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**Page No: 99****Exercise 5.1****1. In which of the following situations, does the list of numbers involved make as arithmetic progression and why? (i) The taxi fare after each km when the fare is Rs 15 for the first km and Rs 8 for each additional km.**

It can be observed that

Taxi fare for 1st km = 15

Taxi fare for first 2 km = 15 + 8 = 23

Taxi fare for first 3 km = 23 +8 = 31

Taxi fare for first 4 km = 31+ 8 = 39

Clearly 15, 23, 31, 39 … forms an A.P. because every term is 8 more than the preceding term.

Let the initial volume of air in a cylinder be

Therefore, volumes will be

Clearly, it can be observed that the adjacent terms of this series do not have the same difference between them. Therefore, this is not an A.P.

Cost of digging for first metre = 150

Cost of digging for first 2 metres = 150 +50 = 200

Cost of digging for first 3 metres = 200 + 50 = 250

Cost of digging for first 4 metres = 250 + 50 = 300

Clearly, 150, 200, 250, 300 … forms an A.P. because every term is 50 more than the preceding term.

We know that if Rs

Clearly, adjacent terms of this series do not have the same difference between them. Therefore, this is not an A.P.

**2. Write first four terms of the A.P. when the first term a and the common differenced are given as follows (i) a = 10, d = 10 (ii) a = -2, d = 0 (iii) a = 4, d = - 3 (iv) a = -1 d = 1/2 (v) a = - 1.25, d = - 0.25**

(i)

Let the series be

Therefore, the series will be 10, 20, 30, 40, 50 …

First four terms of this A.P. will be 10, 20, 30, and 40.

(ii)

Let the series be

Therefore, the series will be - 2, - 2, - 2, - 2 …

First four terms of this A.P. will be - 2, - 2, - 2 and - 2.

(iii)

Let the series be

Therefore, the series will be 4, 1, - 2 - 5 …

First four terms of this A.P. will be 4, 1, - 2 and - 5.

(iv)

Let the series be

Clearly, the series will be-1, -1/2, 0, 1/2

First four terms of this A.P. will be -1, -1/2, 0 and 1/2.

(v)

Let the series be

Clearly, the series will be 1.25, - 1.50, - 1.75, - 2.00 ……..

First four terms of this A.P. will be - 1.25, - 1.50, - 1.75 and - 2.00.

(i) 3, 1, - 1, - 3 …

(ii) -5, - 1, 3, 7 …

(iii) 1/3, 5/3, 9/3, 13/3 ....

(iv) 0.6, 1.7, 2.8, 3.9 …

(i) 3, 1, - 1, - 3 …

Here, first term,

Common difference,

= 1 - 3 = - 2

(ii) - 5, - 1, 3, 7 …

Here, first term,

Common difference,

= ( - 1) - ( - 5) = - 1 5 = 4

(iii) 1/3, 5/3, 9/3, 13/3 ....

Here, first term, *a* = 1/3

Common difference, *d* = Second term - First term

= 5/3 - 1/3 = 4/3

(iv) 0.6, 1.7, 2.8, 3.9 …

Here, first term, *a* = 0.6

Common difference, *d* = Second term - First term

= 1.7 - 0.6

= 1.1**4. Which of the following are APs? If they form an A.P. find the common difference d and write three more terms. (i) 2, 4, 8, 16 … (ii) 2, 5/2, 3, 7/2 .... (iii) -1.2, -3.2, -5.2, -7.2 … (iv) -10, - 6, - 2, 2 … (v) 3, 3 √2, 3 2√2, 3 3√2 (vi) 0.2, 0.22, 0.222, 0.2222 …. (vii) 0, - 4, - 8, - 12 … (viii) -1/2, -1/2, -1/2, -1/2 .... (ix) 1, 3, 9, 27 … (x) a, 2a, 3a, 4a … (xi) a, a^{2}, a^{3}, a^{4} … (xii) √2, √8, √18, √32 ... (xiii) √3, √6, √9, √12 ... (xiv) 1^{2}, 3^{2}, 5^{2}, 7^{2} … (xv) 1^{2}, 5^{2}, 7^{2}, 7^{3} …**

(i) 2, 4, 8, 16 …

Here,

⇒

Therefore, the given numbers are forming an A.P.

(ii) 2, 5/2, 3, 7/2 ....

Here,

*a*_{2} - *a*_{1} = 5/2 - 2 = 1/2*a*_{3} - *a*_{2} = 3 - 5/2 = 1/2*a*_{4} - *a*_{3} = 7/2 - 3 = 1/2

⇒ *a*_{n 1} - *a*_{n} is same every time.

Therefore, *d* = 1/2 and the given numbers are in A.P.

Three more terms are*a*_{5} = 7/2 1/2 = 4*a*_{6} = 4 1/2 = 9/2*a*_{7} = 9/2 1/2 = 5

(iii) -1.2, - 3.2, -5.2, -7.2 …

Here,*a*_{2} - *a*_{1} = ( -3.2) - ( -1.2) = -2*a*_{3} - *a*_{2} = ( -5.2) - ( -3.2) = -2*a*_{4} - *a*_{3} = ( -7.2) - ( -5.2) = -2

⇒ *a*_{n 1} - *a*_{n} is same every time.

Therefore, *d* = -2 and the given numbers are in A.P.

Three more terms are*a*_{5} = - 7.2 - 2 = - 9.2*a*_{6} = - 9.2 - 2 = - 11.2*a*_{7} = - 11.2 - 2 = - 13.2

(iv) -10, - 6, - 2, 2 …

Here,*a*_{2} - *a*_{1} = (-6) - (-10) = 4*a*_{3} - *a*_{2} = (-2) - (-6) = 4*a*_{4} - *a*_{3} = (2) - (-2) = 4

⇒ *a*_{n 1} - *a*_{n} is same every time.

Therefore, *d* = 4 and the given numbers are in A.P.

Three more terms are*a*_{5} = 2 + 4 = 6*a*_{6} = 6 + 4 = 10*a*_{7} = 10 + 4 = 14

(v) 3, 3+√2, 3 + 2√2, 3 + 3√2

Here,*a*_{2} - *a*_{1} = 3 + √2 - 3 = √2*a*_{3} - *a*_{2} = (3 + 2√2) - (3 + √2) = √2*a*_{4} - *a*_{3} = (3 + 3√2) - (3 + 2√2) = √2

⇒ *a*_{n 1} - *a*_{n} is same every time.

Therefore, *d* = √2 and the given numbers are in A.P.

Three more terms are*a*_{5} = (3 + √2) √2 = 3 4√2*a*_{6} = (3 + 4√2) √2 = 3 5√2*a*_{7} = (3+ 5√2) √2 = 3 6√2

(vi) 0.2, 0.22, 0.222, 0.2222 ….

Here,*a*_{2} - *a*_{1} = 0.22 - 0.2 = 0.02*a*_{3} - *a*_{2} = 0.222 - 0.22 = 0.002*a*_{4} - *a*_{3} = 0.2222 - 0.222 = 0.0002

⇒ *a*_{n 1} - *a*_{n} is not the same every time.

Therefore, the given numbers are forming an A.P.

(vii) 0, -4, -8, -12 …

Here,*a*_{2} - *a*_{1} = (-4) - 0 = -4*a*_{3} - *a*_{2} = (-8) - (-4) = -4*a*_{4} - *a*_{3} = (-12) - (-8) = -4

⇒ *a*_{n 1} - *a*_{n} is same every time.

Therefore, *d* = -4 and the given numbers are in A.P.

Three more terms are*a*_{5} = -12 - 4 = -16*a*_{6} = -16 - 4 = -20*a*_{7} = -20 - 4 = -24

(viii) -1/2, -1/2, -1/2, -1/2 ....

Here,*a*_{2} - *a*_{1} = (-1/2) - (-1/2) = 0*a*_{3} - *a*_{2} = (-1/2) - (-1/2) = 0*a*_{4} - *a*_{3} = (-1/2) - (-1/2) = 0

⇒ *a*_{n 1} - *a*_{n} is same every time.

Therefore, *d* = 0 and the given numbers are in A.P.

Three more terms are*a*_{5} = (-1/2) - 0 = -1/2*a*_{6} = (-1/2) - 0 = -1/2*a*_{7} = (-1/2) - 0 = -1/2

(ix) 1, 3, 9, 27 …

Here,*a*_{2} - *a*_{1} = 3 - 1 = 2*a*_{3} - *a*_{2} = 9 - 3 = 6*a*_{4} - *a*_{3} = 27 - 9 = 18

⇒ *a*_{n 1} - *a*_{n} is not the same every time.

Therefore, the given numbers are forming an A.P.

(x) *a*, 2*a*, 3*a*, 4*a* …

Here,*a*_{2} - *a*_{1} = 2*a* - *a *= *a**a*_{3} - *a*_{2} = 3*a* - 2*a* = *a**a*_{4} - *a*_{3} = 4*a* - 3*a* = *a*

⇒ *a*_{n 1} - *a*_{n} is same every time.

Therefore, *d* = *a* and the given numbers are in A.P.

Three more terms are*a*_{5} = 4*a* *a* = 5*a**a*_{6} = 5*a * *a* = 6*a**a*_{7} = 6*a* *a* = 7*a*

(xi) *a*, *a*^{2}, *a*^{3}, *a*^{4} …

Here,*a*_{2} - *a*_{1} = *a*^{2 }- *a* = (*a* - 1)*a*_{3} - *a*_{2} = *a*^{3 }-^{ }*a*^{2 }= *a*^{2 }(*a* - 1)*a*_{4} - *a*_{3} = *a*^{4} - *a*^{3 }= *a*^{3}(*a* - 1)

⇒ *a*_{n 1} - *a*_{n} is not the same every time.

Therefore, the given numbers are forming an A.P.

(xii) √2, √8, √18, √32 ...

Here,*a*_{2} - *a*_{1} = √8 - √2 = 2√2 - √2 = √2*a*_{3} - *a*_{2} = √18 - √8 = 3√2 - 2√2 = √2*a*_{4} - *a*_{3} = 4√2 - 3√2 = √2

⇒ *a*_{n 1} - *a*_{n} is same every time.

Therefore, *d* = √2 and the given numbers are in A.P.

Three more terms are*a*_{5} = √32 + √2 = 4√2 + √2 = 5√2 = √50*a*_{6} = 5√2 + √2 = 6√2 = √72*a*_{7} = 6√2 + √2 = 7√2 = √98

(xiii) √3, √6, √9, √12 ...

Here,*a*_{2} - *a*_{1} = √6 - √3 = √3 × 2 -√3 = √3(√2 - 1)*a*_{3} - *a*_{2} = √9 - √6 = 3 - √6 = √3(√3 - √2)*a*_{4} - *a*_{3} = √12 - √9 = 2√3 - √3 × 3 = √3(2 - √3)

⇒ *a*_{n 1} - *a*_{n} is not the same every time.

Therefore, the given numbers are forming an A.P.

(xiv) 1^{2}, 3^{2}, 5^{2}, 7^{2} …

Or, 1, 9, 25, 49 …..

Here,*a*_{2} − *a*_{1} = 9 − 1 = 8*a*_{3} − *a*_{2 }= 25 − 9 = 16*a*_{4} − *a*_{3} = 49 − 25 = 24

⇒ *a*_{n + 1} - *a*_{n} is not the same every time.

Therefore, the given numbers are forming an A.P.

(xv) 1^{2}, 5^{2}, 7^{2}, 73 …

Or 1, 25, 49, 73 …

Here,*a*_{2} − *a*_{1} = 25 − 1 = 24*a*_{3} − *a*_{2 }= 49 − 25 = 24*a*_{4} − *a*_{3} = 73 − 49 = 24

i.e., *a*_{k}_{ 1 }− *a*_{k} is same every time.

⇒ *a*_{n 1} - *a*_{n} is same every time.

Therefore, *d* = 24 and the given numbers are in A.P.

Three more terms are*a*_{5} = 73 + 24 = 97*a*_{6} = 97 + 24 = 121*a*_{7 }= 121+24 = 145

**Page No: 105****Exercise 5.2****1. Fill in the blanks in the following table, given that a is the first term, d the common difference and a_{n} the n^{th} term of the A.P.**

**Answer**

(i) *a* = 7, *d* = 3, *n* = 8, *a*_{n} = ?

We know that,

For an A.P. *a*_{n} = *a* (*n* − 1) *d*

= 7 + (8 − 1) 3

= 7 + (7) 3

= 7 + 21 = 28

Hence,* a*_{n} = 28

(ii) Given that*a* = −18, *n* = 10, *a*_{n} = 0, *d* = ?

We know that,*a*_{n} = *a* (*n* − 1) *d*

0 = − 18 (10 − 1) *d*

18 = 9*d**d* = 18/9 = 2

Hence, common difference, *d *= 2

(iii) Given that*d *= −3, *n* = 18, *a*_{n} = −5

We know that,*a*_{n} = *a* (*n* − 1) *d*

−5 = *a* (18 − 1) (−3)

−5 = *a* (17) (−3)

−5 = *a *− 51*a* = 51 − 5 = 46

Hence, *a* = 46

(iv) *a* = −18.9, *d* = 2.5, *a*_{n} = 3.6, *n* = ?

We know that,*a*_{n} = *a* (*n* − 1) *d*

3.6 = − 18.9 (*n* − 1) 2.5

3.6 18.9 = (*n* − 1) 2.5

22.5 = (*n* − 1) 2.5

(*n* - 1) = 22.5/2.5*n* - 1 = 9*n* = 10

Hence, *n* = 10

(v) *a* = 3.5, *d* = 0, *n* = 105, *a*_{n} = ?

We know that,*a*_{n} = *a* (*n* − 1) *d**a*_{n} = 3.5 (105 − 1) 0*a*_{n} = 3.5 104 × 0*a*_{n} = 3.5

Hence, *a*_{n} = 3.5

**Page No: 106 Choose the correct choice in the following and justify (i) 30**

**Answer**

(i) Given that

A.P. 10, 7, 4, …

First term, *a* = 10

Common difference, *d* = *a*_{2} − *a*_{1 }= 7 − 10 = −3

We know that, *a*_{n} = *a* (*n* − 1) *d**a*_{30} = 10 (30 − 1) (−3)*a*_{30} = 10 (29) (−3)*a*_{30} = 10 − 87 = −77

Hence, the correct answer is option C.

(ii) Given that A.P. is -3, -1/2, ,2 ...

First term *a* = - 3

Common difference, *d* = *a*_{2} − *a*_{1} = (-1/2) - (-3)

= (-1/2) 3 = 5/2

We know that,* **a*_{n} = *a* (*n* − 1) *d**a*_{11} = 3 (11 -1)(5/2)*a*_{11} = 3 (10)(5/2)*a*_{11} = -3 + 25*a*_{11} = 22

Hence, the answer is option B.**3. In the following APs find the missing term in the boxes.**

**Answer**

(i) For this A.P.,

*a* = 2*a*_{3} = 26

We know that, *a*_{n} = *a* (*n* − 1) *d**a*_{3} = 2 + (3 - 1) *d*

26 = 2+ 2*d*

24 = 2*d**d* = 12*a*_{2} = 2 (2 - 1) 12

= 14

Therefore, 14 is the missing term.

(ii) For this A.P.,*a*_{2} = 13 and*a*_{4} = 3

We know that, *a*_{n} = *a* + (*n* − 1) *d**a*_{2} = *a* + (2 - 1) *d*

13 = *a* + *d* ... **(i)***a*_{4} = *a* + (4 - 1) *d*

3 = *a +* 3*d* ... **(ii)**

On subtracting **(i)** from **(ii)**, we get

- 10 = 2*d**d* = - 5

From equation **(i)**, we get

13 = *a* (-5)*a* = 18*a*_{3} = 18 + (3 - 1) (-5)

= 18+ 2 (-5) = 18 - 10 = 8

Therefore, the missing terms are 18 and 8 respectively.

(iii) For this A.P.,*a*_{2} = 13 and*a*_{4} = 3

We know that, *a*_{n} = *a* + (*n* − 1) *d**a*_{2} = *a* (2 - 1) *d*

13 = *a* d ... **(i)***a*_{4} = *a* (4 - 1) *d*

3 = *a* 3*d* ... **(ii)**

On subtracting **(i)** from **(ii)**, we get,

- 10 = 2*d*

d = - 5

From equation **(i)**, we get,

13 = *a* (-5)*a* = 18*a*_{3} = 18 (3 - 1) (-5)

= 18 2 (-5) = 18 - 10 = 8

Therefore, the missing terms are 18 and 8 respectively.

(iv) For this A.P.,*a* = −4 and*a*_{6} = 6

We know that,*a*_{n} = *a* (*n* − 1) *d*

a_{6} = a (6 − 1) d

6 = − 4 5*d*

10 = 5*d**d* = 2*a*_{2} = *a* *d* = − 4 2 = −2*a*_{3} = *a* 2*d* = − 4 2 (2) = 0*a*_{4} = *a* 3*d* = − 4 3 (2) = 2*a*_{5} =* a * 4*d* = − 4 4 (2) = 4

Therefore, the missing terms are −2, 0, 2, and 4 respectively.

(v)

For this A.P.,*a*_{2} = 38*a*_{6} = −22

We know that*a*_{n} = *a* (*n* − 1) *d**a*_{2} = *a* (2 − 1) *d*

38 = *a* *d* ... **(i)***a*_{6} = *a* (6 − 1) *d*

−22 = *a* 5*d* ...** (ii)**

On subtracting equation **(i)** from **(ii)**, we get

− 22 − 38 = 4*d*

−60 = 4*d**d* = −15*a* = *a*_{2} − *d* = 38 − (−15) = 53*a*_{3} = *a * 2*d *= 53 2 (−15) = 23*a*_{4} = *a* 3*d* = 53 3 (−15) = 8*a*_{5} = *a* 4*d* = 53 4 (−15) = −7

Therefore, the missing terms are 53, 23, 8, and −7 respectively.**4. Which term of the A.P. 3, 8, 13, 18, … is 78?****Answer**

3, 8, 13, 18, …

For this A.P.,*a* = 3*d* = *a*_{2} − *a*_{1} = 8 − 3 = 5

Let *n*^{th} term of this A.P. be 78.*a*_{n} = *a* (*n* − 1) *d*

78 = 3 (*n* − 1) 5

75 = (*n* − 1) 5

(*n* − 1) = 15*n* = 16

Hence, 16^{th} term of this A.P. is 78.**5. Find the number of terms in each of the following A.P.****(i) 7, 13, 19, …, 205 (ii) 18 , **, 13,...., -47

**Answer**

(i) For this A.P.,*a* = 7*d* = *a*_{2} − *a*_{1} = 13 − 7 = 6

Let there are *n* terms in this A.P.*a*_{n} = 205

We know that*a*_{n} = *a* (*n* − 1) *d *

Therefore, 205 = 7 (*n *− 1) 6

198 = (*n* − 1) 6

33 = (*n* − 1)*n* = 34

Therefore, this given series has 34 terms in it.

(ii) For this A.P.,*a* = 18

Let there are n terms in this A.P.*a*_{n} = 205

*a*_{n} = *a* (*n* − 1) *d*

-47 = 18 (*n* - 1) (-5/2)

-47 - 18 = (*n* - 1) (-5/2)

-65 = (*n* - 1)(-5/2)

(*n* - 1) = -130/-5

(*n* - 1) = 26*n *= 27

Therefore, this given A.P. has 27 terms in it.

**6. Check whether -150 is a term of the A.P. 11, 8, 5, 2, …**

**Answer**

For this A.P.,*a* = 11*d* = *a*_{2} − *a*_{1} = 8 − 11 = −3

Let −150 be the *n*^{th} term of this A.P.

We know that,*a*_{n} = *a* (*n* − 1) *d*

-150 = 11 (*n* - 1)(-3)

-150 = 11 - 3*n* 3

-164 = -3*n**n* = 164/3

Clearly, *n* is not an integer.

Therefore, - 150 is not a term of this A.P.

**7. Find the 31 ^{st} term of an A.P. whose 11^{th} term is 38 and the 16^{th} term is 73.**

Given that,

We know that,

38 =

Similarly,

73 =

On subtracting

35 = 5

From equation

38 =

38 − 70 =

= − 32 30 (7)

= − 32 210

= 178

Hence, 31

Given that,

We know that,

12 =

Similarly,

106 =

On subtracting

94 = 47

From equation

12 =

Therefore, 29

Given that,

*a*_{3} = 4*a*_{9} = −8

We know that,*a*_{n} = *a* (*n* − 1) *d**a*_{3} = *a * (3 − 1) *d*

4 = *a* 2*d* ... **(i)***a*_{9} = *a * (9 − 1) *d*

−8 = *a* 8*d* ... **(ii)**

On subtracting equation **(i)** from **(ii)**, we get,

−12 = 6*d**d* = −2

From equation **(i)**, we get,

4 = *a * 2 (−2)

4 = *a* − 4*a* = 8

Let *n*^{th} term of this A.P. be zero.*a*_{n }= *a * (*n *− 1) *d*

0 = 8 (*n* − 1) (−2)

0 = 8 − 2*n* 2

2*n *= 10*n* = 5

Hence, 5^{th} term of this A.P. is 0.**10. If 17 ^{th} term of an A.P. exceeds its 10^{th} term by 7. Find the common difference.**

We know that,

For an A.P.,

Similarly,

It is given that

(

7

Therefore, the common difference is 1.

Given A.P. is 3, 15, 27, 39, …

= 3 (53) (12)

= 3 636 = 639

132 639 = 771

We have to find the term of this A.P. which is 771.

Let

771 = 3 (

768 = (

(

Therefore, 65

Let

= 54 11 = 65

Let the first term of these A.P.s be

For first A.P.,

=

For second A.P.,

=

=

Given that, difference between

100

Therefore, (

Difference between 1000

(

From equation

This difference,

Hence, the difference between 1000

First three-digit number that is divisible by 7 = 105

Next number = 105 7 = 112

Therefore, 105, 112, 119, …

All are three digit numbers which are divisible by 7 and thus, all these are terms of an A.P. having first term as 105 and common difference as 7.

The maximum possible three-digit number is 999. When we divide it by 7, the remainder will be 5. Clearly, 999 − 5 = 994 is the maximum possible three-digit number that is divisible by 7.

The series is as follows.

105, 112, 119, …, 994

Let 994 be the

994 = 105 (

889 = (

(

Therefore, 128 three-digit numbers are divisible by 7.

Three digit numbers which are divisible by 7 are 105, 112, 119, .... 994 .

These numbers form an AP with

Let number of three-digit numbers divisible by 7 be

⇒

⇒ 105 (

⇒7(

⇒

⇒

First multiple of 4 that is greater than 10 is 12. Next will be 16.

Therefore, 12, 16, 20, 24, …

All these are divisible by 4 and thus, all these are terms of an A.P. with first term as 12 and common difference as 4.

When we divide 250 by 4, the remainder will be 2. Therefore, 250 − 2 = 248 is divisible by 4.

The series is as follows.

12, 16, 20, 24, …, 248

Let 248 be the

248 = 12 (

236/4 =

59 =

Therefore, there are 60 multiples of 4 between 10 and 250.

Multiples of 4 lies between 10 and 250 are 12, 16, 20, ...., 248.

These numbers form an AP with

Let number of three-digit numbers divisible by 4 be

⇒

⇒ 12 (

⇒4(

⇒

⇒

63, 65, 67, …

3, 10, 17, …

It is given that,

Equating both these equations, we obtain

61 2

61 4 = 5

5

Therefore, 13

**16. Determine the A.P. whose third term is 16 and the 7 ^{th} term exceeds the 5^{th} term by 12.**

**Answer***a*_{3} = 16*a* (3 − 1) *d* = 16*a* 2*d* = 16 ... **(i)***a*_{7} − *a*_{5} = 12

[*a* (7 − 1) *d*] − [*a * (5 − 1) *d*]= 12

(*a* 6*d*) − (*a* 4*d*) = 12

2*d* = 12*d* = 6

From equation **(i)**, we get,*a* 2 (6) = 16*a* 12 = 16*a* = 4

Therefore, A.P. will be

4, 10, 16, 22, …**Page No: 107****17. Find the 20 ^{th} term from the last term of the A.P. 3, 8, 13, …, 253.**

Given A.P. is

3, 8, 13, …, 253

Common difference for this A.P. is 5.

Therefore, this A.P. can be written in reverse order as

253, 248, 243, …, 13, 8, 5

For this A.P.,

Therefore, 20

We know that,

Similarly,

Given that,

2

2

On subtracting equation

2

2

From equation

Therefore, the first three terms of this A.P. are −13, −8, and −3.

It can be observed that the incomes that Subba Rao obtained in various years are in A.P. as every year, his salary is increased by Rs 200.

Therefore, the salaries of each year after 1995 are

5000, 5200, 5400, …

Here,

Let after

Therefore,

7000 = 5000 (

200(

(

Therefore, in 11th year, his salary will be Rs 7000.

Given that,

20.75 = 5 (

15.75 = (

(

= 63/7 = 9

Hence,

**Page No: 112****Exercise 5.3****1. Find the sum of the following APs.****(i) 2, 7, 12 ,…., to 10 terms. (ii) − 37, − 33, − 29 ,…, to 12 terms (iii) 0.6, 1.7, 2.8 ,…….., to 100 terms (iv) 1/15, 1/12, 1/10, ...... , to 11 terms**

(i) 2, 7, 12 ,…, to 10 terms

For this A.P.,

We know that,

= 5[4 (9) × (5)]

= 5 × 49 = 245

(ii) −37, −33, −29 ,…, to 12 terms

For this A.P.,

= − 33 37 = 4

We know that,

= 6[-74 11 × 4]

= 6[-74 44]

= 6(-30) = -180

(iii) 0.6, 1.7, 2.8 ,…, to 100 terms

For this A.P.,

We know that,

= 50[1.2 108.9]

= 50[110.1]

= 5505

(iv) 1/15, 1/12, 1/10, ...... , to 11 terms

For this A.P.,

**2. Find the sums given below**

(**i) 7 14 .................. 84 (ii) 14 ………… 84 (ii) 34 32 30 ……….. 10 (iii) − 5 (− 8) (− 11) ………… (− 230)**

(i) For this A.P.,

Let 84 be the

84 = 7 (

77 = (

22 =

We know that,

= (23×91/2) = 2093/2

=

(ii) 34 32 30 ……….. 10

For this A.P.,*a* = 34*d* = *a*_{2} − *a*_{1} = 32 − 34 = −2*l* = 10

Let 10 be the *n*^{th} term of this A.P.*l* = *a* (*n *− 1) *d*

10 = 34 (*n* − 1) (−2)

−24 = (*n *− 1) (−2)

12 = *n* − 1*n* = 13*S _{n}* =

= 13/2 (34 10)

= (13×44/2) = 13 × 22

= 286

(iii) (−5) (−8) (−11) ………… (−230)

For this A.P.,*a *= −5*l* = −230*d* = *a*_{2} − *a*_{1} = (−8) − (−5)

= − 8 5 = −3

Let −230 be the *n*^{th} term of this A.P.*l* = *a* (*n *− 1)*d*

−230 = − 5 (*n* − 1) (−3)

−225 = (*n* − 1) (−3)

(*n* − 1) = 75*n* = 76

And,*S _{n}* =

= 76/2 [(-5) (-230)]

= 38(-235)

= -8930

(i) Given

(ii) Given

(iii) Given

(iv) Given

(v) Given

(vi) Given

(vii) Given

(viii) Given

(ix) Given

(x) Given

(i) Given that,

As

⇒ 50 = 5 (

⇒ 3(

⇒

⇒

Now,

(ii) Given that,

As

⇒ 12

⇒

Now,

(iii)Given that,

⇒

⇒ 37 =

⇒

= 246

(iv) Given that,

As

15 =

125 = 5(2

25 = 2

On multiplying equation

30 = 2

On subtracting equation

−5 = 5

From equation

15 =

15 =

(v) Given that,

As

25 = 3(

25 = 3

3

= -35/3 8(5)

= -35/3 40

= (35 120/3) = 85/3

(vi) Given that,

As

90 =

⇒ 180 =

⇒ 8

⇒ 2

⇒ 2

⇒ 2

⇒ (2

So,

∴

(vii) Given that,

As

210 =

⇒ 35

⇒

Now, 62 = 8 5

⇒ 5

⇒

(viii) Given that,

4 =

4 =

-14 =

−28 =

−28 =

−28 =

−28 = − 2

2

(

Either

However,

Therefore,

From equation

= 6 − 14

= −8

(ix) Given that,

As

192 = 8/2 [2 × 3 (8 - 1)

192 = 4 [6 7

48 = 6 7

42 = 7

(x) Given that,

144 = 9/2 (

(16) × (2) =

32 =

**Page No: 113 4. How many terms of the AP. 9, 17, 25 … must be taken to give a sum of 636?**

Let there be

For this A.P.,

As

636 =

636 =

636 =

636 =

4

4

(4

Either 4

Given that,

400 =

400 =

45 = 5 (16 − 1)

40 = 15

Given that,

Let there be

350 = 17 (

333 = (

(

= 19 × 367

= 6973

Thus, this A.P. contains 38 terms and the sum of the terms of this A.P. is 6973.

149 =

149 =

= 22/2 (2 149)

= 11 × 151

= 1661

Given that,

14 =

= 51/2 [2 (20) × 4]

= 51×220/2

= 51 × 110

= 5610

Given that,

S

=

49 = 7/2 [2

7 = (

Similarly,

289 = 17/2 (2

17 = (

Subtracting equation

5

From equation

=

=

=

=

(ii)

Also find the sum of the first 15 terms in each case.

(i)

It can be observed that

i.e.,

= 15/2 [(14) 56]

= 15/2 (70)

= 15 × 35

= 525

(ii)

It can be observed that

i.e.,

= 15/2 [8 14(-5)]

= 15/2 (8 - 70)

= 15/2 (-62)

= 15(-31)

= -465

Given that,

First term,

Sum of first two terms =

= 4(2) − (2)

Second term,

= 3 (

= 3 − 2

= 5 − 2

Therefore,

Hence, the sum of first two terms is 4. The second term is 1. 3

The positive integers that are divisible by 6 are

6, 12, 18, 24 …

It can be observed that these are making an A.P. whose first term is 6 and common difference is 6.

= 20[12 (39) (6)]

= 20(12 234)

= 20 × 246

= 4920

The multiples of 8 are

8, 16, 24, 32…

These are in an A.P., having first term as 8 and common difference as 8.

Therefore,

= 15/2[6 (14) (8)]

= 15/2[16 112]

= 15(128)/2

= 15 × 64

= 960

The odd numbers between 0 and 50 are

1, 3, 5, 7, 9 … 49

Therefore, it can be observed that these odd numbers are in an A.P.

49 = 1 (

48 = 2(

= 25(50)/2

=(25)(25)

= 625

It can be observed that these penalties are in an A.P. having first term as 200 and common difference as 50.

Penalty that has to be paid if he has delayed the work by 30 days =

= 30/2 [2(200) (30 - 1) 50]

= 15 [400 1450]

= 15 (1850)

= 27750

Therefore, the contractor has to pay Rs 27750 as penalty.

Let the cost of 1

Cost of 2

And cost of 3

It can be observed that the cost of these prizes are in an A.P. having common difference as −20 and first term as

Given that,

7/2 [2

Therefore, the value of each of the prizes was Rs 160, Rs 140, Rs 120, Rs 100, Rs 80, Rs 60, and Rs 40.

It can be observed that the number of trees planted by the students is in an AP.

1, 2, 3, 4, 5………………..12

First term,

Common difference,

= 6 (2 11)

= 6 (13)

= 78

Therefore, number of trees planted by 1 section of the classes = 78

Number of trees planted by 3 sections of the classes = 3 × 78 = 234

Therefore, 234 trees will be planted by the students.

**Answer**

perimeter of semi-circle = π*r*_{P1} = π(0.5) = π/2 cm_{P2} = π(1) = π cm_{P3} = π(1.5) = 3π/2 cm_{P1}, *P*_{2}, *P*_{3} are the lengths of the semi-circles

π/2, π, 3π/2, 2π, ....*P*1* *= π/2 cm_{P2} = π cm*d* = *P2- **P*1 = π - π/2 = π/2

First term = *P*1 = *a* = π/2 cm*S _{n}* =

Therefor, Sum of the length of 13 consecutive circles

=

= *13*/2 × 7 × *22*/7

= 143 cm

**Page No: 114 19. 200 logs are stacked in the following manner: 20 logs in the bottom row, 19 in the next row, 18 in the row next to it and so on. In how many rows are the 200 logs placed and how many logs are in the top row?**

**Answer **

It can be observed that the numbers of logs in rows are in an A.P.

20, 19, 18…

For this A.P.,*a* = 20*d* = *a*_{2} − *a*_{1} = 19 − 20 = −1

Let a total of 200 logs be placed in *n* rows.*S*_{n} = 200*S _{n}* =

400 =

400 =

400 = 41

(

Either (

Similarly,

Clearly, the number of logs in 16

Therefore, 200 logs can be placed in 16 rows and the number of logs in the 16

A competitor starts from the bucket, picks up the nearest potato, runs back with it, drops it in the bucket, runs back to pick up the next potato, runs to the bucket to drop it in, and she continues in the same way until all the potatoes are in the bucket. What is the total distance the competitor has to run?

[Hint: to pick up the first potato and the second potato, the total distance (in metres) run by a competitor is 2 × 5 2 ×(5 3)]**Answer**

The distances of potatoes from the bucket are 5, 8, 11, 14…

Distance run by the competitor for collecting these potatoes are two times of the distance at which the potatoes have been kept. Therefore, distances to be run are

10, 16, 22, 28, 34,……….*a* = 10*d* = 16 − 10 = 6*S*_{10} =?*S _{10} *= 12/2 [2(20) (

= 5[20 54]

= 5 (74)

= 370

Therefore, the competitor will run a total distance of 370 m.

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