NCERT Solutions Exercise 9.1 , 9.2 , 9.3, 9.4 Sequences and Series for Class 11

Write the first five terms of each of the sequences in Exercises 1 to 6 whose nth terms are:

Q1. an = n (n + 2) 

Given,

nth term of a sequence an = n (n + 2) 

On substituting n = 1, 2, 3, 4, and 5, we get the first five terms

a1 = 1(1 + 2) = 3

a2 = 2(2 + 2) = 8

a3 = 3(3 + 2) = 15

a4 = 4(4 + 2) = 24

a5 = 5(5 + 2) = 35

Hence, the required terms are 3, 8, 15, 24, and 35.

Q2. an = n/n+1

Given the nth term, an = n/n+1

On substituting n = 1, 2, 3, 4, 5, we get

Q3. an = 2n

Given the nth term, an = 2n

On substituting n = 1, 2, 3, 4, 5, we get

a1 = 21 = 2

a2 = 22 = 4

a3 = 23 = 8

a4 = 24 = 16

a5 = 25 = 32

Q4. an = (2n – 3)/6

Given the nth term, an = (2n – 3)/6

On substituting = 1, 2, 3, 4, 5, we get

Q5.  an = (-1)n-1 5n+1

Given the nth term, an = (-1)n-1 5n+1

On substituting = 1, 2, 3, 4, 5, we get

Q6.

On substituting n = 1, 2, 3, 4, 5, we get the first 5 terms.

Find the indicated terms in each of the sequences in Exercises 7 to 10 whose nth terms are:

Q7.  an = 4n – 3; a17, a24

Given,

The nth term of the sequence is an = 4n – 3

On substituting n = 17, we get

a17 = 4(17) – 3 = 68 – 3 = 65

Next, on substituting n = 24, we get

a24 = 4(24) – 3 = 96 – 3 = 93

Q8. an = n2/2n ; a7

Given,

The nth term of the sequence is an = n2/2n

Now, on substituting n = 7, we get

a7 = 72/27 = 49/ 128

Q9. an = (-1)n-1 n3; a9

Given,

The nth term of the sequence is an = (-1)n-1 n3

On substituting n = 9, we get

a9 = (-1)9-1 (9)3 = 1 x 729 = 729

Q10. 

On substituting n = 20, we get

Write the first five terms of each of the sequences in Exercises 11 to 13 and obtain the corresponding series:

Q11. a1 = 3, an = 3an-1 + 2 for all n > 1

Solution:

Given, an = 3an-1 + 2 and a1 = 3

Then,

a2 = 3a1 + 2 = 3(3) + 2 = 11

a3 = 3a2 + 2 = 3(11) + 2 = 35

a4 = 3a3 + 2 = 3(35) + 2 = 107

a5 = 3a4 + 2 = 3(107) + 2 = 323

Thus, the first 5 terms of the sequence are 3, 11, 35, 107 and 323.

Hence, the corresponding series is

3 + 11 + 35 + 107 + 323 …….

Q12. a1 = -1, an = an-1/n, n ≥ 2

Given,

an = an-1/n and a1 = -1

Then,

a2 = a1/2 = -1/2

a3 = a2/3 = -1/6

a4 = a3/4 = -1/24

a5 = a4/5 = -1/120

Thus, the first 5 terms of the sequence are -1, -1/2, -1/6, -1/24 and -1/120.

Hence, the corresponding series is

-1 + (-1/2) + (-1/6) + (-1/24) + (-1/120) + …….

Q13. a1 = a= 2, an = an-1 – 1, n > 2

Given,

a1 = a2, an = an-1 – 1

Then,

a3 = a2 – 1 = 2 – 1 = 1

a4 = a3 – 1 = 1 – 1 = 0

a5 = a4 – 1 = 0 – 1 = -1

Thus, the first 5 terms of the sequence are 2, 2, 1, 0 and -1.

The corresponding series is

2 + 2 + 1 + 0 + (-1) + ……

Q14. The Fibonacci sequence is defined by

1 = a1 = a2 and an = an – 1 + an – 2, n > 2

Find an+1/an, for n = 1, 2, 3, 4, 5 

Given,

1 = a1 = a2

an = an – 1 + an – 2, n > 2

So,

a3 = a2 + a1 = 1 + 1 = 2

a4 = a3 + a2 = 2 + 1 = 3

a5 = a4 + a3 = 3 + 2 = 5

a6 = a5 + a4 = 5 + 3 = 8

Thus,

 

Q1. Find the sum of odd integers from 1 to 2001.

Solution:

The odd integers from 1 to 2001 are 1, 3, 5, …1999, 2001.

It clearly forms a sequence in A.P.

Where the first term, a = 1

The common difference, d = 2

Now,

a + (n -1)d = 2001

1 + (n-1)(2) = 2001

2n – 2 = 2000

2n = 2000 + 2 = 2002

n = 1001

We know,

Sn = n/2 [2a + (n-1)d]

Q2. Find the sum of all natural numbers lying between 100 and 1000, which are multiples of 5.

The natural numbers lying between 100 and 1000, which are multiples of 5, are 105, 110, … 995.

It clearly forms a sequence in A.P.

Where the first term, a = 105

The common difference, d = 5

Now,

a + (n -1)d = 995

105 + (n – 1)(5) = 995

105 + 5n – 5 = 995

5n = 995 – 105 + 5 = 895

n = 895/5

n = 179

We know,

Sn = n/2 [2a + (n-1)d]

Q3. In an A.P, the first term is 2, and the sum of the first five terms is one-fourth of the next five terms. Show that the 20th term is –112.

Given,

The first term (a) of an A.P = 2

Let’s assume d is the common difference of the A.P.

So, the A.P. will be 2, 2 + d, 2 + 2d, 2 + 3d, …

Then,

Sum of first five terms = 10 + 10d

Sum of next five terms = 10 + 35d

From the question, we have

10 + 10d = ¼ (10 + 35d)

40 + 40d = 10 + 35d

30 = -5d

d = -6

a20 = a + (20 – 1)d = 2 + (19) (-6) = 2 – 114 = -112

Therefore, the 20th term of the A.P. is –112.

Q4. How many terms of the A.P. -6, -11/2, -5, …. are needed to give the sum –25?

Let’s consider the sum of n terms of the given A.P. as –25.

We known that,

Sn = n/2 [2a + (n-1)d]

where n = number of terms, a = first term, and d = common difference

So here, a = –6

d = -11/2 + 6 = (-11 + 12)/ 2 = 1/2

Thus, we have

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 ½ (pq + 1) where p ≠ q.  

Q6. If the sum of a certain number of terms of the A.P. 25, 22, 19, … is 116. Find the last term.

Solution:

Given A.P.,

25, 22, 19, …

Here,

First term, a = 25 and

Common difference, d = 22 – 25 = -3

Also given, the sum of a certain number of terms of the A.P. is 116.

The number of terms is n.

So, we have

Sn = n/2 [2a + (n-1)d] = 116

116 = n/2 [2(25) + (n-1)(-3)]

116 x 2 = n [50 – 3n + 3]

232 = n [53 – 3n]

232 = 53n – 3n2

3n2 – 53n + 232 = 0

3n2 – 24n – 29n+ 232 = 0

3n(n – 8) – 29(n – 8) = 0

(3n – 29) (n – 8) = 0

Hence,

n = 29/3 or n = 8

As n can only be an integral value, n = 8

Thus, the 8th term is the last term of the A.P.

a8 = 25 + (8 – 1)(-3)

= 25 – 21

= 4

Q7. Find the sum to n terms of the A.P., whose kth term is 5k + 1.

Solution:

Given, the kth term of the A.P. is 5k + 1.

kth term = ak = + (k – 1)d

And,

+ (k – 1)d = 5k + 1

a + kd – d = 5k + 1

On comparing the coefficient of k, we get d = 5

– = 1

a – 5 = 1

⇒ a = 6

Q8.

NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.2-Q8.1

Q9. The sums of n terms of two arithmetic progressions are in the ratio 5n + 4 : 9n + 6. Find the ratio of their 18th terms.

Q10. If the sum of first p terms of an A.P. is equal to the sum of the first q terms, then find sum of the first (p + q) terms.

ncert-solutions-class-11-mathematics-chapter-9-ex-9-2-9

Q11.  Sum of the first p, q and r terms of an A.P. are a, b and c, respectively.

ncert-solutions-class-11-mathematics-chapter-9-ex-9-2-11   Prove that  

Let a1 and d be the first term and the common difference of the A.P., respectively.

Then, according to the question, we have

Now, subtracting (2) from (1), we get

Q12. The ratio of the sums of m and n terms of an A.P. is m2n2. Show that the ratio of the mth and the nth term is (2m – 1): (2n – 1).

Let’s consider that a and b are the first term and the common difference of the A.P., respectively.

Then, from the question, we have

ncert-solutions-class-11-mathematics-chapter-9-ex-9-2-15

Hence, the given result is proved.

Q13. If the sum of n terms of an A.P. is 3n2 + 5n and its mth term is 164, find the value of m.

Let’s consider a and b to be the first term and the common difference of the A.P., respectively.

am = a + (m – 1)d = 164 … (1)

The sum of the terms is given by,

Sn = n/2 [2a + (n-1)d]

Q14. Insert five numbers between 8 and 26 such that the resulting sequence is an A.P.

Let’s assume A1, A2, A3, A4, and A5 to be five numbers between 8 and 26 such that 8, A1, A2, A3, A4, A5, 26 are in an A.P.

Here, we have,

= 8, = 26, n = 7

So,

26 = 8 + (7 – 1) d

6d = 26 – 8 = 18

= 3

Now,

A1 = a + d = 8 + 3 = 11

A2 = a + 2d = 8 + 2 × 3 = 8 + 6 = 14

A3 = a + 3d = 8 + 3 × 3 = 8 + 9 = 17

A4 = a + 4= 8 + 4 × 3 = 8 + 12 = 20

A5 = a + 5d = 8 + 5 × 3 = 8 + 15 = 23

Therefore, the required five numbers between 8 and 26 are 11, 14, 17, 20, and 23.

Q15. If is the A.M. between a and b, then find the value of n.

The A.M between a and b is given by (a + b)/2

Then, according to the question,

Q16. Between 1 and 31, m numbers have been inserted in 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.

Let’s consider a1, a2, … am be m numbers such that 1, a1, a2, … am, 31 is an A.P.

And here,

a = 1, b = 31, n = m + 2

So, 31 = 1 + (m + 2 – 1) (d)

30 = (m + 1) d

d = 30/ (m + 1) ……. (1)

Now,

a1 = a + d

a2 = a + 2d

a3 = a + 3d …

Hence, a7 = a + 7d

am–1 = a + (m – 1) d

According to the question, we have

Q17. A man starts repaying a loan as the first instalment of Rs. 100. If he increases the instalment by Rs 5 every month, what amount will he pay in the 30th instalment?

Given,

The first instalment of the loan is Rs 100.

The second instalment of the loan is Rs 105, and so on as the instalment increases by Rs 5 every month.

Thus, the amount that the man repays every month forms an A.P.

And then, A.P. is 100, 105, 110, …

Where the first term, a = 100

Common difference, d = 5

So, the 30th term in this A.P. will be

A30  = a + (30 – 1)d

= 100 + (29) (5)

= 100 + 145

= 245

Therefore, the amount to be paid in the 30th instalment will be Rs 245.

Q18. The difference between any two consecutive interior angles of a polygon is 5°. If the smallest angle is 120°, find the number of the sides of the polygon.

It’s understood from the question that the angles of the polygon will form an A.P. with a common difference d = 5° and first term a = 120°.

And we know that the sum of all angles of a polygon with n sides is 180° (n – 2).

Thus, we can say

 

 

Q1. Find the 20th and nth terms of the G.P. 5/2, 5/4, 5/8, ………

Given G.P. is 5/2, 5/4, 5/8, ………

Here, a = First term = 5/2

r = Common ratio = (5/4)/(5/2) = ½

Thus, the 20th term and nth term

Q2. Find the 12th term of a G.P. whose 8th term is 192, and the common ratio is 2.

Given,

The common ratio of the G.P., r = 2

And, let a be the first term of the G.P.

Now,

a8 = ar 8–1 = ar7

ar7 = 192

a(2)7 = 192

a(2)7 = (2)6 (3)

Q3. The 5th, 8th and 11th terms of a G.P. are pq and s, respectively. Show that q2 = ps.

Let’s take a to be the first term and r to be the common ratio of the G.P.

Then, according to the question, we have

a5 = a r5–1 a r4 = p … (i)

aa r8–1 a r7 = q … (ii)

a11 = a r11–1 a r10 = … (iii)

Dividing equation (ii) by (i), we get

Q4. The 4th term of a G.P. is the square of its second term, and the first term is –3. Determine its 7th term.

Let’s consider a to be the first term and r to be the common ratio of the G.P.

Given, a = –3

And we know that,

an = arn–1

So, aar3 = (–3) r3

a2 = a r1 = (–3) r

Then, from the question, we have

(–3) r3 = [(–3) r]2

⇒ –3r3 = 9 r2

⇒ r = –3

a7 = a r 7–1 a r6 = (–3) (–3)6 = – (3)7 = –2187

Therefore, the seventh term of the G.P. is –2187.

Q5. Which term of the following sequences:

(a) 2, 2√2, 4,… is 128 ? (b) √3, 3, 3√3,… is 729 ?

(c) 1/3, 1/9, 1/27, … is 1/19683 ?

(a) The given sequence, 2, 2√2, 4,…

We have,

a = 2 and r = 2√2/2 = √2

Taking the nth term of this sequence as 128, we have

Therefore, the 13th term of the given sequence is 128.

(ii) Given the sequence, √3, 3, 3√3,…

We have,

a = √3 and r = 3/√3 = √3

Taking the nth term of this sequence to be 729, we have

Therefore, the 12th term of the given sequence is 729.

(iii) Given sequence, 1/3, 1/9, 1/27, …

a = 1/3 and r = (1/9)/(1/3) = 1/3

Taking the nth term of this sequence to be 1/19683, we have

Q6. For what values of x, the numbers -2/7, x, -7/2 are in G.P?

Solution:

The given numbers are -2/7, x, -7/2

Common ratio = x/(-2/7) = -7x/2

Also, common ratio = (-7/2)/x = -7/2x

Q7. Find the sum to 20 terms in the geometric progression 0.15, 0.015, 0.0015 …

Given G.P., 0.15, 0.015, 0.00015, …

Here, a = 0.15 and r = 0.015/0.15 = 0.1 

Q8. Find the sum to n terms in the geometric progression √7, √21, 3√7, ….

The given G.P. is √7, √21, 3√7, ….

Here,

a = √7 and

Q9.  Find the sum to n terms in the geometric progression 1, -a, a2, -a3 …. (if a ≠ -1)

The given G.P. is 1, -a, a2, -a3 ….

Here, the first term = a1 = 1

And the common ratio = r = – a

We know that,

Q10. Find the sum to n terms in the geometric progression x3, x5, x7, … (if x ≠ ±1 )

Given G.P. is x3, x5, x7, …

Here, we have a = x3 and r = x5/x3 = x2

Q11. Evaluate:

Q12. The sum of the first three terms of a G.P. is 39/10, and their product is 1. Find the common ratio and the terms.

Let a/r, a, ar be the first three terms of the G.P.

a/r + a + ar = 39/10 …… (1)

(a/r) (a) (ar) = 1 …….. (2)

From (2), we have

a3 = 1

Hence, a = 1 [Considering real roots only]

Substituting the value of a in (1), we get

1/r + 1 + r = 39/10

(1 + r + r2)/r = 39/10

10 + 10r + 10r2 = 39r

10r2 – 29r + 10 = 0

10r2 – 25r – 4r + 10 = 0

5r(2r – 5) – 2(2r – 5) = 0

(5r – 2) (2r – 5) = 0

Thus,

r = 2/5 or 5/2

Therefore, the three terms of the G.P. are 5/2, 1 and 2/5.

Q13. How many terms of G.P. 3, 32, 33, … are needed to give the sum 120?

Given G.P. is 3, 32, 33, …

Let’s consider that n terms of this G.P. be required to obtain the sum 120.

We know that,

Here, a = 3 and r = 3

Equating the exponents, we get n = 4

Therefore, four terms of the given G.P. are required to obtain the sum 120.

 

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