# Samacheer Kalvi 10th Maths Solutions Chapter 2 Pdf

## Samacheer Kalvi 10th Maths Solutions Chapter 2 Pdf

TN Samacheer Kalvi 10th Maths Solutions Chapter 2: hello students are You looking for Samacheer Kalvi 10th Maths Solutions Chapter 2 – Numbers and Sequences on internet. If yes, here we have given TNSCERT Class 10 Maths Solution Chapter 2 Pdf Guide for Samacheer Kalvi 10th Mathematics Solution for Exercise 2.1, 2.2 ,2.3 ,2.4 ,2.5 ,2.6 ,2.7 ,2.8 ,2.9 ,2.10 ,2.11

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### TN Samacheer Kalvi 10th Maths Solutions Chapter 2 – Numbers and Sequences

 Board TNSCERT Class 10th Maths Class 10 Class Subject Maths Chapter 2 (Exercise 2.1, 2.2 ,2.3 ,2.4 ,2.5 ,2.6 ,2.7 ,2.8 ,2.9 ,2.10 ,2.11) Chapter Name Numbers and Sequences

### All Exercise Solution

#### Exercise – 2.1

(1) Find all positive integers , when divided by 3 leaves remainder 2 .

=> By Euclid’s division algorithm

Let a, b are positive integers such that

a = bq+ e , 0 ≤ r < b

Now, b = 3, and r = 2 then,

a = 3q + 2 , q ∈ (ℤ+ +{0})

∴ Therefore, the positive integers are 2, 5, 8, 11, 14, ……

(2) A man has 532 flower pots. He wants to arrange then in rows such that each row contains 21 flower pots , Find the number of completed rows and how mary flower pots are left over .

=> By Euclid’s division algorithm .

Let a, b , ∈ ℤ+ + then

a = bq + r , 0   r < b

∴ Now, a = 532 and p = 21 then ,

532 = 21 .q + r , 0 ≤ r< 21

Putting  q = 25 then,

532 =  21×25+ r [∵0≤ r < 21]

r = 532 – 525

r = 7

Therefore, Numbers of completed

rows = 25 and left

over flower pots = 7

(3) Prove that the product of two consecutive positive integers in divisible by 2.

=> Let two consecutive positive integers are a and (a + 1).

∴ Product = a (a+1) = a2+a   p is integers

Let a is even then, a = 2p

Product = a2 + a

= (2p)2 + 2p

= 4p2 + 2p

= 2(2p2 + p), divisible by 2 .

∴ product is divisible by 2 .

Now, a is odd, then, a = 2q + 1, q is integers

∴ product = a2 + a

= (2q + 1)2 + 2q + 1

= 4q2 + 4q + 1+2q +1

= 4q2 + 6q + 2

= 2(2q2 + 3q + 1) divisible by 2

∴ product is divisible by 2 .

∴ Therefore The product of two consecutive positive integer is divisible by 2 . [proved]

(4) When the positive integers a, b and c are divied by 13, the respective remainders are 9, 7 and 10 . Show that a + b + c is divisible by 13 .

=> By Euclid’s division algorithm

a = 13x + 9

b = 13y + 7   where x, y z are integers .

c = 13z + 10

Now, a + b + c

= 13x + 9 + 13y +7+13z + 10

= 13(x +y + z) + 26

= 13 (x + y + z + 2),

This is divisible by 13.

∴ Therefore, (a +b + c) is divisible by 13. [proved]

(5) Prove that square of any integer leaves the remainder either 0 or 1 when divided by 4 .

=> Let a is any integer .

Now, let a is even then a = 2x , x ∈ integer

∴ a2 = (2x)2

= 4x2, is divisible by 4.

∴ a2 is divisible by 4 and remainder 0 .

Let a is odd , then a = 2y + 1 , y ∈ z

a2 = (2y +1)2

= 4y2 +4y + 1

= 4(y2+y) +1 .

Now, a2 is divided by 4 then remainder is 1.

∴ square of any integer leaves the remainder either 0 or 1 when divided by 4 . [proved]

(6) Use Euclid’s division algorithm to find the highest common factor (HCF) of

(i) 340 and 412

(ii) 867 and 255

(iii) 10224 and 9648

(iv) 84,90 and 120 .

=> (i) 340 and 412

412 = 340 × 1 + 72 [by Euclid’s division Algorithm]

340 = 72 × 4 + 52 [by . E.D. A] [∴ E . D. A = Euclid’s division algorithm]

72 = 52 × 1 + 20 [ by E. D. A]

52 = 20 × 2 + 12 [ by E. D. A]

20 = 12 × 1+ 8 [by E. D. A]

12 = 8 × 1 + 4 [by E. D. A]

8 = 4 × 2 + 0 [by E. D. A]

Now, remainders is zero.

∴  Therefore H. C. F. of 340 and 412 is 4 .

(ii) 867 and 255.

Now, 867 = 255 × 3 + 102 [by E. D. A]

255 = 102 × 2 + 51 [by E. D. A]

102 = 51 × 2 + 0 [ by E. D. A]

Now, remainder is zero.

∴ Thus, 867 and 255 H. C. F is 51.

(iii) 10224 and 9648

Now, 10224 = 9648 × 1 + 576 [by E. D. A]

9648 = 576 × 16 + 432 [by E. D. A]

576 = 432 × 1 + 144 [ by E. D. A]

432 = 144 × 3 + 0 [ by E. D. A] .

Now, remainder is zero.

∴ Therefore, 10224 and 9648 H.C. F is 144 .

(iv) 84 , 90 and 120 .

Now, 90 = 84 × 1 + 6 [by E. D. A]

84 = 6 × 14 + 0 [by E. D. A]

Now, remainder is zero

∴ 84 and 90 H.C.F is 6.

Now, find out 6 and 120 H.C.F.

∴ 120 = 6 × 20 + 0 [ by E.D.A]

∴ remainder is zero

∴ 6 and 120 H.C.F is 6 .

∴ Therefore, 84,90 and 120 H.C.F is 6 .

[NOTE:- E.D.A = Euclid’s division algorithm]

(7) Find the largest number which divided 1230 and 1926 leaving remainder 12 in each ease .

=> Given that 1230 and 1926 leaving remainder 12 in each ease .

∴ 1230 – 12 = 1218 and 1926- 12 = 1914

Now, required number is H.C.F of 1218 and 1914 .

∴ 1914 = 1218 × 1 + 696 [ by E.D.A]

∴ 1218 = 696 × 1 + 522 [ by E.D.A]

∴ 696 = 522 ×  1 + 174 [ by E.D.A]

522 = 174 × 3 + 0 [ by E.D.A]

Now, remainder  is zero .

∴ 1218 and 1914 H.C.F is 174

Therefore, 174 divides 1230 and 1926 leaving remainder 12 in each ease .

(8)  If d is the highest common factor of 32 and 60 find x and y satisfying d = 32x + 60y .

=> Now, Find out H.C.F of 32 and 60 then ,

60 = 32× 1 + 28 [ by E.D.A] …….(i)

32 = 28 × 1+ 4 [ by E.D.A] …..(ii)

28 = 4 × 7 + 0 [ by E.D.A]

∴ remainder is zero .

∴ 32 and 60 H.C.F is 4

∴ Now, 4 = 32x + 60y .

from (ii), 32 = 28 × 1 + 4

∴ 4 = 32 – 28

4 = 32 – 28

4 = 32 –(60- 32) [∵ by (i)]

4 = 32 – 60 + 32

4 = 32 × 2 – 60

4 = 60 × (-1) + 32 × 2

4 = 32 × 2 + 60 × (-1)

∴ x =2 , y= -1 .

x =2 and y = -1 satisfying d = 32x + 60y .

(9) A positive integer when divided by 88 gives the remainder 61 . What will be the remainder when the same number is divided by 11 ?

=> By Euclid’s division algorithm,

a = 88b +61 ,

where a is positive integer and b ∈ ℤ .

Now, 88 is a divisible by 11.

∴ 88b is a divisible by 11.

Then, 61 = 11× 5 + 6 [by E.D.A]

∴ a is divided by 11  then remainder is 6.

∴ Therefore the remainder is 6 .

(10) Prove that two consecutive positive integers are always co-prime .

=> Let a and a+1 be two consecutive positive integers .

any two positive integers is called co-prime if H.C.F of two numbers is 1 .

Now, a+1 = a × 1 + 1 [by E.D.A]

a = 1 × a + 0 [by E.D.A]

∴ remainder is zero .

∴ a and a+1 H.C.F is 1 .

so,  a and a+1 is co-prime .

∴ Two consecutive positive integers are always co-prime .[proved]

#### Exercise – 2.2

(1) For what values of natural number n, 42 can end with the digit 6 ?

=> 4n = (2×2)n = 2n×2n

Now, n = 1 , then 4n = 21 × 21 = 4

n = 2 , then  4n = 22 × 22 = 4 × 4 = 16 (last digit 6)

n = 3 , then  4n = 23 × 23 = 8 × 8 = 64

n = 4 , then  4n = 24× 24 = 16 × 16 = 256 (last digit 6)

n = 5, then  4n = 25 × 25 = 32 × 32 = 1024

n = 6, then 4n = 26 × 26 = 64 × 64 = 1316 ( last digit 6)

…….

Now, n = 2, 4, 6, …. then 4n last digit 6 .

∴ Therefore n is even then 4n last digit 6 .

(2) If m, n are natural numbers , for what values of m, does 2m× 5m end in 5 ?

=>  ∴ 2n × 5m

∴ 2n is every time even for all values of n .

∴ for all values of m , 5m is odd because last digit always is 5 .

∴ 2n × 5m

for all values of m , n

2n × 5m is always end digit is 0 .

so, 2n×2m can not end digit 5 , for all values of m.

∴ Thus, m has no values 2n × 2m end in 5 .

(3)  Find the H C F of 25 25 25 and 36 36 36 .

=>  25 25 25 and 36 36 36

using by fundamental theorem of arithmetic .

25 25 25 = 5 × 5 × 10101

and 36 36 36 = 2 × 2 × 3 × 3 × 10101

∴ common factor of 25 25 25 and 36 36 36 is

= 10101

∴ H.C.F of 25 25 25 and 36 36 36

is 10101 .

(4) If 13824 = 2a × 3b then find a and b .

=>  Here , 13824 = 2a × 3b ……(i)

∴ 13824 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3

13824 = 29 × 33……..(ii)

Now, (i) and (ii) compose, we get

∴ a = 9 , b = 3 .

(5) If p1x1 × p2x2× p3x3 × p4x4 =113400 where p1 , p2, p3, p4 are primes in ascending order and x1, x2, x3,x4 are integers . find the values of p1, p2 ,p3, p4 and x1, x2, x3, x4 .

=> Given, 113400 = p1x1 × p2x2× p3x3 × p4x4 ……(i)

Now, 113400 = 2 × 2  × 2 × 3  × 3 × 3 × 3  × 5  × 5 × 7

= 23 × 34  × 52  × 7 [where, 2,3,5, 7 are prime ]

∴ 113400 = 23 × 34× 52× 7 ….. (ii)

Now, (i) and (ii) compare , we get

∴ p1 = 2 , p2 =3, p3= 5 , p4= 7

and x1 = 3, x2= 4 , x3= 2 , x4= 1

(6) Find the LCM and HCF of 408 and 170 by applying fundamental theorem of arithmetic.

=> Now, 408 = 2 × 2 × 2 × 3 × 17 = 23 × 3 × 17 and 170 = 2 × 5 × 17 = 2 × 5 × 17

∴ L.C.M of 408 and 170

= 23 × 3 × 5 × 17 [∵ take for all [prime highest power] .

= 8 × 3 × 5 × 17

= 2040

∴ H.C.F of 408 and 170

= 2 × 17 [take for common prime factor]

= 34

Thus, L C M = 2040 and HCF = 34.

(7) Find the greatest number consisting of 6 digits which is exactly divisible by 24, 15 , 36 ?

=> Now, 24 = 2 × 2 × 2 × 3 = 23 × 3

15 = 3 × 5 = 3 × 5

and 36 = 2 × 2 × 3 × 3 = 22 × 32

∴ LCM of 24, 15 and 36

= 2× 32 × 5 = 8 × 9 × 5

= 360

we know that 6 digit greatest number is 999999

Now, 999999 = 360 × 2777 + 279.

∴ remainder is 279

∴ Now , 999999 – 279 = 999720 .

∴ Now, 999720 is divisible by 360.

∴ Therefore 999720 greatest 6 digit number is divisible by 24, 15 and 36 .

(8) What is the smallest number that when divided by three numbers such as 35, 56 and 91 leaves remainder 7 in each case.

=> Now, 35 = 5 × 7

56 = 2 × 2 × 2 × 7 = 23 × 7

and 91 = 7 × 13

∴ L C M of 35, 56, 91

= 23 × 5 × 7 ×13 [take for all prime highest power]

= 3640

Now, 3640 + 7

= 3647

∴ 3647 is divided three number such as 35, 56 and 91 then leaves remainder 7 in ease.

∴ Therefore, the required smallest number is 3647.

(9) Find the least number that is divisible by the first ten natural numbers.

=> Now, 1 = 1

2 = 1 × 2

3 = 1 × 3

4 = 1 × 2 × 2 = 1 × 22

5 = 1 × 5

6 = 1 × 2 × 3

7 = 1 × 7

8 = 1 × 2 × 2 × 2 = 1 × 23

9 = 1 × 3 × 3 = 1 × 32

10 = 1 × 2 × 5

Now, L C M of first ten natural number is

= 1 × 23 × 32 × 5 × 7 [take all prime factor highest power]

= 8 × 9 × 5 × 7

= 2520

∴ 2520 is least number that is divisible by the first ten natural numbers .

∴ Therefore required number is 2520 .

#### Exercise – 2.3

(1) Find the least positive value of x such that

(i) 71 ≡ x (mod 8)

(ii) 78 + x ≡ 3 (mod 5)

(iii) 89 ≡ (x + 3) (mod 4)

(iv) 96 ≡ x/7 (mod 5)

(v) 5x ≡ 4 (mod 6)

=> 71 ≡ x (mod 8)

71 – x = 8p for some integer p .

∴ Now, (71- x) is a multiple of 8.

Now, 8× 8 = 71 – x [∵ Let p = 8

x = 71 – 64

x = 7

∴ x = 7

(ii) 78 + x ≡ 3 (mod 5)

78 + x – 3 = 5p, for some integer p.

75 + x = 5p

75 + x =  5 × 16 [putting p = 15]

x = 80 – 75

x = 5 .

∴ x = 5 .

(iii) 89 ≡ (x + 3) (mod 4)

∴ 89 – x – 3 = 4p , for some integer p .

86 – x = 4p

86 – x = 21 × 4 [putting p = 21]

86 – x = 84

-x = 84 – 86

x = 2

∴ x = 2

(iv) 96 ≡ x/7 (mod 5)

∴ 96 – x/7 = 5p , for some integer p .

672 – x = 35p .

672 – x = 35 × 19

672 – x = 665

-x = 665 – 672

– x = – 7

x = 7

∴ x = 7

(v) 5x ≡ 4 (mod 6)

5x – 4 = 6p , for some integer  p .

5x – 4 = 6 [putting p = 1]

5x = 6 + 4

5x = 10

x = 2

∴ x = 2.

(2) If x is congruent to 13 module 17 then 7x – 3 is congruent to which number module 17?

=> Given, x is congruent to 13 module 17

∴ x – 13 = 17p [for some integer p]

Now, find out x least values then

putting p = 1

∴ x – 13 = 17

x = 17 + 13 = 30 .

∴ Now, 7x – 3

= 7 × 30 – 3

= 210 – 3 = 207

∴ Now, 207 = 17×12 + 3

∴ 207 ≡ 3 (mod 17) .

∴ Therefore required number is 3 .

(3) Solve 5x ≡ 4 (mod 6)

=> 5x ≡ 4 (mod 6)

5x – 4 = 6p, for some integer p.

5x = 6p + 4

x = 6p + 4 /5

Putting p = 1 then, x = 6 + 4 /5 = 10/5 = 2 .

Putting p = 6 , then x = 6 × 6 + 4/5

= 40/5 = 8

putting = p = 6 then x = 6 × 11 + 4/5

….  = 70/5 = 14

∴ The solutions are

x = 2, 8, 14, 20 , ….

(4) solve 3x – 2  ≡ 0 (mod 11)

=> 3x – 2  ≡ 0 (mod 11)

3x – 2 – 0 = 11p for some integer p.

3x = 11p + 2

x = 11p + 2/ 3

putting p = 2 then, x = 11×2 +2/ 3 = 24/3 = 8

putting p = 5 then, x = 11×5 + 2/ 3 = 57/3 = 19

putting p = 11 then x = 11 × 11 + 2/ 3 = 123/3 = 41

∴ The solution are

x = 8, 19, 41, …..

(5) what is the time 100 hours after 7 am ?

=> Now, every 12 hours comes 7 clock .

∴ 100 ≡ 4 (mod 12)

∴ required time is 7 + 4 = 11 clock

∴ The time 100 hours after 7 am is 11 am

(6) what is the time 15 hours before 11 pm ?

=> Now, every 12 hours before comes 11 pm

∴ 15 ≡ 3 (mod 12)

∴ required time is = 11 – 3 = 8 clock

∴ The time 15 hours before 11 pm is 8 pm .

(7) Today is Tuesday my uncle will come after 45 days. In which day my uncle will be coming ?

=> Now, every 7 days comes Tuesday .

∴ 45 ≡ 3 (mod 7).

∴ uncle will be coming is = Tuesday + 3 days

= Friday.

Thus, Friday my uncle will be coming .

(8) Prove that 2n+6 × 9n is always divisible by 7 for any positive integer n .

=> Now, 2n + 6 × 9n, for any positive integer n.

∴ n = 1, then, 2 + 6 × 9 = 2 + 54 = 56 = 7 × 8, is divisible by 7.

∴ Let n = p then , 2p + 6 × 9p , is divisible by 7 .

∴ 2p + 6 × 9p = 7a for any z+ in a … (i)

∴ Now, n = p + 1 then , 2p+1 + 6 × 9p+1

= 2p.2 + 6 × 9p.9

=  2p. 2 + 9 × (7a-2p) [by (i)]

= 2.2p + 9 × 7a – 9 × 2p

= 9 × 7a – 7 × 2p

= 7 (9a – 2p) , is divisible by 7 .

Now, Let n = p

∴ 2p + 6 × 9p is divisible by 7.

then also n = p+ 1

2p+1+ 6 × 9p+ is divisible by 7.

∴ 2n + 6 × 9n + n1 is always divisible by 7 for any positive integer n.

(9) Find the remainder when 281 is divisible by 17

=> we know that

24 ≡ -1 (mod 17)

(2)4×20 ≡ (-1)20 (mod 17)

=> 280 ≡ 1 (mod 17)

=> 280 × 2 ≡ 1 ×  2(mod 17)

=> 281  ≡ 2 (mod 17) .

∴ 281 is divisible by 17 then remainder is 2 .

(10) The duration of flight travel from Chennai to London through British Airlines is approximately 11 hours. The airplane beings its journey on Sunday at 23:30 hours. If the time at Chennai is four and half hours ahead to that of London’s time, then find the time at London, when will the flight  lands at London Airport .

=> Given that journey beings on Sunday at 23:30 hours.

Total journey time taken by 11 hours.

Now, (23:30 + 11)

= 34: 30

= 10 : 30

∴ Now, Time difference is 4 ½ hours ahead to London.

∴ (10: 30 – 4: 30)

= 6

∴ Next day Monday 6 a.m flight lands at London Airport .

#### Exercise – 2.4

(1) Find the next three terms of the following sequence.

(i) 8, 24, 72,….

(ii) 5, 1, -3, …..

(iii) ¼, 2/9, 3/16,….

=> (i) Given sequence

8, 24, 72 ,….

Now, 8 × 3 = 24

24 × 3 = 72

similarly, 4th term 72  ×  3 = 216

5th term 216 × 3 = 648

6th term 648 × 3 = 1944

Thus, 8, 24, 72, 216, 648, 1944, …..

(ii) Given sequence –

5, 1, -3 ,….

Now, 1 – 5 =-4

-3 -1 = -4

or

5 – 4 = 1

1 – 4 = -3

∴ similarly 4th term = -3 -4 = -7

5th term = -7 -4 = -11

6th term = -11 -4 = -15

∴ Thus the sequence

5, 1, -3, -7, -11, -15 ,…..

(iii) Given sequence ,

¼, 2/9 , 3/16, ……

Now, ∴ 1/4 = 1/(2)2 = 1/(1+1)2

∴ 2/9= 2/(3)2 = 2/(2+1)2

∴ 3/16= 3/(4)2= 3/(3+1)2

∴ given sequence are n/(n+1)2 [nth term]

∴ given sequence 4th term

∴ n = 4, then , 4/(4+1)2 = 4/(5)2 = 4/25

∴ 5th term , n = 5 then ,

5/(5+1)2 = 5/(6)2 = 5/36

∴ 6th term , n = 6 then ,

6/(6+1)2 = 6/(7)2 = 6/49 .

∴ Thus the sequence ,

¼, 2/9, 3/16, 4/25, 5/36, 6/49, …..

(2) Find the first four terms of the sequences whose nth terms are given by

(i) an = n3 – 2

(ii) an = (-1)n+1 n(n+1)

(iii) an = 2n2 – 6

=> (i) Given sequence nth term

an = n3– 2

if n = 1, then a1= (1)3 -2 = 1-2 = -1

if n =2 , then , a2= (2)3 – 2= 8-2 = 6

if n = 3 , then a3 = (3)3 – 2 = 27 – 2 = 25

if n = 4 , then a4 = (4)3 – 2 = 64 – 2 = 62

∴ given sequence first four term are

-1, 6, 25, 62 ,

(ii) Given sequence nth term

an = (-1)n+1 n (n+1)

if n = 1 , then a1 = (-1)1+1 1(1+1) = (-1)2.2 = 1.2 = 2

if n = 2 , then a2= (-1)2+1 2(2+1) = (-1)3.6 = -6

if n = 3 , then a3 = (-1)3+1 3(3+1) = (-1)4.12= 12

if n = 4, then a4 = (-1)4+1 4(4+1) = (-1)5 . 20 = -20

∴ given sequence first four terms are

2, -6, 12, -20

(iii) Given sequence nth term

an = 2n2 -6

if n = 1, then a1 = 2.(1)2 -6 = 2-6 = -4

if n = 2, then a2 = 2.(2)2 -6 = 8-6 = 2

if n = 3, then a3 = 2.(3)2 – 6 = 18.6 = 12

if n = 4, then a4 = 2.(4)2 -6 = 32 -6 = 26

∴ given sequence first four terms are

-4, 2, 12, 26 .

(3) Find the nth  term of the following sequences

(i) 2, 5, 10, 17,…..

(ii) 0, ½, 2/3, ….

(iii) 3, 8, 13, 18, …..

=> (i) Given sequence

2, 5, 10, 17,….

∴ 1st term a1= 2 = (1)2 +1

2nd term a2 = 5 = (2)2 + 1

3rd term a3 = 10 = (3)2 + 1

4th term a4 = 17 = (4)2+1

∴ similarly , nth term an = (n)2 +1

Thus given sequence nth term is (n)2+1.

(ii) Given sequence

0, ½, 2/3,….

1st term, a1 = 0= 1-1/1

2nd term, a2 = ½ = 2-1/2

3rd term, a3 = 2/3 = 3-1/3

similarly nth  term an = (n-1)/2

∴ given sequence nth term is (n-1)/n

(iii) Given sequence

3, 8, 13, 18 .

1st term, a1 = 3 = 5 × 1 – 2

2nd term, a2 = 8 = 5× 2 – 2

3rd term, a3 = 13 = 5 × 3 – 2

4th term, a3 = 18 = × 4 – 2

∴ similarly, nth term, an = 5n – 2

∴ Thus the given sequence nth term is

an = 5n – 2 .

(4) Find the indicated terms of the sequence whose nth terms are given by

(i) an = 5n/n+2; a6 and a13

(ii) an = – (n2 – 4) : a4 and a11

=> (i) Given that

an = 5n/n+2

∴ if n = 6 then a6 = 5× 6 / 6+2 = 30/8

if n = 13 , then a13 = 5× 13/13+2 = 5× 13/ 18 = 13/3

Thus , a6 = 30/8 and a13 = 13/3

(ii) Given that

an = – (n2 -4)

if n = 4 , then a4 = – ((4)2 – 4)

= -(16 – 4) = -12

if n = 11 , then a11 = -((11)2 – 4)

= – (121- 4) = -117

Thus ∴ a4 = -12 and a11 = -117

(5) Find a8 and a15 whose nth term is

an = {n2-1/n +3 ; n is even , n ∈ ℕ

n2/ 2n +1 ; n is odd , n ∈ ℕ

=> Given that

an = {n2-1/n +3 ; n is even , n ∈ ℕ

n2/ 2n +1 ; n is odd , n ∈ ℕ

∴ Now, 8 is even then

a8 = (8)2 -1/8+3 = 64-1/11 = 63/11

∴ 15 is odd then ,

a15 = (15)2/2× 15+1 = 225/31

∴ a8 = 63/11 and a15 = 225/31

(6) If a1 = 1, a2= 1 and an= 1an-1+ an-2 , n≥3, n ∈ ℕ , then find the first six term of the sequence .

=> Given sequence nth term

an = 2 an-1 + an-2 , n≥ 3 , n ∈ ℕ

if n = 3, then a3 = 2a2 + a1

= 2 × 1 + 1 [∵a1 = 1 and a2 = 1]

= 3

if n = 4 , then a4 = 2a3+ a2

= 2 × 3 + 1 [∵ a3 = 3 and a2 = 1 ]

= 6 +1

= 7

if n = 5 , then a5 = 2a4 + a3

= 2 × 7 + 3 [∵ a4 = 7 and a3 = 3 ]

= 14 + 3

= 17

if n = 6 , then a6 = 2a5 + a4

= 2 × 17 + 7 [∵ a5 = 17 and a4 = 7 ]

= 34 + 7

= 41

if n = 7, then a7 = 2a6 + a5

= 2  × 41 + 17 [∵ a6 = 41 and a5 = 17 ]

= 82 + 17

= 99

if n = 8 , then a8 = 2a7 + a6

= 2 × 99 + 41 [∵ a7 = 99 and a6 = 41 ]

= 198 + 41

= 239

∴ Thus the first six terms are

a3 = 3

a4 = 7

a5 = 17

a6 = 41

a7 = 99

a8 = 239

#### Exercise – 2.5

(1) Check whether the following sequences are in A.P.

(i) a-3, a-5 , a-7, …..

(ii) ½ , 1/3, ¼, 1/5,….

(iii) 9, 13, 17, 21, 25,….

(iv) -1/3, 0 , 1/3 , 2/3 ,….

(v) 1, -1, 1, -1, 1,-1 ,…..

=>  (i) Given that sequence

a-3, a-5, a-7, ….

Now, t1 = a- 3 , t2 = a- 5 , t3 = a- 7

∴ t2 – t1 = a- 5 – a + 3 = -2

∴ t3 – t2 = a – 7 – a + 5 = -2

Now, t2 – t1 = t3 – t2

∴ Difference between consecutive terms are equal.

∴ Therefore the sequence a -3 , a- 5 , a – 7…… is in  A.P.

(ii) since the sequence

½, 1/3, ¼, 1/5, …..

∴ t1 = ½ , t2 = 1/3, t3= ¼ , t4 = 1/5

Now, t2 – t1 = 1/3 – ½ = 1/6

t3 – t2 = ¼ – 1/3 = 1/12

t4 t3 = 1/5 – ¼ = – 1/20

Now, t2 – t1 ≠  t3  – t2 ≠ t4  – t3

∴ Difference between consecutive terms are not equal .

so, ½ , 1/3, ¼, 1/5 , ……. are not A.P.

(iii) Given the sequence

9, 13, 17, 21, 25 , ……

t1 = 9 , t2= 13 , t3 = 17 , t4 = 21 , t5 = 25

Now, t2 – t1 = 13 -9 = 4

t3 – t2 = 17 -13 = 4

t4 – t3 = 21 – 17 = 4

t5 – t4 = 25 – 21 = 4

Now, t2 – t1 = t3 – t2 = t4 – t3 = t5 – t4

∴ Difference between consecutive terms are equal

so, 9, 13, 17, 21, 25 ,…… are in A.P.

(iv) Given the sequence ,

– 1/3 , 0 , 1/3, 2/3, ….

∴ t1 = -1/3 , t2 = 0 , t3 = 1/3 , t4 = 2/3

Now, t2 – t1 = 0 – (-1/3) = 1/3

t3 – t2 = 1/3 – 0 = 1/3

t4 – t3 = 2/3 – 1/3 = 1/3

Now, t2 – t1 = t3 – t2 = t4 – t3

Thus , differences between consecutive term are equal

so, -1/3 , 0 , 1/3 , 2/3, ….. are in A.P.

(v) Given the sequence ,

1, -1 , 1, -1, 1, -1 ,……

t1 = 1 , t2 = -1 , t3 = 1 , t4 = -1 , t5 = 1 , t6 = -1

Now, t2 – t1 = -1-1 = -2

t3 – t2 = 1 – (-1)  = 2

t4 – t3 = -1 -1 = -2

t5 – t4 = 1 – (-1) = 2

t6 – t5 = -1 – (- 1) = -2

Now, t2 – t1 ≠ t3 – t2

Thus, the difference between consecutive terms are not equal .

∴ Therefore, 1, -1, 1, -1,1, -1, ……. are not in A.P.

(2) First term a and common difference d are given below . Find the corresponding A.P.

(i) a = 5, d = 6

(ii) a = 7 , d = -5

(iii) a = ¾ , d = ½

=>  (i) given that

First term (a) = 5

and common difference (d) = 6

∴ We know that ,

In an A.P. nth term is tn = a+ (n -1) d [a = first term d = common difference ]

if n = 2 , then  t2 = a + 1 . d

= 5 + 6 [∵ a = 5, d = 6]

= 11

if n = 3 , then t3 = a+2.d

= 5 + 2. 6 [ a = 5 , d = 6 ]

= 5 + 12

= 17

if n = 4 , then t4 = a+3d = 5 + 3. 6 [∵ a= 5 , d = 6]

= 5 + 18

= 23

Thus, the required A.P is , 5, 11, 17, 23 ,…..

(ii) given that

a = 7 , d = -5

∴ we know that ,

In an A.P., nth term is  tn = a + (n -1) d [∵ a = first term d = common difference ]

if n = 2 , then t2 = a+ 1 . d = 7-5 [∵ a = 7 , d = -5]

= 2

if n =3 , then t3 = a + 2d = 7 – 10 [∵ a = 7 , d = -5]

= -3

if n = 4 , then t4 = a +3d = 7 -15 [∵ a = 7 , d=-5]

= -8

Therefore, the required A.P is 7, 2, -3, -8, ……

(iii) Given that

a = ¾ , d = ½

we know that ,

In an A.P, nth term is tn = a + (n -1)d [ a = 1st term d= common difference ]

if n = 2 , then t2 = a+1.d

= ¾ + ½ [∵a = ¾ , d = ½ ]

= 5/4

if n = 3 , then t3  = a + 2d = ¾ + 2.1/2 [∵ a = ¾ ,d = ½ ]

= ¾ +1

= 7/4

if n = 4 , then t4 = a + 3d = ¾ + 3/2 [∵ a = ¾ , d = ½]

= 3+6/4 = 9/4

Thus, the required A.P is ¾ , 5/4, 7/4, 9/4,……

(3) Find the first term and common difference of the arithmetic progressions whose nth terms are given below

(i) tn = -3 + 2n

(ii) tn = 4 – 7n  .

=>  (i) tn = -3 + 2n

if n = 1 , then t1 = -3 +2 = -1

if n = 2, then t2 = -3 +4 = 1

Now, t2 – t1 = 1 –(-1)  = 2

∴ first term (a) = -1

and common difference (d) = 2

(ii) tn = 4 – 7n

if n = 1 , then t1 = 4 – 7 = -3

if n = 2 , then t2 = 4 – 14 = -10

t2 – t1 = -10 –(-3) = -10 + 3 = -7

∴ First term (a) = -3

and common difference (d) = -7

(4) Find the 19th term of an A.P. -11 , -15 , -19, ……

=> Given A.P. is

-11 , -15, -19

∴ a = -11 and d = -15 – (-11)

= -15 + 11

= -4

we know that, in an A.P

nth term tn = a + (n -1) d [ a = first term d = common difference]

if n = 19 , then t19 = a + 18d

= -11 + 18 (-4) [∵ a = -11 , d = -4]

= -11 – 72

= – 83

∴ Thus, required 19th term is -83

(5) Which term of an A.P. 16, 11, 6, 1, …. is – 54 ?

=> Given A.P is

16, 11, 6, 1,….

∴ a = 16 and d = 11-16 = -5

We know that , In an A.P nth term is

tn = a + (n -1) d [ a = first term d = common difference ]

Now, given that tn = -54 ,

∴ tn = a + (n-1) d

– 54 = 16 + (n -1) (-5) [∵ tn = -54 , a = 16 d = -5]

-54 = 16 – 5n + 5

-54 = 21 – 5n

– 5n = -54 – 21

– 5n = -75

n = 15

∴ Thus 15th term is – 54 .

(6) Find the middle term (s) of an A.P

9, 15. 21, 27,…… , 183 .

=>  we know that

In a finite A.P whose first term is a and last term L, then the number of terms in the A.P  is n = L-a/d + 1 , where d = common difference .

∴ given A.P is

a = 9 , d = 15 -9 =6

and    L = 183

∴ no of terms (n) = L-a/d + 1

= 183 – 9/ 6 + 1

= 174/6 + 1

= 29 + 1 = 30 .

∴ given A.P total term 30 .

so, 15th and 16th are the middle terms .

∴ t15 = a + 14d = 9+14.6 [∵ a = 9, d = 6]

= 9 + 84

= 93

and t16 = a + 15d = 9 + 15 . 6 [∵ a = 9 , d = 6]

= 9 + 90

= 99

∴ Thus, the required middle terms are 93, 99

(7) If nine times ninth term is equal to the fifteen times fifteenth term , show that six times  twenty fourth term is zero .

=> Given that ,

9 t9 = 15 t15

15 t15 = 9 t9

15 t15 – 9 t9 = 0

15 (a + 14d) -9 (a + 8d) = 0 [∵ t15 = a + 14d  and t9 = a + 2d , where a = first term and d = common difference]

15a – 9a + 210d – 72 d = 0

6a + 138 d = 0

6(a+ 23d) = 0

6 t24 = 0 [∵ t24 = a + 23d]

∴ 6 t24 = 0 [proved]

(8) If 3+ k , 18 – k , 5k+ 1 are in A.P then find k .

=> Let t1 = 3+ k , t2 = 18 –k , t3= 5k + 1

Now, given terms are in A.P

So, t2 – t1 = t3 – t2

18 – k – 3 – k = 5k + 1 – 18 + k

15 – 2k = 6k -17

8k = 32

k = 4

∴ k = 4

(9) Find x, y and z, given  that the numbers x , 10 , y , 24, z are in A.P.

=> Let t1 = x , t2 = 10 , t3 = y , t4 = 24 , t5 = z

Now, given terms are in A.P

so, t2– t1 = t3– t2 =t4 – t3 = t5 – t5

Now, t2 – t1 = t3 – t2

10 – x = y -10

-x – y = -10 – 10

x + y = 20 …..(i)

Now,

∴ t3 – t2 = t4 – t3

y – 10 = 24 – y

y + y = 24 + 10

2y  = 34

y = 17

∴ y = 17

Now, t4 – t3 = t5 – t4

24 – y = z -24

-y – z = – 24 -24

y + z = 48 ……(ii)

from (i) , y = 17, we get

x + y = 20

x + 17 = 20

x = 20 – 17 = 3

x = 3

from (ii) , y = 17 , we get

y + z = 48

17 + z = 48

z = 48 – 17

z = 31

∴ Thus, x = 3 , y = 17 , z = 31 .

(10) In a theatre, there are 20 seats in the front row and 30 rows were allotted. Each successive row contains two additional seats than its front row. How many seats are there in the last row?

=> first row contains seats is

a = 20

Now, Each successive row contains two additional seats that its front row.

∴ common difference (d) = 2

Now, total 30 rows contains.

we know that tn = a + (n-1) d [ a = first term d = common difference]

if  n = 30 then t30 = a + 29d

= 20 + 29× 2 [∵ a = 20 d = 2]

= 20 + 58

= 78

∴ Hence , 78 seats are there in the last row .

(11) The sum of three consecutive terms that are in A.P is 27 and their product is 288 . Find the three terms.

=> Let three consecutive terms are

a – d , a, a + d

Now, a – d + a + a + d = 27

3a = 27

∴ a = 9 ……(i)

and (a –d) . a. (a + d) = 288

(a2 – d2). a = 288

{(9)2 – (d)2} . 9 = 288 [by (i)]

(81 – d2) = 32

d2 = 81 – 32

d2 = 49

d = ± 7

Now, a = 9 and d = ± 7.

Hence, the three consecutive terms are

2, 9, 16

(12) The ratio of 6th and 8th term of and A.P is 7: 9 . Find the ratio of 9th term to  13th term .

=> Given that ,

t6 / t8 = 7/9

a + 5d/a+ 7d = 7/9 [∵ t6 = a + 5d and t8 = a + 7d

a= first term , d = common difference .]

9a + 45d = 7a + 49d

2a = 4d

a = 2d …..(i)

Now, t9 : t13

a + 8d : a + 12d [∵ t9 = a + 8d , t13 = a + 12d ]

2d + 8d : 2d + 12d  [ by equation (i) ]

10d : 14d

5 : 7

∴ The ratio of 9th term to 13th term is 5 : 7 .

(13) In a winter season let us take the temperature of Uoty from Monday to Friday to be  in A.P . The sum of temperatures from Monday to Wednesday is 0°c and the sum of the temperatures from Wednesday to Friday is 18°c . Find the te mperature on each of the five days .

=>  Let Monday to Friday temperatures are

a – 2d , a – d , a , a + d , a + 2d

Now, a – 2d + a – d + a = 0

3a – 3d = 0

a – d = 0 ….(i)

and a + a + d + a + 2d = 18

3a + 3d = 18

a + d = 6 ….. (ii)

Now, (i) – (ii) , we get

a – d – a – d = 0 – 6

– 2d = – 6

∴ d = 3

from (i), a – d = 0

a -3 = 0 [∵ d = 3]

a = 3

∴ a = 3 and  d= 3

Therefore Monday to Friday temperatures are

-3, 0 , 3, 6, 9

(14) Priya earned ₹ 15,000 in the first month. there after her salary increased by ₹ 1500 per year . Her expenses are ₹ 13,000 during the first month and the expenses increases by ₹ 900 per year. How long will it take for her to save ₹ 20, 000 per month .

=> first year monthly save is = (15,000 – 13000)

= 2000

second year monthly earned is = (15000 + 1500)

= 16500

second year monthly expenture is = (13000 + 900 )

= 13900

second year monthly save is = (16500 – 13900)

= 2600

similarly, 3rd year monthly save is = (18000 – 14800)

= 3200

∴ every year difference (d) = 2600 – 200

= 600

1st year save (a) = 2000

∴ Let n years to save ₹20,000 per month.

Now, tn = a + (n – 1) d

∴ tn = 20,000 , a = 2000 and d = 600

∴ 2000 + (n – 1) 600 = 20,000

600n – 600 = 18, 000

600n = 18600

n = 18600/600

n = 31

∴ Therefore 31 years to save ₹ 20,000 per month .

#### Exercise – 2.6

(1) Find the sum of the following

(i) 3, 7, 11, ….. up to 40 terms .

(ii) 102, 97 , 92, …. up to 27 terms .

(iii) 6 + 13+ 20 + ….. + 97 .

=>  (i) 3 , 7 , 11, …. up to 40 terms .

∴ first term(a) = 3

∴ common difference (d) = 7 – 3 = 4

Now, 40th term (tn) = a + (n -1) d

= 3 + 39 × 4 [∵ a = 3 , d = 4 ]

= 3 + 156

= 159

required sum (sn) n/2 [a + L] [n= total term number , a = first term , L = last term ]

= 40/2 [ 3 + 159]

= 20 × 162

= 3240

∴ s40 = 3240

(ii) 102, 97, 92 , …. up to 27 terms .

Now, a = 102

d = 97 – 102 = -5

∴ required sum (sn) = n /2 [ a+ a + (n -1) d] [∵ L= a + (n-1)d , a = first term , d = common difference ]

s27 = 27/2 [102 + 102 + 26 × (-5) ]

s27  = 27 / 2 [ 204 – 130 ]

= 27/2 × 74

= 27 × 37

∴ s27 = 999

(iii) 6 + 13 + 20 + …… + 97

∴ first term (a) = 6

common difference (d) = 13 – 6 = 7

Now, Let 97 is the nth term.

∴ tn  = a + (n-1)d

97 = 6 + (n-1) 7

97 = 6 + 7n – 7

7n = 98

n = 14

∴ 97 is the 14th term.

∴ total terms (n) = 14

∴ required sum (sn) = n/2 [ a + L ]

s14 = 14/2 [ 6 + 97]

= 7× 103

= 721

∴ 6 + 13 + 20 + …… + 97 = 721

(2) How many consecutive odd integers begining with 5 will sum to 480 ?

=> Given that

First term (a) = 5

∴ 7 is the next term.

∴ Common difference (d) = 7 – 5 = 2

Now, Let n numbers consecutive odd integers

∴ Now, sn = 480

∴ n/2 [a + a+(n-1)d] = 480 [∵sn = n/2 [a+L], L= a+ (n-1)d]

n/2 [ 5+5+(n-1) 2] = 480

n [10 + 2n – 2] = 480 × 2

2n2 + 8n = 480 × 2

n2 + 4n – 480 = 0

n2 + 24n – 20n – 480 = 0

n (n + 24) – 20 (n + 24) = 0

(n + 24) (n – 20) = 0

n + 24 = 0   n – 20 = 0

n = – 24       n = 20

∴ n ≠ – 24

(Numbers of terms cannot be negative)

∴ 20 numbers consecutive odd integers beginning with 5 will sum to 480.

(3) Find the sum of first 28 terms of an A.P. whose nth term is 4n – 3 .

=> Given that,

nth term tn = 4n – 3

if n = 1, then t1 = 4.1 – 3 = 1

∴ first term (a) = 1

if n = 28 , then t28 = 4×28 – 3

= 112 – 3 = 109

∴ first 28 terms sum = 28/2 [ 1+ 109]

= 14 × 110

= 1540

∴ required sum = 1540

(4) The sum of first n terms of a certain series is given as 2n2 – 3n . Show that the series in an A.P.

=> Given that

sn = 2n2 – 3n

∴ similarly sn -1 = 2(n-1)2 – 3 (n-1)

= 2 (n2 -2n + 1) – 3n + 3

= 2n2 – 4n + 2 – 3n + 3

= 2n2 – 7n + 5

∴ given sequence nth term

tn = sn – sn -1

= 2n2 – 3n – 2n2 + 7n – 5

= 4n – 5

∴ tn = 4n – 5

if n = 1, then t1 = 4 × 1 – 5 = -1

if n = 2 , then t2 = 4 × 2 – 5 = 3

if n = 3 , then t3 = 4 × 3 -5 = 7

…….

∴ Now, t2 – t1 = 3 – (-1) = 4

t3 – t2 = 7 – 3= 4

∴ t2 – t1 = t3 – t2

∴ Difference between consecutive term are equal.

∴ Given series in an A .P.

(5) The 104th term and 4th term of an A.P. are 125 and 0 . Find the sum of first 35 terms .

=> Given that

t104 = 125 and t4 = 0

∴ a+ 103d = 125 ….(i) and a + 3d = 0 …(ii) [∵a= first term , d = common difference ]

Now, (i) – (ii), we get ,

a + 103d – a – 3d = 125 – 0

100d = 125

d = 125/100

d = 5 /4

from (ii) , we get ,

a + 3d = 0

a = -3d

a = – 3 × 5/4 [∵ d = 5/4]

a = – 15/4

we know that ,

sn = n/2 [ a + a+(n-1)d] [∵L = a+ (n-1)d]

sn = n /2 [2a + (n-1) d]

if n = 35 , then

s35 = 35/2 [2 × (-15/4) + 34 × 5/4]

s35 = 35/2 [-30/4 + 170/4]

= 35 × 35/2

= 1225/ 2

612.5

∴ Thus required sum is 612. 5 .

(6) Find the sum of all odd positive integers less than 450.

=> Total numbers of all odd positive integers less that 450 is = last term – first term/difference + 1

= 449 –1 / 2 + 1

= 448/2 + 1

= 224 + 1 = 225

∴ first term (a) = 1 and last term(L) = 449

∴ sum = n/2 (a + L)

= 225/2 (1 + 449)

= 225 × 450/2

= 225 × 225

= 50625

∴ Thus required sum is 50625.

(7) Find the sum of all natural numbers between 602 and 902 which are not divisible by 4 .

=> Now, find the sum of all natural numbers between 602 and 902 which are divisible by 4 .

∴ first number (a) = 604

and last number (L) = 900

∴ common difference (d) = 4

∴ Total numbers (n) = L – a/ d + 1

= 900 – 604 / 4 + 1

= 296 /4 + 1

= 75

∴ sum (sn) = 75/2 (604 + 900) [∵ sn = n/2 (a +L)]

= 75/2 × 1504

= 75 × 752

∴ Now, find the sum of all natural numbers between 602 and 902 .

∴ first number(a1) = 603

last number (L1) = 902

and difference (d) = 1

∴ Total numbers (n2) = L – a1/ d1 + 1

= 901 – 603/1 + 1

= 298 + 1= 299

∴ sum (sn1)= 299/2 (603 + 901)

= 299/2 × 1504

= 299 × 752

Now, sn1 – sn = 299 × 752 – 75 × 752

= 224 × 752 = 168448

∴ sum of all natural numbers between 602 and 902 which are not divisible by 4 is 168448 .

(8) Raghu wish to buy a laptop . He can buy it by paying  ₹ 40,000 cash or by giving it in 10 installments  as  ₹ 4800 in the first month , ₹ 4750 in the second month, ₹ 4700 in the third month and so on . if he pays the money in this fashion . find

(i) total amount paid in 10 installments .

(ii) how much extra amount that he has to pay than the cost ?

=>  (i) first month pay (a) = 4800

and second month pay = 4750

∴ last month pay (L) = t10 = (a + 9d)

= 4800 + 9 × (-50)

= 4800 – 450

= 4350

∴ total amount paid in 10 installments

= 10/2 (4800 + 4350)

= 5 × 9150

= 45750

(ii) The extra amount he pays in installments

= 45750 – 40,000

= 5750

(9) A man repays a loan of ₹ 65,000 by paying ₹ 400 in the first month and then increasing the payment by ₹ 300 every month. How long will it take for him to clear the loan?

=> first month pay (a) = 400

and per month money difference (d) = 300

∴ Now, Let n month clear the loan

∴ tn = (a + (n -1 )d)

= 400 + (n-1) 300

= 400 + 300n – 300

= 300n + 100

∴ sn = n/2 ( 400 + 300n + 100)

= n /2 (500 + 300n)

∴ Now, n/2 (500 + 300n) = 65,000

500n + 300n2 = 1300, 000

3n2 + 5n – 1300 = 0

3n2 + 65n – 60n – 1300 = 0

n (3n + 65) – 20 (3n + 65) = 0

(3n + 65) (n – 20) = 0

3n + 65 = 0   n -20 = 0

3n = -65          n = 20 .

n ≠ -65/3

(month don’t negative)

∴ 20 month loan clears.

(10) A brick staircase  has a total of 30 steps . The bottom step requires 100 bricks such successive

step requires two bricks less than the previous step .

(i) How many bricks are required for the top most step.

(ii) How many bricks are required to build the stair case ?

=> Given that total steps (n) = 30

first step requires bricks (a) = 100

∴ every bricks difference (d) = -2

(i) ∴ top most step = t30

∴ t30 = a + (30 -1)d

= 100 + 29 × (-2) [∵ a=100, d = -2]

= 100 – 58

= 42

Thus, numbers of bricks required for the top most step are 42 .

(ii) Total bricks are required to build the stair case = sn.

sn = 30/2 (100 + 42) [a = 100, L = 82]

= 15 × 142

= 2130

Thus numbers of required of to build the stair case are 2130 .

(11) If s1 , s2 , s3, ….., sm are the sums of n terms of m A.P ‘s whose first terms are 1, 2, 3,…., (2m -1) respectively , then show that s1 + s2 + s3 + … + sm = ½ m2 (mn + 1) .

=> Now, given that

a1 = 1 , a2 = 2 , a3 = 3 …. am = m

and d1 = 1, d2= 3, d3 = 5 , …. dm = 2m -1

∴ we know that sn = n/2 (2a + (n-1)d)

∴ s1 = n/2 (2.1 + (n-1).1) [∵ a1 = 1, d1 = 1]

s1 = n/2 (2 + n -1)

= n/2(1 + n) = n/2 (n+1)

similarly , we get,

s2 = n /2 (3n + 1)

s3 = n/2 (5n + 1)

sm = n/2 (n(2m-1)+1)

∴ Now, s1 + s2 + s3 + ….. + sm

= n/2 (n+1) + n/2 (3n+ 1) + n/2 (5n + 1) + …. + n/2 (n(2m-1)+1)

= n/2 (n + 1+ 3n + 1+ 5n + 1 + …. + n ( 2m -1) +1 )

= n/2 [(n + 3n + 5n + ….. + n (2m -1)] + m.1] [∵ total m terms]

= n/2 [ n { 1+3 + 5 + …. + (2m -1 )} + m]

= n/2 [{n × m/2(1 + 2m -1)} + m] [∵ terms of m sm = m/2(a+L)]

= n/2[n × m × m × m]

= mn/2 [mn + 1] [proved]

(12) Find the sum [a-b/a+b + 3a – 2b/a+b + 5a -3b/a+b + …. to 12 terms]

=> ∴ first term (a1) = a-b/a+b

∴ common difference (d) = 3a – 2b/ a+b – a-b/a+b

= 3a – 2b – a + b/a+ b

= 2a – b / a+b

∴ 12 terms = t12

t12 = a1 + 11d [ tn = a1 + (n-1) d]

= a-b/a+b + 11 × 2a-b/a+b [∵ a = a-b/a+b ∴ d = 2a –b / a+b]

= a-b + 22a – 11b/ a + b

= 23a – 12b/ a + b

∴ we know that sn = n/2 (a1 + L)

∴ s12 = 12/2 (a1 + L)

= 6 (a-b / a+b +23a – 12b/ a+b [∵a1 = a-b / a+b ∴ L = 23a -12b / a+b

= 6 × a – b + 23a – 12b/a + b

= 6 × 24a – 13b/a + b

∴ required sum = 6 × 24a – 13b / a + b

#### Exercise – 2.7

(1) Which of the following sequences are in G.P ?

(i) 3, 9, 27, 81, ….

(ii) 4, 44, 444, 4444, …..

(iii) 0.5, 0.05, 0.005 , …..

(iv) 1/3, 1/6 , 1/12, …..

(v) 1, -5 , 25, -125 ,…..

(vi) 120, 60, 30, 18, ….

(vii) 16, 4, 1, ¼, ……

=>  (i) 3, 9, 27, 81,…..

∴ a1 = 3 , a2 = 9 , a3 = 27 , a4 = 81 ,……

∴ Now, a2/a1 = 9/3 = 3

a3/a2 = 27/9 = 3

a4/a3 = 81/27 = 3

∴ The ratio between  successive terms are equal .

the sequence 3, 9, 27, 81 , in G.P.

(ii) 4, 44, 444, 4444,…

Now, t1= 4 , t2= 44 , t3 = 444 , t4 = 4444

∴ t2/t1 = 44/4 = 11

t3/t2 = 444/44 = 10.09

t4/t3 = 4444/444 = 10.009 .

∴ The ratio between successive terms are not equal .

The sequence 4, 44, 444, 4444,….. is not G.P.

(iii) 0.5, 0.05, 0.005,….

t1= 0.5 , t2 = 0.05, t3= 0.005

∴ t2/t1 = 0.05× 10/0.5 ×100 = 1/10

t3/t2 = 0,005× 1/0.05 × 10 = 1/10

∴ The ratio between successive terms are equal .

∴ The sequence 0.5, 0.05, 0.005 is G.P.

(iv) 1/3 , 1/6 , 1/12 , ….

t1=1/3 , t2 = 1/6 , t3 = 1/12

∴ t2/t1 = 3/6 = ½

∴ t3/ t2 = 6/12 = ½

∴ The ratio between successive terms are equal.

∴ The sequence 1/3 , 1/6 , 1/12 , ….. in G.P.

(v) 1, -5 , 25 , -125 ,….

t1= 1 , t2 = -5 , t3 = 25 , t4 = -125

∴ t2/t1 = -5/1 = -5

t3/t2 = 25/-5 = -5

t4 / t3 = -125/25 = -5

∴ The ratio between successive  terms are equal .

∴ The sequence 1 , -5, 25, -125 , ….. in G. P .

(vi) 120 , 60, 30, 18, ……

t1 = 120 , t2 = 60 , t3 = 30 , t4 = 18

t2/ t1 = 60/120 = ½

t3 / t2 = 30/60 = ½

t4 / t3 = 18 /30 = 3/5

∴ The ratio between successive terms are not equal .

∴ The sequence 120, 60, 30, 18 ,….. in not G.P .

(vii) 16, 4, 1, ¼, …..

t1 = 16 , t2 = 4 , t3 = 1 , t4 = ¼

∴ t2 / t1 = 4/16 = ¼

∴ t3 / t2 = ¼ = ¼

∴ t4 / t3 = ¼ /1 = ¼

∴ The ratio between successive terms are equal.

∴ The sequence 16 , 4, 1, 1/4 ,…… in G . P .

(2) Write the first three terms of the G.P.  whose first term and the common ratio are given below .

(i) a = 6 , r = 3 ,

(ii) a = √2 , r = √2 ,

(iii) a = 1000 , r = 2/5

=> (i) Given that

a = 6 , r = 3 .

We know that nth term of a G .P.  is

tn = arn-1

if n = 1 , then t1 = ar1-1 = 6.3° = 6

if n = 2, then t2 = ar2-1 = ar = 6 × 3 = 18

if n = 3 , then t3 = ar3-1 = ar2 = 6 × 9 = 54

∴ required the sequence is 6, 18 , 54 , ….

(ii) Given that a = √2 , r = √2

we know that nth term of a G . P. is

tn = arn-1

if n = 1 then t1 = ar1-1 = ar°  = √2 (√2) = √2

if n =2 then t2 = ar2-1 = ar = √2 × √2 = 2

if n = 3 then t3 = ar3-1 = ar2 =  √2 × (√2)2 = 2 √2

∴ required the sequence is √2 , 2 , 2 √2, ……

(iii) Given that

a = 1000, r = 2/5

we know that nth term of a G.P. is

tn = arn-1

if n = 1, then t1 = ar1-1 = ar° = 1000 × (2/5) ° = 1000

if n = 2 , then  t2 = ar2-1 = ar = 1000 × 2/5

= 200 × 2

= 400

∴ if n = 3, then t3 = ar3-1 = ar2

= 1000 × (2/5)2

= 1000 × 4/24

= 40 × 4

= 160

∴ required the sequence is

1000, 400 , 160 , …..

(3) In a G .P.  729 , 243, 81, ….. find t7 .

=> Given sequence 729, 243, 81 , ….

first term (a) = 729

∴ common ratio (r) = 243 /729 = 1/3

∴ we know that nth term of a G . P is

tn = arn-1

if n = 7 , then t7 = ar7-1 = ar6

= 729 × (1/3)6

= 729  × 1 / 729

= 1

∴ required t7 = 1.

(4) Find x so that x + 6 , x +12 and x + 15 are consecutive terms of a geometric progression .

=> Given that

t1 = x +6, t2 = x+12 , t3 = x + 15

Now, given sequence in G.P.

So, t2/t1 = t3/ t2

x + 12 /x + 6 = x+ 15 / x + 12

(x + 12)2 = (x + 6) (x + 15)

x2 24x + 144 = x2 + 6x + 15x + 90

24x + 144 = 21x + 90

3x = 90 -144

x =  – 54 /3

x = -18

∴ x = -18.

(5) Find the number of terms in the following G .P.

(i) 4, 8, 16, ….., 8192 ?

(ii) 1/3 , 1/9 , 1/27 , …., 1/ 2187 .

=> (i) 4 , 8 , 16 , ….., 8192 .

∴ a = 4 , r = 8/4 = 2

∴ Let total terms of this sequence is n .

∴ tn = arn-1 [∵ given sequence in G.P. ]

Now, 8192 = 4 × (2)n-1

(2)n-1 = 2048

(2)n-1 = (2)11

n -1 = 11

n = 11+ 1 = 12

∴ This sequence total term is 12 .

(ii) 1/3 , 1/9 , 1/27 , … 1/2187 .

∴ a = 1/3 , r = 1/9 / 1/3 = 3/9 = 1/3

∴ Let total terms of given sequence is n .

∴ tn = arn-1 [∵ given sequence in G . P ]

Now, 1 / 2187 = 1/3 × (1/3)n-1

1/2187 = (1/3)n

(1/3)n = (1/3)7

n = 7

∴ This sequence total terms is 7 .

(6) In a G.P. the 9th term is 32805 and 6th term is 1215 . Find the 12th term .

=> Let first term = a and common ratio = r

∴ Now , t9 = 32805 and t6 = 1215

ar8 = 32805 …. (i) and ar5 = 1215 …. (ii)

Now, (i)/(ii) , we get

ar8/ar5 = 32805/1215 = 27

r3 = 27

r = 3

from (ii)  equating putting r = 3 , we get .

ar5 = 1215

a = 1215/r5 = 1215/3 × 3 × 3 × 3 × 3

a = 1215 / 243 = 5

∴ Now, a = 5 , and r = 3

∴ t12 = ar11 = 5 × (3)11

= 5 × 177147

= 885735 .

∴ required t12 = 885735 .

(7) Find the 10th term of a G.P. whose 8th term is 768 and the common ratio is 2 .

=> Let first term = a and r = 2

given 8th term

∴ t8 = 768

ar7 = 768

a × (2)7 = 768 [∵ r = 2]

a = 768 / 128 = 6

∴ 10th term of this sequence is

t10 = ar9 = 6 × (2)9

= 6 × 512

= 3072

∴ required term is 3072 .

(8) If a , b, c, are in A . P. then show that 3a, 3b , 3c are in G.P.

=> Given that a , b, c are in A.P.

∴ b – a = c – b ….(i)

Now, Let t1 = 3a , t2 = 3b , t3 = 3c

∴ t2/t1 = 3b/3a = 3b-a

∴ t3 / t2 = 3c / 3b = 3c-b = 3b –a  [ by equation (i) ] .

∴ common ratio are equal .

∴ is a , b, c are in A.P.  then 3a , 3b , 3c are in G.P. [proved]

(9) In a G.P. the product of three consecutive terms is 27 and the sum of the product of two terms taken at a time is 57/2 . Find the three terms.

=> Let the three terms are a/r , a, ar

Now, a/r × a × ar = 27 and a/r × a + a × ar + a/r × ar = 57/2

=> a3 = 27

a = 3 …(i)

a2(1/r + r + 1 ) = 57/2

a2 (r2+r+1/r) = 57/2 ….(ii)

Now, (ii) equation putting a = 3 , we get

(3)2 (r2 + r + 1 / r) = 57/2

18 (r2 + r + 1 / r) = 57

18r2 + 18r + 18 = 57r

18r2 – 39r + 18 = 0

6r2 – 13r  + 6 = 0

6r2 – 9r – 4r + 6 = 0

3r (2r – 3) -2 (2r -3 ) = 0

(2r – 3) (3r -2) = 0

2r – 3 = 0

r = 3/2

3r -2 = 0

r = 2/3

∴ a = 3 and r = 3/2 or 2/3

when a = 3 and r = 3/2 , then three terms are

2, 3, 9/2

when a = 3 and r = 2/3 , then three terms are

9/2 , 3, 2

Thus required three terms are

2, 3, 9/2 or 9/2, 3, 2 .

(10) A man joined a company as assistant manager . The company gave him a starting salary of ₹ 60,000 and agreed to increase his salary 5% annually . What will be his salary after  5 years ?

=> Given that

starting salary (a) = 60, 000

2nd year salary = 60, 000 + 5/100 × 60,000 [∵ increase his salary 5% annually ]

= 60,000 + 3000

= 63000

∴ every year common ratio (r) = 63000/60000 = 63/60 = 21/20

∴ 5th year salary = t5

t5 = ar5-1 [∵ tn = arn-1]

t5 = ar4

t5 = 60,000 × (21/20)4

= 3 ×  21× 21 × 21 ×21/ 2 × 2 × 2

= 3 × 194481 / 8

= 583443/ 8

= 72930.37

after 5 year salary = 72930.37 + 5/100 × 72930.37

= 72930.37 + 3646.51

= 76576.88

= 76577 (approximately)

∴ Thus after 5 year salary is 76577 .

(11) Sivamani is attending an interview for a job and the company gave two offers to him . offer A : ₹20,000 to start with followed by a guaranteed annual increase of 6% for the first 5 years .

Offer B : ₹ 22,000 to start with followed by a guaranteed annual increase of 3% for the first 5 years.

What is his salary in the 4th year with respect to the offers A and B ?

=>   For A:-

Starting salary (a) = 20,000

2nd year salary = 20,000 + 6/100 × 20000

= 20000 + 1200

= 21200

∴ every year common ratio = 212/200

= 106/100 = 53/50

4th year salary = t4

t4 = ar4-1 [∵ tn = arn-1]

= 20,000 × (53/50)3

= 4 ×   53 × 53 × 53/ 5 × 5

= 4 × 148877 / 25

= 595508 / 25

= 23820.32

For b:-

Starting salary (a) = 22000

2nd year salary = 22000+ 3/1 × 220

= 22000 + 660 = 22660

∴ every year common ratio = 2266/2200 = 103/100

4th year salary = t4

t4 = ar4-1 [∵ tn = arn-1 ]

= 22000 × (103/100)3

= 22 ×  103×103×103 / 1 × 10 × 100

=  24039.99

Therefore 4th year salary for  A is 23820.32 and 4th year salary for B is 24039.99 .

(12) If a, b, c are three consecutive terms of an A.P. and x , y, z are three consecutive terms of a G.P.  then prove that xb-c × yc-a× za-b = 1 .

=> Given that a, b, c are in A.P.

so, b – a = c – b …… (i)

2b = a + c ….(iii)

and x , y, z are in G.P.

so, y/x = z/y

y2 = xz ……(ii)

Now, xb-c × yc-a × za-b

= xa-b × yc-a × za-b [ by equation (i)]

= yc-a × (xz)a-b

= yc-a × (y2)a-b [ by equation (ii) ]

= yc – a + 2a – 2b

= yc – a + 2a – a – c [by equation (iii) ]

= y0 = 1

∴ xb –c × yc – a × za – b  = 1 [proved]

#### Exercise – 2.8

(1) Find the sum of first n terms of the G.P.

(i) 5, -3, 9/5, -27/25 ,….

(ii) 256 , 64, 16 , ….

=> we know that, sum of first n terms of a G.P. is

sn = a(rn -1) / r – 1 . r ≠ 1 [ a = first term , r = common ratio ]

∴ (i) 5, -3, 9/5, -27/25 , ….

first term (a) = 5

common ratio (r) = -3/5 = – 3/5

∴ sn = a(rn-1)/ r -1 , r ≠ 1

= 5 (- 3/5)n -1 / -3/5 -1

= 25/8 [1 – (-3/5)n ]

(ii) 256, 64, 16 , …..

first term (a) = 256

common ratio (r) = 64/256 = ¼

∴ sn = a(rn -1) / r – 1, r ≠ 1

= 256 ((1/4)n – 1) / ¼ -1

= 1024 / 3 [1 – (1/4)n]

(2) Find the sum of first six terms of the G.P. 5, 15, 45, ……

=> Given sequence

5, 15, 45, …..

∴ first term (a) = 5

∴ common ratio (r) = 15/5 = 3

∴ we know that sn = a(rn – 1) / r -1  r ≠ 1 .

if n = 6 , then s6 = a(r6 – 1) / r -1  r ≠ 1 .

= 5  × (36 -1) / 3-1

= 5  × (729 -1) / 2

= 5  × 728 / 2

= 5  × 364

= 1820

This required sum is 1820 .

(3) Find the first term of the G.P. whose common ratio 5 and whose sum to first 6 terms is 46872.

=> Given that

r = 5

and s6 = 46872

a(r6-1) / r – 1 = 46872 [∵ sn = a(rn – 1) / r-1 a = first term]

a (56 -1 )/ 5 – 1 = 46872

a = 46872 × 4 / 15624

a = 12

∴ Therefore first term is 12 .

(4) Find the sum to infinity of

(i) 9 + 3 + 1 + …..

(ii) 21 + 14 + 28/3 + ……..

=> we know that ,

The sum of infinite terms of a G.P. is given by

s = a / 1- r , -1 < r < 1 .

(i)  9 + 3 + 1 + …..

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

∴ s =  a / 1- r = 9 / 1 – 1/3

= 9 / 2/3

= 27/2

∴ required sum = 27/2

(ii) 21 + 14 + 28/3 + …..

∴ a = 21, r = 14/21 = 2/3

s = a / 1- r = 21 / 1 – 2/3 = 63

∴ required sum = 63 .

(5) If the first term of an infinite G .P. is 8 and its sum to infinity is 32/3 then find the common ratio .

=> Let common ratio = r.

∴ Given that

a = 8 and s= 32/3

∴ s = 32/3

a / 1-r = 32 /3 [ s = a / 1-r , a<r<1]

8 / 1-r = 32/3

1/1-r = 4/3

4 – 4r = 3

4r = 4 -3 = 1

r = ¼

∴ common ratio (r) = 1/4 .

(6) Find the sum to n terms of the series

(i) 0.4 + 0.44 + 0.444 + …… to n terms .

(ii) 3 + 33 + 333 + ….. to n terms .

=>(i) 0.4 + 0.44 + 0. 444 + …….. to n terms .

sn = 4/10 + 44/100 + 444/ 1000 + ….. n terms

sn = 4/9 [ 9/10 + 99/100 + 999/1000 + ….. n terms ]

sn = 4/9 [ 1- 1/10 + 1- 1/100 + 1- 1/1000 + …. n terms ]

sn = 4/9 [ (1 + 1+…..n terms ) – (1/10 + 1/100+ 1/1000 + …. n terms )]

sn = 4/9 [ n – (1/10 + 1/100 + 1/1000+ ….. n terms )] …….(i)

Now, 1/10 + 1/100+ 1/1000+ …… n terms .

∴ a = 1/10 and r = 1/10

∴ sn1 = 1/10 + 1/100+ 1/1000+ ….. n terms

sn1 = a(rn-1)/(r-1) [n1= n]

sn1 = 1/10((1/10)n -1)/ 1/10 -1

= 1/10((1/10)n -1) / -9/10

= 1/9(1-(1/10)n)

from equation (i) we get

sn = 4/9 [ n -1/9 (1-(1/10)n]

sn = 4n/9 – 4/81 (1-(1/10)n)

Thus the required sum is 4n/9 – 4/81 (1-(1/10)n) .

(ii) 3 + 33 + 333 + ….. to n terms .

sn = 3 (1+ 11 + 111+ ……. n terms )

sn = 3/9 ( 9 + 99 + 999 + …. n terms)

sn = 1/3 (10 -1 + 100 -1 + 1000 -1 + …. n terms)

sn = 1/3 [ (10 + 100 + 1000 + …. n terms) – (1+ 1+ 1+ …… n terms )]

sn = 1/3 [(10 + 100 + 1000+ ….. n terms ) – n]

Now , 10 + 100 + 1000 + …… n terms is series is G .P.

so, a =10 , t = 10

∴ sn = 10(10n -1)/ 10 – 1

= 10/9(10n -1)

∴ sn = 1/3 ]10/9 (10n -1 ) –n]

sn = 10/27 (10n -1) – n/3

∴ required sum is 10/27 (10n -1 ) – n/3 .

(7) Find the sum of the geometric series .

3 + 6 + 12+ …. + 1536.

=> Let given series n terms .

∴ tn = 1536

arn-1 = 1536

3rn-1 = 1536 [∵ a = 3 ]

r (2)n-1 = 512 [∵ r = 6/3 = 2 ]

(2)n-1 = (2)9

n-1 = 9

n = 10

∴ sn = a(rn– 1) / r – 1

if n = 10 , then

s10  = a (r10 – 1) / r -1

= 3(210 – 1) / 2- 1

= 3 × (1024 – 1)

= 3 × 1023

= 3069

∴ Thus, required sum is 3069 .

(8) Kumar writes a letter to four of his friends . He asks each one of them to copy the letter and mail to four different persons with the instruction that they continue the process similarly. Assuming that the process is unaltered and it costs 2 to mail one letter . Find the amount spent on postage when 8th set of letters is mailed .

=>   ∴ first term = 4

2nd term = 16

3rd term = 64

∴ a = 4 and r = 16/4 = 4

∴ The last term = arn-1 =  4(4)8-1

= 4(4)7

Now, s8 = a(rn-1) / r – 1

= 4 (48 – 1) / 4-1

= 4(65536 -1)/3

= 4 × 65535 / 3

= 87380

Therefore amount spent to mail the letter = 2 × 87380 = 174760

(9) Find the rational form of the number 0. 123.

=> Now, 0.123

= 0.123123123

= 0.123 + 0. 000123 + 1.000000123 + ….

Now, This is an infinite geometric series.

∴ a = 0.123 and r = 0.000123 × 1000 / . 123 × 1000000

= 0.001

we know that

The sum of infinite terms of a G.P. is given by

s = a/ 1-r, -1 < r < 1

s = 0.123/ 1- 0.001

= 0.123/ 0.999

= 41 / 333

∴ s = 41/333

∴ 0.123 is rational form is 41/333

(10) If sn = (x + y) + (x2+ xy + y2) + (x3 + x2y + xy2+ y3) + ……. n terms then prove that

(x-y) sn = [ x2(xn-1)/x-1 – y2(yn-1)/y – 1 ]

=> Given that

sn = (x + y) + (x2+ xy + y2) + (x3 + x2y + xy2+ y3) + ……. both side multiplying  on ‘x’, we get

x sn = (x + y ) x + (x2 + xy + y2 ) x + (x3 + x2y + xy2 + y3)x + ……

x sn= x2 + xy + x3 + x2y + xy2 + x4 + x3y + x2y2+ xy3 + ….. .(i)

similarly both side multiplying  on ‘y’ we get .

y sn = xy + y2 + x2y + xy2 + y3 + x3y + x2y2 + xy3 + y4 + ……. (ii)

Now, (i) – (ii) , we get ,

x sn – y sn = (x2 + xy + x3 + x2y + xy2 + x4 + x3y + x2y2 + xy3 + ……) – (xy + y2 + x2y + xy2 + y3 + x3y + x2y2 + xy3 + y4 + ……)

(x – y) sn = (x2 + x3 + x4 + ….. ) n terms – (y2+ y3 + y4 + …..) n terms

(x – y) sn = x2 (xn-1)/ x-1 – y2(yn-1) / y – 1 [∵ sn = a(rn-1) / r -1 ]

(x – y) sn = [ x2(xn-1)/x-1 – y2(yn – 1)/ y-1 ]  [proved] .

#### Exercise – 2.9

(1) Find the sum of the following series

(i) 1 + 2+ 3 + …… + 60

(ii) 3 + 6 + 9 + …… + 96

(iii) 51 + 52 + 53 + ……. + 92

(iv) 1 + 4 + 9 + 16 + …… + 225

(v) 62 + 72 + 82 + ….. + 212

(vi) 103 + 113 + 123 + …….. + 203

(vii) 1 + 3 + 5 + ….. + 71

=> (i) 1 + 2 + 3 + …… + 60

we know that sum of first n natural numbers is

sn = n (n+1) / 2

Now, s60 = 60(60 + 1) / 2

= 30 × 61

= 1830

∴ Therefore the required sum is 1830 .

(ii)  3 + 6 + 9 + ……. + 96

= 3 ( 1+ 2 + 3 + …….. + 32)

∴ sum = 3 × 32(32 + 1) / 2 [∵ sn = n(n+1)/2 n = natural number ]

= 3 × 16 × 33

= 99 × 16

= 1584

∴ Thus the required sum is 1584

(iii) 51 + 52 + 53 + ……. + 92

= (1 + 2 + 3 + …. 92) – ( 1+ 2 + 3 + …… + 50)

=  92 (92 + 1) / 2 – 50(50 + 1)/ 2 [∵n = natural number ∴ sn = n(n+1) / 2 ]

= 42 × 93 / 2 – 50 × 51 / 2

= 46 × 93 – 25 × 51

= 4278 – 1275

= 3003

∴ Thus 51 + 52 + 53 + …… + 92 = 3003

(iv) 1 + 4 + 9 + 16+ …… + 225

= 12 + 22 + 32 + 42 + ….. + 152

we know that

sum of square of first n natural numbers is

n(n+1) (2n + 1) / 6 .

∴ 12 + 22 + 32 + 42 + …….. + 152

= 15(15+1) (2 × 15 +1) / 6

= 5 × 8 × 31

= 40 × 31

= 1240  ∴ Thus required sum is 1240 .

(v) 62 + 72 + 82 + ……. + 212

= (12 + 22 + 32 + …… + 212) – (12 + 22 + 32 + 42+ 52)

= 21(21+1) (2 × 21 + 1) /6 – 5(5+1)(2× 5 + 1)/ 6

= 7 × 11 × 43 – 5 × 11

= 11 (301 -5)

= 11 × 296

= 3256

∴ 62 + 72 + 82 + ……. 212 = 3256.

(vi) 103 + 113 + 123 + …… + 203

= (13 + 23 + 33 + …. + 202 ) – (13 + 23 + 33 + ….. 93)

we know that sum of cube first n natural numbers is (n(n+1)/2)2 .

= (13 + 23 + 33 + ……. + 203 ) – (13 + 23 + 33 + ….+ 93)

= (20(20+1)/2)2 – (9(9+1)/2)2

= (10 × 21)2 – (9× 5)2

= (210)2 – (45)2

= 44100 – 2025

= 42075 .

∴ Thus 103 + 113 + 123 + …… + 203 = 42078 .

(vii) 1 + 3 + 5 + ….. + 71

and given the series total terms

n = (71 – 1)/2 + 1 [∵ a = 1 , L = 71 and d = 5 – 3 = 2 ]

= 70/2 + 1

= 35 +1

= 36

1 + 3 + 5 + ……. + 71

we know that sum of first n odd natural numbers is = n2

1 + 3 + 5 + ……. + 71

= (36)2 [∵ n = 36]

= 1296 .

∴ Therefore, 1 + 3 + 5 + …. 71 = 1296 .

(2) If 1 + 2 + 3 + ….. + k = 325 , then find

13 + 23 + 33 + ……. + k3 .

=> Now, 1 + 2 + 3 + … + k = 325

∴ k (k + 1) / 2 = 325 [∵ sum of first n natural number is n (n+1)/ 2 ]

∴ k(k+1) / 2 = 325 …….. (i)

∴ 13 + 23 + 33 + ……. + k3

= (k (k+1)/2)2 [∵ sum of cube first n natural number = (n(n+1)/2)2 ]

= (325)2 [ by equation (i) ]

= 105625 .

∴ Therefore, 13 + 23 + 33 + …… + k3 = 105625

when 1+ 2 + 3 + ……. + k = 325 .

(3) If 13+ 23 + 33 + …… + k3 = 44100 then find 1+ 2+ 3 + …. + k .

= Now, 13 + 23 + 33 + ….. + k3 = 44100

(k(k+1)/2)2 = 44100 [∵ sum of cube first n natural no = (n (n+1) /2)2 ]

[ k(k+1) /2]2 = (210)2

∴ k(k+1) / 2 = 210 … (i)

∴ 1 + 2 + 3 + ….. + k

= k(k+1) /2 [∵ sum of first n natural numbers is = n(n+ 1) / 2 ]

= 210 [ by equation (i) ]

∴ 1 + 2 + 3 + ……. + k = 210 when 13 + 23 + 33 + ……. + k3 = 44100

(4) How many terms of the series 13 + 23 + 33 + …… should be taken to get the sum 14400?

=> Let n terms taken to get the sum 14400 .

∴ 13 + 23 + 33 + ….. + n3 = 14400

(n(n+1)/2)2 = 14400 [∵ sum of cube first n natural numbers is = (n (n + 1) /2)2 ]

(n (n+1)/2)2 = (120)2

n(n + 1)/2 = 120

n2 + n = 240

n2 + n – 240 = 0

n2 + 16n -15n – 240 = 0

n (n + 16) – 15 (n + 16) = 0

(n + 16 ) (n -15) = 0

n + 16 =0

n ≠ -16 [ terms is not negative]

n – 15 = 0

n = 15 .

∴ Thus, the number of terms taken is 15 .

(5) The sum of the cubes of the first n natural numbers is 2025 , then find the value of n .

=> we know that

The sum of the cubes of the first n natural numbers is = (n(n + 1)/2)2

∴ (n(n+1)/2)2 = 2025

(n(n+1) / 2)2 = (45)2

n(n+1) = 90

n2 + n – 90 = 0

n2 + 10n – 9n – 90 = 0

n (n+10) -9 (n+10) = 0

(n + 10) (n -9) = 0

n + 10 = 0

n ≠ -10 (term is not negative )

n – 9 = 0

n = 9

Thus n = 9 .

(6) Rekha has 15 square colour papers of sizes 10cm , 11cm , 12cm , …… , 24cm . How much area can be decorated with these colour papers ?

=> Area of colour paper decorated

= 102 + 112 + 122 + …… + 242

= (12 + 22 + 32 + ……. + 242 ) – (12 + 22 + …… 92)

= 24 (24 + 1) (2 × 24 + 1) / 6  – 9 (9 + 1) (2 × 9 +1) / 6 [∵ sum of square first n natural number is = n (n + 1) (2n + 1)/ 6 ]

= 4 × 25 × 49 – 3 × 5 × 19  / 2

= 4900 – 285

= 4615 .

∴ Area can be decorated with colour papers is 4615 cm2 .

(7) Find the sum of the series

(23 – 13) + (43 – 33) + (63 – 53) + ….. to

(i) n terms

(ii) 8 terms .

=> (23 -13) + (43 – 33) + (63 – 53) + ……

= (23 + 43 + 63 + ……) – (13 + 33 + 53 + ….. )

= 23 (13 + 23 + 33 + …. ) – (13 + 33 + 53 + ….)

(i) n terms .

∴ 23 (13 + 23 + 33 +  ….. + n3) – (13 + 33 + 53 + …. + (2n -1)3)

= 23(n(n+1)/2)2 – [{n(2n+1)}2 -2 {n(n+1}2]

= 8 (n(n+1)/2)2 – [ n2(2n+1)2 -2 n2(n+1)2]

= [ 2 n2 (n+1)2 – n2 (2n + 1)2 + 2n2 (n +1 )2]

= 4n2 (n+1)2 – n2 (2n+1)2

= n2 [ 4n2 + 8n + 4 – 4n2 – 4n -1 ]

= n2 [ 4n + 3 ]

= 4n3 + 3n2

(ii) 8 terms

when , n = 8 then ,

= 4(8)3 + 3.(8)2

= 4 × 512 + 3 × 64

= 2948 + 192

= 2240

Samacheer Kalvi 10th Maths Solutions Chapter 2
Updated: December 6, 2021 — 4:23 pm