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NCERT solutions of class 11 maths exercise 9.2 of chapter 9-Sequence and Series

Sequences and Series exercise 9.2

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.

Sequences and Series exercise 9.2

 

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.

nth  term of an AP is represented by an

The sum of n terms of AP is represented by Sn

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,an= 2001

Applying the formula for the sum of an AP

Where, Sn= 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(an)

an= a + (n -1) d

Putting the values of a,d and an

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, an=995

Finding the value of n from the following formula

an=  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 Sn = -25

Putting the values of a, d and of Sn

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 pth term is 1/q and qth 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

an=  a + (n -1) d

where a is the first term and d is the common difference of AP

We are given

pth term=  1/q

qth 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 Sn= 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

an= a + (n-1)⇒a5 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 Kth term is 5K + 1.

Ans. We are given the Kth term is 5K + 1

ak= 5K + 1

Substituting k=1,2,3….etc we get the first, second, third and other terms of the AP

a1= 5×1 + 1= 6,

a2= 5×2+ 1 =11

a3= 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

Sn = pn + qn²

Substituting n=1,2,3…we get the sum of the 1 term,2 terms,3 terms etc.

S1 = p×1 + q×1² =p + q, sum of 1 term =first term=p +q

S2 = 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 18th 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

Sn = 3n² +5n

Substituting the value of n =1,2,3.. we get the sum of 1 terms, 2 terms, 3 terms..etc

S1 = 3×1² +5×1 = 3 +5 = 8, sum of 1 terms = First term of AP

S2 = 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 A1 , A2, A3, A4 and A5

The sequence including 5 inserted numbers is

8 A1 , A2, A3, A4 ,A5 26

Where n = 7, 7 th term = 26, a = 8

So, a= a + (n -1)d

26 = 8 + 6d

6d =18

d = 3

Therefore

A1=  a +d = 8 +3 = 11, A2= 8 + 2d = 8 +6 =14, A3= 14 + d= 14 +3= 17,A4=17+3=20,A5=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

2an+ 2bn = an+ b.an-1+ a.bn-1+bn

2an+ 2bn – an-bn=b.an-1+ a.bn-1

an+bn=b.an-1+ a.bn-1

an-b.an-1=a.bn-1-bn

an-1(a – b) = bn-1(a -b)

an-1= bn-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 7th 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 nth then it is the (n+1)th term of the series

∴ inserted 7th term = 8th term of the series = 1 + 7d = 1 +(7×30)/(m+1) =(m+211)/(m+1)

inserted (m-1)th number = mth 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 30th installment?

Ans. According to the question, the sequence of instalments every month is

100, 105, 110,115…….

Where a = 100, d = 5

30th installment = 30th term of the above series = a + ( 30 – 1)d = 100 +29×5 =100+145=245

Hence the 30th 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

Sn= 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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