural numbers is:
Σn^{3} = 1^{3} + 2^{3} + 3^{3} + .. + n^{3} = (n(n+1)(2n+1)/6)^{2}.
Free download NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series for CBSE Exam.
Sequences and series class 11 questions and answers - Exercise: 9.1
Question:1 Write the first five terms of each of the sequences in Exercises 1 to 6 whose nth terms are:
Answer:
Given :
Therefore, the required number of terms =3, 8, 15, 24, 35
Question:2 Write the first five terms of each of the sequences in Exercises 1 to 6 whose nthterms are:
Answer:
Given :
Therefore, the required number of terms
Question:3 Write the first five terms of each of the sequences in Exercises 1 to 6 whose nthterms are:
Answer:
Given :
Therefore, required number of terms
Question:4 Write the first five terms of each of the sequences in Exercises 1 to 6 whose nth terms are:
Answer:
Given :
Therefore, the required number of terms
Question:5 Write the first five terms of each of the sequences in Exercises 1 to 6 whose nth terms are:
Answer:
Given :
Therefore, the required number of terms
Question:6 Write the first five terms of each of the sequences in Exercises 1 to 6 whose nth terms are:
Answer:
Given :
Therefore, the required number of terms
Question:7 Find the indicated terms in each of the sequences in Exercises 7 to 10 whose nth terms are:
Answer:
Put
Put n=24,
Hence, we have
Question:8 Find the indicated terms in each of the sequences in Exercises 7 to 10 whose nth terms are:
Answer:
Given :
Put n=7,
Heence, we have
Question:9 Find the indicated terms in each of the sequences in Exercises 7 to 10 whose nth terms are:
Answer:
Given :
Put n =9,
The value of
Question:10 Find the indicated terms in each of the sequences in Exercises 7 to 10 whose nth terms are:
Answer:
Given :
Put n=20,
Hence, value of
Answer:
Given :
Hence, five terms of series are
Series
Answer:
Given :
Hence, five terms of series are
Series
Answer:
Given :
Hence, five terms of series are
Series
Question:14 The Fibonacci sequence is defined by
Answer:
Given : The Fibonacci sequence is defined by
Class 11 maths chapter 9 question answer - Exercise: 9.2
Question:1 Find the sum of odd integers from 1 to 2001.
Answer:
Odd integers from 1 to 2001 are
This sequence is an A.P.
Here , first term =a =1
common difference = 2.
We know ,
The , sum of odd integers from 1 to 2001 is 1002001.
Question:2 Find the sum of all natural numbers lying between 100 and 1000, which are multiples of 5.
Answer:
Numbers divisible by 5 from 100 to 1000 are
This sequence is an A.P.
Here , first term =a =105
common difference = 5.
We know ,
The sum of numbers divisible by 5 from 100 to 1000 is 98450.
Answer:
First term =a=2
Let the series be
Sum of first five terms
Sum of next five terms
Given : The sum of the first five terms is one-fourth of the next five terms.
To prove :
L.H.S :
Hence, 20th term is –112.
Question: 4 How many terms of the A.P. are needed to give the sum –25?
Answer:
Given : A.P. =
Given : sum = -25
Answer:
Given : In an A.P., if pth term is 1/q and qth term is 1/p
Subtracting (2) from (1), we get
Putting value of d in equation (1),we get
Hence,the sum of first pq terms is 1/2 (pq +1), where .
Question:6 If the sum of a certain number of terms of the A.P. 25, 22, 19, … is 116. Find the last term.
Answer:
Given : A.P. 25, 22, 19, ….....
a=25 , d = -3
n could not be so n=8.
Last term
The, last term of A.P. is 4.
Question:7 Find the sum to n terms of the A.P., whose term is 5k + 1.
Answer:
Given :
Comparing LHS and RHS , we have
and
Putting value of d,
Question:8 If the sum of n terms of an A.P. is , where p and q are constants, find the common difference
Answer:
If the sum of n terms of an A.P. is ,
Comparing coefficients of on both side , we get
The common difference of AP is 2q.
Answer:
Given: The sums of n terms of two arithmetic progressions are in the ratio.
There are two AP's with first terms = and common difference =
Substituting n=35,we get
Thus, the ratio of the 18th term of AP's is
Answer:
Let first term of AP = a and common difference = d.
Then,
Given :
Now,
Thus, sum of p+q terms of AP is 0.
Question:11 Sum of the first p, q and r terms of an A.P. are a, b and c, respectively. Prove that
Answer:
To prove :
Let and d be the first term and the common difference of AP, respectively.
According to the given information, we have
Subtracting equation (2) from (1), we have
Subtracting equation (3) from (2), we have
Equating values of d, we have
Dividing both sides from pqr, we get
Hence, the given result is proved.
Question:12 The ratio of the sums of m and n terms of an A.P. is . Show that the ratio of mth and nth term is .
Answer:
Let a and b be the first term and common difference of a AP ,respectively.
Given : The ratio of the sums of m and n terms of an A.P. is .
To prove : the ratio of mth and nth term is .
Put , we get
From equation (1) ,we get
Hence proved.
Question:13 If the sum of n terms of an A.P. is and its term is 164, find the value of m.
Answer:
Given : If the sum of n terms of an A.P. is and its term is 164
Let a and d be first term and common difference of a AP ,respectively.
Sum of n terms =
Comparing the coefficients of n on both side , we have
Also ,
m th term is 164.
Hence, the value of m is 27.
Question:14 Insert five numbers between 8 and 26 such that the resulting sequence is an A.P.
Answer:
Let five numbers be A,B,C,D,E.
Then
Here we have,
Thus, we have
Thus, the five numbers are 11,14,17,20,23.
Question:15 If is the A.M. between a and b, then find the value of n.
Answer:
Given : is the A.M. between a and b.
Thus, value of n is 1.
Answer:
Let A,B,C.........M be m numbers.
Then,
Here we have,
Given : the ratio of and numbers is 5 : 9.
Putting value of d from above,
Thus, value of m is 14.
Answer:
The first instalment is of Rs. 100.
If the instalment increase by Rs 5 every month, second instalment is Rs.105.
Then , it forms an AP.
We have ,
Thus, he will pay Rs. 245 in the 30th instalment.
Answer:
The angles of polygon forms AP with common difference of and first term as .
We know that sum of angles of polygon with n sides is
Sides of polygon are 9 or 16.
Class 11 maths chapter 9 question answer - Exercise: 9.3
Question:1 Find the and terms of the G.P.
Answer:
G.P :
first term = a
common ratio =r
the nth term of G.P
Question:2 Find the term of a G.P. whose term is 192 and the common ratio is 2.
Answer:
First term = a
common ratio =r=2
term is 192
is the term of a G.P.
Question:3 The terms of a G.P. are p, q and s, respectively. Show that
Answer:
To prove :
Let first term=a and common ratio = r
Dividing equation 2 by 1, we have
Dividing equation 3 by 2, we have
Equating values of , we have
Hence proved
Question:4 The term of a G.P. is square of its second term, and the first term is -3. Determine its term.
Answer:
First term =a= -3
term of a G.P. is square of its second term
Thus, seventh term is -2187.
Question:5(a) Which term of the following sequences:
Answer:
Given :
n th term is given as 128.
The, 13 th term is 128.
Question:5(b) Which term of the following sequences:
Answer:
Given :
n th term is given as 729.
The, 12 th term is 729.
Question:5(c) Which term of the following sequences:
Answer:
Given :
n th term is given as
Thus, n=9.
Question:6 For what values of x, the numbers are in G.P.?
Answer:
Common ratio=r.
Thus, for ,given numbers will be in GP.
Answer:
geometric progressions is 0.15, 0.015, 0.0015, ... .....
a=0.15 , r = 0.1 , n=20
Question:9 Find the sum to indicated number of terms in each of the geometric progressions in
Answer:
The sum to the indicated number of terms in each of the geometric progressions is:
Question:12 The sum of first three terms of a G.P. is and their product is 1. Find the common ratio and the terms.
Answer:
Given : The sum of first three terms of a G.P. is and their product is 1.
Let three terms be .
Product of 3 terms is 1.
Put value of a in equation 1,
The three terms of AP are .
Question:13 How many terms of G.P. , … are needed to give the sum 120?
Answer:
G.P.= , …............
Sum =120
These terms are GP with a=3 and r=3.
Hence, we have value of n as 4 to get sum of 120.
Answer:
Let GP be
Given : The sum of first three terms of a G.P. is 16
Given : the sum of the next three terms is128.
Dividing equation (2) by (1), we have
Putting value of r =2 in equation 1,we have
Question:15 Given a G.P. with a = 729 and term 64, determine
Answer:
Given a G.P. with a = 729 and term 64.
(Answer)
Question:16 Find a G.P. for which sum of the first two terms is – 4 and the fifth term is 4 times the third term
Answer:
Given : sum of the first two terms is – 4 and the fifth term is 4 times the third term
Let first term be a and common ratio be r
If r=2, then
If r= - 2, then
Thus, required GP is or
Question:17 If the terms of a G.P. are x, y and z, respectively. Prove that x,y, z are in G.P.
Answer:
Let x,y, z are in G.P.
Let first term=a and common ratio = r
Dividing equation 2 by 1, we have
Dividing equation 3 by 2, we have
Equating values of , we have
Thus, x,y,z are in GP
Question:18 Find the sum to n terms of the sequence, 8, 88, 888, 8888… .
Answer:
8, 88, 888, 8888… is not a GP.
It can be changed in GP by writing terms as
to n terms
Answer:
Here, is a GP.
first term =a=4
common ratio =r
Question:20 Show that the products of the corresponding terms of the sequences form a G.P, and find the common ratio.
Answer:
To prove : is a GP.
Thus, the above sequence is a GP with common ratio of rR.
Answer:
Let first term be a and common ratio be r.
Given : the third term is greater than the first term by 9, and the second term is greater than the by 18.
Dividing equation 2 by 1 , we get
Putting value of r , we get
Thus, four terms of GP are
Question:22 If the terms of a G.P. are a, b and c, respectively. Prove that
Answer:
To prove :
Let A be the first term and R be common ratio.
According to the given information, we have
L.H.S :
=RHS
Thus, LHS = RHS.
Hence proved.
Answer:
Given : First term =a and n th term = b.
Common ratio = r.
To prove :
Then ,
P = product of n terms
Here, is a AP.
Put in equation (2),
Hence proved .
Question:24 Show that the ratio of the sum of first n terms of a G.P. to the sum of terms from term is
Answer:
Let first term =a and common ratio = r.
Since there are n terms from (n+1) to 2n term.
Sum of terms from (n+1) to 2n.
Thus, the required ratio =
Thus, the common ratio of the sum of first n terms of a G.P. to the sum of terms from term is .
Question:25 If a, b, c and d are in G.P. show that
Answer:
If a, b, c and d are in G.P.
To prove :
RHS :
Using equation (1) and (2),
= LHS
Hence proved
Question:26 Insert two numbers between 3 and 81 so that the resulting sequence is G.P.
Answer:
Let A, B be two numbers between 3 and 81 such that series 3, A, B,81 forms a GP.
Let a=first term and common ratio =r.
For ,
The, required numbers are 9,27.
Question:27 Find the value of n so that may be the geometric mean between a and b.
Answer:
M of a and b is
Given :
Squaring both sides ,
Question:28 The sum of two numbers is 6 times their geometric mean, show that numbers are in the ratio
Answer:
Let there be two numbers a and b
geometric mean
According to the given condition,
.............................................................(1)
Also,
.......................................................(2)
From (1) and (2), we get
Putting the value of 'a' in (1),
Thus, the ratio is
Question:29 If A and G be A.M. and G.M., respectively between two positive numbers, prove that the numbers are
Answer:
If A and G be A.M. and G.M., respectively between two positive numbers,
Two numbers be a and b.
...................................................................1
...........................................................................2
We know
Put values from equation 1 and 2,
..................................................................3
From 1 and 3 , we have
Put value of a in equation 1, we get
Thus, numbers are
Answer:
The number of bacteria in a certain culture doubles every hour.It forms GP.
Given : a=30 and r=2.
Thus, bacteria present at the end of the 2nd hour, 4th hour and nth hour are 120,480 and respectively.
Answer:
Given: Bank pays an annual interest rate of 10% compounded annually.
Rs 500 amounts are deposited in the bank.
At the end of the first year, the amount
At the end of the second year, the amount
At the end of the third year, the amount
At the end of 10 years, the amount
Thus, at the end of 10 years, amount
Answer:
Let roots of the quadratic equation be a and b.
According to given condition,
We know that
Thus, the quadratic equation =
Class 11 maths chapter 9 question answer - Exercise: 9.4
Question:1 Find the sum to n terms of each of the series in
Answer:
the series =
n th term =
Question:2 Find the sum to n terms of each of the series in
Answer:
the series =
n th term =
Thus, sum is
Question:3 Find the sum to n terms of each of the series
Answer:
the series
nth term =
Thus, the sum is
Question:4 Find the sum to n terms of each of the series in
Answer:
Series =
.................................
Hence, the sum is
Question:5 Find the sum to n terms of each of the series in
Answer:
series =
n th term =
16th term is
Hence, the sum of the series is 2840.
Question:6 Find the sum to n terms of each of the series
Answer:
series =
=(n th term of 3,6,9,...........) (nth terms of 8,11,14,..........)
n th term =
Hence, sum is
Question:8 Find the sum to n terms of the series in Exercises 8 to 10 whose nth terms is given by
Answer:
nth terms is given by
Question:9 Find the sum to n terms of the series in Exercises 8 to 10 whose nth terms is given by
Answer:
nth terms are given by
This term is a GP with first term =a =2 and common ratio =r =2.
Thus, the sum is
Question:10 Find the sum to n terms of the series in Exercises 8 to 10 whose nth terms is given by
Answer:
nth terms is given by .
Class 11 maths chapter 9 ncert solutions - Miscellaneous Exercise
Question:1 Show that the sum of and terms of an A.P. is equal to twice the term.
Answer:
Let a be first term and d be common difference of AP.
Kth term of a AP is given by,
Hence, the sum of and terms of an A.P. is equal to twice the term.
Question:2 If the sum of three numbers in A.P., is 24 and their product is 440, find the numbers.
Answer:
Let three numbers of AP are a-d, a, a+d.
According to given information ,
When d=3, AP= 5,8,11 also if d=-3 ,AP =11,8,5.
Thus, three numbers are 5,8,11.
Question:3 Let the sum of n, 2n, 3n terms of an A.P. be , respectively, show that
Answer:
Let a be first term and d be common difference of AP.
Subtract equation 1 from 2,
Hence, the result is proved.
Question:4 Find the sum of all numbers between 200 and 400 which are divisible by 7.
Answer:
Numbers divisible by 7 from 200 to 400 are
This sequence is an A.P.
Here , first term =a =203
common difference = 7.
We know ,
The sum of numbers divisible by 7from 200 to 400 is 8729.
Question:5 Find the sum of integers from 1 to 100 that are divisible by 2 or 5.
Answer:
Numbers divisible by 2 from 1 to 100 are
This sequence is an A.P.
Here , first term =a =2
common difference = 2.
We know ,
Numbers divisible by 5 from 1 to 100 are
This sequence is an A.P.
Here , first term =a =5
common difference = 5.
We know ,
Numbers divisible by both 2 and 5 from 1 to 100 are
This sequence is an A.P.
Here , first term =a =10
common difference = 10
We know ,
Thus, the sum of integers from 1 to 100 that are divisible by 2 or 5 is 3050.
Question:6 Find the sum of all two digit numbers which when divided by 4, yields 1 as remainder.
Answer:
Numbers divisible by 4, yield remainder as 1 from 10 to 100 are
This sequence is an A.P.
Here , first term =a =13
common difference = 4.
We know ,
The sum of numbers divisible by 4 yield 1 as remainder from 10 to 100 is 1210.
Question:7 If f is a function satisfying f (x +y) = f(x) f(y) for all x, y N such that f(1) = 3 and
Answer:
Given : f (x +y) = f(x) f(y) for all x, y N such that f(1) = 3
Taking , we have
is forms a GP with first term=3 and common ratio = 3.
Therefore,
Thus, value of n is 4.
Answer:
Let the sum of some terms of G.P. is 315 whose first term and the common ratio are 5 and 2
Therefore,
Thus, the value of n is 6.
Last term of GP=6th term
The last term of GP =160
Answer:
Given: The first term of a G.P. is 1. The sum of the third term and fifth term is 90.
Thus, the common ratio of GP is .
Answer:
Let three terms of GP be
Then, we have
...............................................1
from an AP.
....................................................................2
From equation 1 and 2, we get
If r=2, GP = 8,16,32
If r=0.2, GP= 32,16,8.
Thus, the numbers required are 8,16,32.
Answer:
Let GP be
Number of terms = 2n
According to the given condition,
Let the be GP as
Thus, the common ratio is 4.
Answer:
Given : first term =a=11
Let AP be
Given: The sum of the first four terms of an A.P. is 56.
Also, The sum of the last four terms is 112.
Thus, the number of terms of AP is 11.
Question:13 If then show that a, b, c and d are in G.P.
Answer:
Given :
Taking ,
Taking,
From equation 1 and 2 , we have
Thus, a,b,c,d are in GP.
Question:14 Let S be the sum, P the product and R the sum of reciprocals of n terms in a G.P. Prove that
Answer:
Ler there be a GP
According to given information,
To prove :
LHS :
Hence proved
Question:15 The pth, qth and rth terms of an A.P. are a, b, c, respectively. Show that
Answer:
Given: The pth, qth and rth terms of an A.P. are a, b, c, respectively.
To prove :
Let the first term of AP be 't' and common difference be d
Subtracting equation 2 from 1, we get
Subtracting equation 3 from 2, we get
Equating values of d, from equation 4 and 5, we have
Hence proved.
Question:16 If are in A.P., prove that a, b, c are in A.P.
Answer:
Given: are in A.P.
Thus, a,b,c are in AP.
Question:17 If a, b, c, d are in G.P, prove that are in G.P.
Answer:
Given: a, b, c, d are in G.P.
To prove: are in G.P.
Then we can write,
Let be in GP
LHS:
Hence proved
Thus, are in GP
Answer:
Given: a and b are the roots of
Then,
Also, c, d are roots of
Given: a, b, c, d form a G.P
Let,
From 1 and 2, we get
and
On dividing them,
When , r=2 ,
When , r=-2,
CASE (1) when r=2 and x=1,
i.e. (q + p) : (q – p) = 17:15.
CASE (2) when r=-2 and x=-3,
i.e. (q + p) : (q – p) = 17:15.
Question:19 The ratio of the A.M. and G.M. of two positive numbers a and b, is m : n. Show that
Answer:
Let two numbers be a and b.
According to the given condition,
...................................................................1
We get,
.....................................................2
From 1 and 2, we get
Putting the value of a in equation 1, we have
Question:20 If a, b, c are in A.P.; b, c, d are in G.P. and 1/c , 1/d , 1/e are in A.P. prove that a, c, e are in G.P.
Answer:
Given: a, b, c are in A.P
Also, b, c, d are in G.P.
Also, 1/c, 1/d, 1/e are in A.P
To prove: a, c, e are in G.P. i.e.
From 1, we get
From 2, we get
Putting values of b and d, we get
Thus, a, c, e are in G.P.
Question:21(i) Find the sum of the following series up to n terms:
Answer:
is not a GP.
It can be changed in GP by writing terms as
to n terms
Thus, the sum is
Question:21(ii) Find the sum of the following series up to n terms: .6 +. 66 +. 666+…
Answer:
Sum of 0.6 +0. 66 + 0. 666+….................
It can be written as
to n terms
Question:22 Find the 20th term of the series terms.
Answer:
the series =
Thus, the 20th term of series is 1680
Question:23 Find the sum of the first n terms of the series: 3+ 7 +13 +21 +31 +…
Answer:
The series: 3+ 7 +13 +21 +31 +…..............
n th term =
Question:24 If are the sum of first n natural numbers, their squares and their cubes, respectively, show that
Answer:
To prove :
From given information,
Here ,
Also,
From equation 1 and 2 , we have
Hence proved .
Question:25 Find the sum of the following series up to n terms:
Answer:
n term of series :
Here, are in AP with first term =a=1 , last term = 2n-1, number of terms =n
Question:26 Show that
Answer:
To prove :
the nth term of numerator
nth term of the denominator
Numerator :
Denominator :
From equation 1,2,3,we have
Hence, the above expression is proved.
Answer:
Given : Farmer pays Rs 6000 cash.
Therefore , unpaid amount = 12000-6000=Rs. 6000
According to given condition, interest paid annually is
12% of 6000,12% of 5500,12% of 5000,......................12% of 500.
Thus, total interest to be paid
Here, is a AP with first term =a=500 and common difference =d = 500
We know that
Sum of AP:
Thus, interest to be paid :
Thus, cost of tractor = Rs. 12000+ Rs. 4680 = Rs. 16680
Answer:
Given: Shamshad Ali buys a scooter for Rs 22000.
Therefore , unpaid amount = 22000-4000=Rs. 18000
According to the given condition, interest paid annually is
10% of 18000,10% of 17000,10% of 16000,......................10% of 1000.
Thus, total interest to be paid
Here, is a AP with first term =a=1000 and common difference =d = 1000
We know that
Sum of AP:
Thus, interest to be paid :
Thus, cost of tractor = Rs. 22000+ Rs. 17100 = Rs. 39100
Answer:
The numbers of letters mailed forms a GP :
first term = a=4
common ratio=r=4
number of terms = 8
We know that the sum of GP is
costs to mail one letter are 50 paise.
Cost of mailing 87380 letters
Thus, the amount spent when the 8th set of the letter is mailed is Rs. 43690.
Answer:
Given : A man deposited Rs 10000 in a bank at the rate of 5% simple interest annually.
Interest in fifteen year 10000+ 14 times Rs. 500
Amount in 15 th year
Amount in 20 th year
Answer:
Cost of machine = Rs. 15625
Machine depreciate each year by 20%.
Therefore, its value every year is 80% of the original cost i.e. of the original cost.
Value at the end of 5 years
Thus, the value of the machine at the end of 5 years is Rs. 5120
Answer:
Let x be the number of days in which 150 workers finish the work.
According to the given information, we have
Series is a AP
first term=a=150
common difference= -4
number of terms = x+8
Since x cannot be negative so x=17.
Thus, in 17 days 150 workers finish the work.
Thus, the required number of days = 17+8=25 days.
9.1 Introduction
9.2 Sequences
9.3 Series
9.4 Arithmetic Progression (A.P.)
9.5 Geometric Progression (G.P.)
9.6 Relationship Between A.M. and G.M.
9.7 Sum to n terms of Special Series
If you are interested in Sequence and Series maths chapter 9 class 11 exercises then these are listed below.
chapter-1 | Sets |
chapter-2 | Relations and Functions |
chapter-3 | Trigonometric Functions |
chapter-4 | Principle of Mathematical Induction |
chapter-5 | Complex Numbers and Quadratic equations |
chapter-6 | Linear Inequalities |
chapter-7 | Permutation and Combinations |
chapter-8 | Binomial Theorem |
chapter-9 | Sequences and Series |
chapter-10 | Straight Lines |
chapter-11 | Conic Section |
chapter-12 | Introduction to Three Dimensional Geometry |
chapter-13 | Limits and Derivatives |
chapter-14 | Mathematical Reasoning |
chapter-15 | Statistics |
chapter-16 | Probability |
Easy Language: The class 11 maths ch 9 is written in simple and easy-to-understand language, making it easy for students to comprehend the concepts.
Comprehensive Coverage: The sequences and series ncert solutions provides a comprehensive coverage of all the topics that are enumerated in the syllabus, making it an ideal reference material for students preparing for exams.
Conceptual Clarity: The class 11 maths ch 9 question answer lays a strong foundation for the concepts of sequences and series, providing students with a clear understanding of the fundamentals.
NCERT solutions for class 11 biology |
NCERT solutions for class 11 maths |
NCERT solutions for class 11 chemistry |
NCERT solutions for Class 11 physics |
There are some important formulas from NCERT solutions for class 11 maths chapter 9 sequences and series which you should remember-
There are 32 questions given in a miscellaneous exercise. To get command on this chapter you need to practice more problems, and you can solve miscellaneous exercise for the same. If you are finding difficulties, you can take help of the NCERT solutions for class 11 maths chapter 9 sequences and series which has solutions of miscellaneous exercise too.
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The sequence and series class 11 solutions includes topics such as basic concepts of sequences and series, arithmetic progression (A.P.), geometric progression (G.P.), relationship between A.M. and G.M., and sum to n terms of special series. students can also refer NCERT Syllabus to understand important topics. Also they can practice class 11 maths chapter 9 solutions to get good hold on these concepts.
NCERT solutions are helpful for the students if they stuck while solving the NCERT problems. Also, it will give them new ways to solve the NCERT problems. after solving them, you can have more than one way to solve a problem as well as your confidence level increase which immensely help you to score well in exam.
A progression is a sequence that follows a particular pattern. In Arithmetic Progression (A.P.), any two consecutive terms have a constant difference. To gain a deeper understanding of these concepts, students are advised to refer to the expertly designed NCERT Solutions available at CAreers360. Although the textbook provides sufficient information, utilizing the best study materials can aid in comprehending their practical applications.
Interested students can find a detailed NCERT solutions for class 11 maths by clicking on the link. they can prcatice these solutions to get indepth understanding of the concepts. The sequence and series class 11 ncert solutions are crucial as these are foundation for other chapters such as statistics also questions for this chapter are easy to moderates.
To minimize conceptual errors in class 11th maths chapter 9 Sequence and Series, it is necessary to practice extensively as it comprises several complex topics. Though the initial understanding of fundamental concepts might pose a challenge, achieving high scores is feasible with proper guidance. To excel in class 11 chapter 9 maths , students must tackle a variety of challenging problems. NCERT Solutions provide quick problem-solving techniques and familiarize students with the types of questions that are likely to appear on exams.
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