NCERT solutions of class 11 maths exercise 9.2 of chapter 9-Sequence and series
NCERT solutions of class 11 maths exercise 9.2 of chapter 9-Sequence are the solutions of chapter 9 of class 11 NCERT maths textbook prescribed by the CBSE board for the students of 11 class maths students. Sequence and series are applied in science and technology, economics, and business in predicting the forthcoming outcome of the implemented inputs, thus study of sequence and series is a compulsory part of class 11 maths of every school board worldwide. Here the NCERT solutions of class 11 maths exercise 9.2 of chapter 9-Sequence and series are solved by an expert of maths by a step by step method, so every student can understand it easily without getting any help from anybody. Future study point has enhanced its activity to provide quality study material in order to fulfill your deficit in the study due to your disconnection from the school induced by the Covid-19 pandemic.
Here the questions are based on the Arithmetic progression in exercise 9.2 of the chapter 9-Sequence and series. The questions are based on two formulas of AP which you already have studied in class 10 NCERT maths textbook. Here the questions of exercise 9.1 are more complex in comparison to the 10 class.
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Here you can recall two formulas used here for the solution of exercise 9.2-Sequences and Series.
n^{th }term of an AP is represented by a_{n}
The sum of n terms of AP is represented by S_{n}
Where a = first term of an AP, d = common difference, which is calculated by subtracting the previous term from next term and n = number of terms
NCERT solutions of class 11 maths of chapter 9-Sequence and series
Exercise 9.1 -Sequence and Series
Exercise 9.3-Sequence and Series
Exercise 9.4-Sequence and Series
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NCERT solutions of class 11 maths exercise 9.2 of chapter 9-Sequence and series
Exercise 9.2
Q1. Find the sum of odd integers from 1 to 2001.
Ans. The series of odd integers from 1 to 2001 is an AP with common difference 2.
1 + 3 + 5 +………2001
From the series , we have
First-term of AP,a = 1, Common difference of AP,d = 2, Final term of the AP,a_{n}= 2001
Applying the formula for the sum of an AP
Where, S_{n}= Sum of the series, Here we don’t have the value of n, so first calculating the value of n from the formula for the n the term(a_{n})
a_{n}= a + (n -1) d
Putting the values of a,d and a_{n}
2001 = 1 + (n -1)2 = 1 + 2n -2 = 2n -1
2n = 2002
n = 1001
Hence, the sum of the integers from 1 to 2001 is = 1002001
Q2.Find the sum of all natural numbers lying between 100 and 1000. which are multiple of 5.
Ans.The natural numbers between 100 and 1000 divisible by 5 is a series of AP, written as bellow
105 + 110 + 115 +……..995
Where, first term, a = 105, common difference, d = 5 and n th term, a_{n}=995
Finding the value of n from the following formula
a_{n}= a + (n -1)d
995 = 105 + (n -1)5
5n – 5 + 105 = 995
5n = 995 -100 = 895
n = 179
Now applying the following formula for sum of the AP
Substituting the value of terms we know
= 179 × 550 = 98450
Hence the required sum of the given AP is =98450
Q3. In an AP, the first term is 2 and the sum of the first five terms is one-fourth of the next five terms. Show that 20 th term is -112.
Ans.Let the common difference of the given AP is = d,First term of the given AP is , a = 2
Therefore AP is written as , 2, 2 + d, 2 + 2d, 2 + 3d……..
Considering the first 5 terms and next 5 terms, total terms =10
The sum of first 5 terms = 2× 5 +(d +2d+3d+4d)= 10 +10d
The sum of next 5 terms =2×5 + (5d +6d +7d+8d+9d) =10 +35d
According to question
10 +10d = 1/4(10 + 35d)
10 + 10 d = 10/4 + 35d/4
10d -35d/4 = 10/4 -10
(40d -35d)/4 = (10 -40)/4
5d = -30
d = -6
Therefore 20 the term = a +(n-1)d = 2 + (20 -1)(-6) = 2 -114 =-112
Q4.How many terms of the A.P.-6 ,-11/2 , -5 ,… are needed to give the sum -25?
Ans.The given AP -6, -11/2, -5……
Let there are n number of terms to give sum of -25,where first term,a = -6, common difference,d = -11/2+6 =1/12
Where S_{n }= -25
Putting the values of a, d and of S_{n}
n(n -25) = -100
n² – 25n +100 = 0
n² – 20n – 5n +100 =0
n(n – 20) – 5(n – 20) = 0
(n – 20)(n – 5) = 0
n = 20, 5
Therefore number of terms of the series to give the sum -25 either are 20 or 5.
Q5.In an A.P., if p^{th} term is 1/q and q^{th} term is 1/p, prove that the sum of first pq terms is 1/2(pq + 1), where p ≠ q.
Ans.The n the term of AP is given as following
a_{n}= a + (n -1) d
where a is the first term and d is the common difference of AP
We are given
p^{th} term= 1/q
q^{th} term = 1/p
Subtracting equation (ii) from (i)
Putting the value of d in equation (i)
Now, applying the following formula for the sum of n the term of AP
Putting the value a = d =1/pq and n = pq
Therefore the sum of pq terms of the series is 1/2(pq +1)
Q6.If a sum of a certain number of terms of the A.P.25, 22, 19, …is 116. Find the last term?
Ans.The sum of the given series is A.P.25, 22, 19, …is 116
a = 25, d = 22-25 = -3 and S_{n}= 116
232 = n(50 -3n+3)
50n -3n² +3n -232= 0
-3n² + 53n -232 = 0
3n² – 53n + 232 = 0
3n² -24n – 29n + 232 = 0
3n(n – 8) – 29(n -8) =0
(n – 8)(3n – 29) =0
n = 8, n = 29/3
n can not equal to 29/3 since it is not a natural number,so neglecting this value of n
Therefore n = 8
Now the last term of the series is
a_{n}= a + (n-1)⇒a_{5} a= 25 +(8-1)(-3) = 25-21 = 4
Hence the last term of the given series is =4
Q7.Find the sum to n terms of the A.P.,whose K^{th }term is 5K + 1.
Ans. We are given the K^{th }term is 5K + 1
a_{k}= 5K + 1
Substituting k=1,2,3….etc we get the first, second, third and other terms of the AP
a_{1}= 5×1 + 1= 6,
a_{2}= 5×2+ 1 =11
a_{3}= 5×3+ 1 =16
…..and so on
Hence , the AP is 6,11,16………
Where a = 6, d= 11 -6 = 5
Applying the formula used for sum of n the of an AP
Substituting the values of a and d
Therefore the sum of the n the term of AP is n/2(5n +7)
Q8.If the sum of n terms of an A.P. is (pn +qn²), where p and q are constants, find the common difference.
Ans. We are given the sum of n terms of an AP= pn +qn²
S_{n} = pn + qn²
Substituting n=1,2,3…we get the sum of the 1 term,2 terms,3 terms etc.
S_{1} = p×1 + q×1² =p + q, sum of 1 term =first term=p +q
S_{2} = p×2 + q×2² =2p +4q, second term =sum of 2 terms – first term=2p+4q -p-q=p+3q
Hence the AP is p+q,p+3q,……
Therefore the common difference of AP = (p+3q) -(p +q) =2q
Q9.The sum of n terms of two arithmetic progressions are in the ratio 5n + 4 : 9n + 6. Find the ratio of their 18^{th} terms.
Ans. We are given the ratio of two arithmetic progression
= 5n + 4 : 9n + 6
Let the first terms of both of the AP are a and a’and common differences are d and d’
The ratio of the sum of their n terms is given as follows
Writing the LHS of the above equation in the form of a +(n -1)d
…….(i)
The ratio between 18 th terms of both AP is
Comparing both sides
We have treated LHS of (i) as the ratio of n th terms of the series,so putting n =35 in RHS of (i) we get the required ratio
Therefore the ratio between 18 th term of both AP is 179 : 321
Q10.If the sum of first p terms of an A.P. is equal to the sum of the first q terms, then find the sum of the first (p + q) terms.
Ans.Let the first term of the AP is a and common difference is d
We are given
Sum of first p terms = Sum of first q terms
2ap +p²d -pd = 2aq +q²d -qd
2a(p-q) +(p² -q²)d-(p -q)d = 0
2a(p-q) +(p -q)(p+q)d-(p -q)d = 0
2a + d(p+q -1) = 0
The sum of first (p + q) terms is given as
Putting the value of d
Therefore the sum of first (p+q) term is 0
Q11. Sum of the first p,q and r terms of an AP are a,b and c respectively .Prove that
Ans. We are given sum of the first p,q and r terms of an AP are a,b and c respectively
Therefore
The sum of q terms = b
The sum of r terms = c
Multiplying (i) by (q-r), (ii) by (r-p) and (iii) by (p -q) and then adding all three equations
= 0 +0 = 0
Hence proved
Q12.The ratio of the sums of the m and n terms of an AP is m² : n² .Show that the ratio of m th and n th term is (2m -1) : (2n -1)
Ans.We are given
Putting m = p and n = q in above equation
Treating denominator and numerator of LHS as a n th and m th term
…..(i)
The ratio of m th term and n th term is
…(ii)
Comparing (i) and (ii)
Putting the value of p 2m -1 and q= 2n -1 in RHS of equation (i)
Therefore the ratio between m th and n th terms is 2m -1 : 2n -1
Q13.If sum of n terms of an AP is 3n² +5n and its m th term is 164.Find the value of m.
Ans. We are given the sum of n terms
S_{n} = 3n² +5n
Substituting the value of n =1,2,3.. we get the sum of 1 terms, 2 terms, 3 terms..etc
S_{1} = 3×1² +5×1 = 3 +5 = 8, sum of 1 terms = First term of AP
S_{2} = 3×2² +5×2 = 12 + 10 = 22, Second term = Sum of two term – first term =22-8 =14
Therefore AP is 8,14,20……
d = 14 – 8 = 6
We are given
m the term = 164
8 + (m -1)6 = 164
8 + 6m -6 = 164
6m = 162
m = 27
Hence, the value of m is 27
Q14. Insert five numbers between 8 and 26 such that resulting sequence is an AP.
Ans. Let 5 numbers inserted between 8 and 26 are A_{1} , A_{2,} A_{3,} A_{4} and A_{5}
The sequence including 5 inserted numbers is
8 A_{1} , A_{2,} A_{3,} A_{4} ,A_{5} 26
Where n = 7, 7 th term = 26, a = 8
So, a_{n }= a + (n -1)d
26 = 8 + 6d
6d =18
d = 3
Therefore
A_{1}= a +d = 8 +3 = 11, A_{2}= 8 + 2d = 8 +6 =14, A_{3}= 14 + d= 14 +3= 17,A_{4}=17+3=20,A_{5}=20+3 = 23
Hence, the 5 inserted numbers are 11, 14, 17,20 and 23
Ans. The AM of a and b is = (a +b)/2
According to question, we have
2a^{n}+ 2b^{n} = a^{n}+ b.a^{n-1}+ a.b^{n-1}+b^{n}
2a^{n}+ 2b^{n} – a^{n}-b^{n}=b.a^{n-1}+ a.b^{n-1}
a^{n}+b^{n}=b.a^{n-1}+ a.b^{n-1}
a^{n}-b.a^{n-1}=a.b^{n-1}-b^{n}
a^{n-1}(a – b) = b^{n-1}(a -b)
a^{n-1}= b^{n-1}
n – 1 = 0
n = 1
Thus, the value of n is 1.
Q16.Between 1 and 31, m numbers have been inserted such a way that the resulting sequence is an A.P.and the ratio of 7^{th} and (m – 1)^{th }numbers is 5: 9. Find the value of m.
Ans. Let the inserted numbers are b,c,d,e…..between 1 and 31
So, the sequence is 1, b,c,d,e…..31
Number of terms in the sequence are = m+2
(m+2)^{th }term is = 31
31 = 1 + (m+ 2-1)d
31 = 1 + (m+1)d
d = 30/(m+1)
Observing the term of sequence we get that if inserted number is n^{th} then it is the (n+1)^{th }term of the series
∴ inserted 7^{th} term = 8^{th }term of the series = 1 + 7d = 1 +(7×30)/(m+1) =(m+211)/(m+1)
inserted (m-1)^{th }number = m^{th }term of the series = 1 +(m -1)d =1+(m-1)×30/(m+1)=(m+1+30m-30)/(m+1) = (31m -29)/(m+1)
According to the question
9m + 1899 = 155m – 145
9m -155m = -145 -1899
-146 m = -2044
m = 14
Hence, m = 14
Q17.A man starts repaying a loan as first installment of Rs.100.If he increases the installment by Rs 5 every month, What amount will he pay in the 30^{th }installment?
Ans. According to the question, the sequence of instalments every month is
100, 105, 110,115…….
Where a = 100, d = 5
30^{th }installment = 30^{th }term of the above series = a + ( 30 – 1)d = 100 +29×5 =100+145=245
Hence the 30^{th } installments is = Rs 245
Q18.The difference between the two consecutive interior angles of a polygon is 5º. If the smallest angle is 120º, find the number of sides of polygon.
Ans. We are given the difference between the two consecutive interior angles of a polygon is = 5º, the smallest angle is =120º, let the number of sides = n
According to question sequence of the interior angles is
120º, 125º, 130º…….
Where a = 120, d = 5
The sum of interior angles of a polygon is given as follows
S_{n}= 180 (n -2)
We also have the sum of AP
Therefore
Putting the value of a and d
235n + 5n² = 360n – 720
5n² – 125n +720 = 0
n² – 25n + 144 = 0
n² – 16n – 9n+ 144 = 0
n( n – 16) – 9(n -16) = 0
(n -16)(n -9) = 0
n = 16, 9
Therefore the polygon will have the number of sides 16 or 9
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Chapter 1-Sets | Chapter 9-Sequences and Series |
Chapter 2- Relations and functions | Chapter 10- Straight Lines |
Chapter 3- Trigonometry | Chapter 11-Conic Sections |
Chapter 4-Principle of mathematical induction | Chapter 12-Introduction to three Dimensional Geometry |
Chapter 5-Complex numbers | Chapter 13- Limits and Derivatives |
Chapter 6- Linear Inequalities | Chapter 14-Mathematical Reasoning |
Chapter 7- Permutations and Combinations | Chapter 15- Statistics |
Chapter 8- Binomial Theorem | Chapter 16- Probability |
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Chapter 1-Relations and Functions | Chapter 9-Differential Equations |
Chapter 2-Inverse Trigonometric Functions | Chapter 10-Vector Algebra |
Chapter 3-Matrices | Chapter 11 – Three Dimensional Geometry |
Chapter 4-Determinants | Chapter 12-Linear Programming |
Chapter 5- Continuity and Differentiability | Chapter 13-Probability |
Chapter 6- Application of Derivation | CBSE Class 12- Question paper of maths 2021 with solutions |
Chapter 7- Integrals | |
Chapter 8-Application of Integrals |
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