__Exercise 5.1 Page: 99__

__Exercise 5.1 Page: 99__

Q. **1. **( Arithmetic Progression )

**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.**

**Solution:**

We can write the given condition as;

Taxi fare for 1 km = 15

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

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

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

And so on……

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

**(ii) The amount of air present in a cylinder when a vacuum pump removes 1/4 of the air remaining in the cylinder at a time.**

**Solution:**

Let the volume of air in a cylinder, initially, be *V* litres.

In each stroke, the vacuum pump removes 1/4th of air remaining in the cylinder at a time. Or we can say, after every stroke, 1-1/4 = 3/4th part of air will remain.

Therefore, volumes will be *V*, 3*V*/4 , (3*V*/4)^{2} , (3*V*/4)^{3}…and so on

Clearly, we can see here, the adjacent terms of this series do not have the common difference between them. Therefore, this series is not an A.P.

**(iii) The cost of digging a well after every metre of digging, when it costs Rs 150 for the first metre and rises by Rs 50 for each subsequent metre.**

**Solution:**

We can write the given condition as;

Cost of digging a well for first metre = Rs.150

Cost of digging a well for first 2 metres = Rs.150+50 = Rs.200

Cost of digging a well for first 3 metres = Rs.200+50 = Rs.250

Cost of digging a well for first 4 metres =Rs.250+50 = Rs.300

And so on..

Clearly, 150, 200, 250, 300 … forms an A.P. with a common difference of 50 between each term.

**(iv) The amount of money in the account every year, when Rs 10000 is deposited at compound interest at 8% per annum.**

**Solution:**

We know that if Rs. P is deposited at *r*% compound interest per annum for n years, the amount of money will be:

P(1+r/100)^{n}

Therefore, after each year, the amount of money will be;

10000(1+8/100), 10000(1+8/100)^{2}, 10000(1+8/100)^{3}……

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

Q. **2. **( Arithmetic Progression )

**Write first four terms of the A.P. when the first term a and the common difference 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

**Solutions:**

(i) *a* = 10, *d* = 10

Let us consider, the Arithmetic Progression series be *a _{1}, a_{2}, a_{3}, a_{4}, a_{5} …*

*a*_{1} = *a* = 10

*a*_{2} = *a*_{1}+*d* = 10+10 = 20

*a*_{3} = *a*_{2}+*d* = 20+10 = 30

*a*_{4} = *a*_{3}+*d* = 30+10 = 40

*a*_{5} = *a*_{4}+*d* = 40+10 = 50

And so on…

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

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

(ii) *a* = – 2, *d* = 0

Let us consider, the Arithmetic Progression series be *a _{1}, a_{2}, a_{3}, a_{4}, a_{5} …*

*a*_{1} = *a* = -2

*a*_{2} = *a*_{1}+*d* = – 2+0 = – 2

*a*_{3} = *a*_{2}+d = – 2+0 = – 2

*a*_{4} = *a*_{3}+*d* = – 2+0 = – 2

Therefore, the A.P. series will be – 2, – 2, – 2, – 2 …

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

(iii) *a* = 4, *d* = – 3

Let us consider, the Arithmetic Progression series be *a _{1}, a_{2}, a_{3}, a_{4}, a_{5} …*

*a*_{1} = *a* = 4

*a*_{2} = *a*_{1}+*d* = 4-3 = 1

*a*_{3} = *a*_{2}+*d* = 1-3 = – 2

*a*_{4} = *a*_{3}+*d* = -2-3 = – 5

Therefore, the A.P. series will be 4, 1, – 2 – 5 …

And, first four terms of this A.P. will be 4, 1, – 2 and – 5.

(iv) *a* = – 1, *d* = 1/2

Let us consider, the Arithmetic Progression series be *a _{1}, a_{2}, a_{3}, a_{4}, a_{5} …*

*a*_{2} = *a*_{1}+*d* = -1+1/2 = -1/2

*a*_{3} = *a*_{2}+*d* = -1/2+1/2 = 0

*a*_{4} = *a*_{3}+*d* = 0+1/2 = 1/2

Thus, the A.P. series will be-1, -1/2, 0, 1/2

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

(v) *a* = – 1.25, *d* = – 0.25

Let us consider, the Arithmetic Progression series be *a _{1}, a_{2}, a_{3}, a_{4}, a_{5} …*

*a*_{1} = *a* = – 1.25

*a*_{2} = *a*_{1} + *d* = – 1.25-0.25 = – 1.50

*a*_{3} = *a*_{2} + *d* = – 1.50-0.25 = – 1.75

*a*_{4} = *a*_{3} + *d* = – 1.75-0.25 = – 2.00

Therefore, the A.P series will be 1.25, – 1.50, – 1.75, – 2.00 ……..

And first four terms of this A.P. will be – 1.25, – 1.50, – 1.75 and – 2.00.

Q.**3(Arithmetic Progression**)

**For the following A.P.s, write the first term and the common difference.(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 …**

**Solutions**

(i) Given series,

3, 1, – 1, – 3 …

First term, *a* = 3

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

⇒ 1 – 3 = -2

⇒ d = -2

**(ii) Given series, – 5, – 1, 3, 7 …**

First term, *a* = -5

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

⇒ ( – 1)-( – 5) = – 1+5 = 4

**(iii) Given series, 1/3, 5/3, 9/3, 13/3 ….**

First term, *a* = 1/3

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

⇒ 5/3 – 1/3 = 4/3

**(iv) Given series, 0.6, 1.7, 2.8, 3.9 …**

First term, *a* = 0.6

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

⇒ 1.7 – 0.6

⇒ 1.1

Q.**4(Arithmetic Progression**)

**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*, 2

*a*, 3

*a*, 4

*a*…(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}…

**Solution**

(i) Given to us,

2, 4, 8, 16 …

Here, the common difference is;

*a*_{2} – *a*_{1} = 4 – 2 = 2

*a*_{3} – *a*_{2} = 8 – 4 = 4

*a _{4}* –

*a*

_{3}= 16 – 8 = 8

Since, *a _{n}*

_{+1}–

*a*or the common difference is not the same every time.

_{n }Therefore, the given series are not forming an A.P.

**(ii) Given, 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

Since, *a _{n}*

_{+1}–

*a*or the common difference is same every time.

_{n}Therefore, *d* = 1/2 and the given series are in A.P.

The next three terms are;

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

*a*_{6} = 4 +1/2 = 9/2

*a*_{7} = 9/2 +1/2 = 5

**(iii) Given, -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

Since, *a _{n}*

_{+1}–

*a*or common difference is same every time.

_{n}Therefore, *d* = -2 and the given series are in A.P.

Hence, next three terms are;

*a*_{5} = – 7.2-2 = -9.2

*a*_{6} = – 9.2-2 = – 11.2

*a*_{7} = – 11.2-2 = – 13.2

**(iv) Given, -10, – 6, – 2, 2 …**

Here, the terms and their difference are;

*a*_{2} – *a*_{1} = (-6)-(-10) = 4

*a*_{3} – *a*_{2} = (-2)-(-6) = 4

*a*_{4} – *a*_{3} = (2 -(-2) = 4

Since, *a _{n}*

_{+1}–

*a*or the common difference is same every time.

_{n}Therefore, *d* = 4 and the given numbers are in A.P.

Hence, next three terms are;

*a*_{5} = 2+4 = 6

*a*_{6} = 6+4 = 10

*a*_{7} = 10+4 = 14

**(v) Given, 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

Since, *a _{n}*

_{+1}–

*a*or the common difference is same every time.

_{n}Therefore, *d* = √2 and the given series forms a A.P.

Hence, next three 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

Since, *a _{n}*

_{+1}–

*a*or the common difference is not same every time.

_{n}Therefore, and the given series doesn’t forms a 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

Since, *a _{n}*

_{+1}–

*a*or the common difference is same every time.

_{n}Therefore, *d* = -4 and the given series forms a A.P.

Hence, next three 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

Since, *a _{n}*

_{+1}–

*a*or the common difference is same every time.

_{n}Therefore, *d* = 0 and the given series forms a A.P.

Hence, next three 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

Since, *a _{n}*

_{+1}–

*a*or the common difference is not same every time.

_{n}Therefore, and the given series doesn’t form a A.P.

**(x) a, 2a, 3a, 4a …**

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*

Since, *a _{n}*

_{+1}–

*a*or the common difference is same every time.

_{n}Therefore, *d* = *a* and the given series forms a A.P.

Hence, next three 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(*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)

Since, *a _{n}*

_{+1}–

*a*or the common difference is not same every time.

_{n}Therefore, the given series doesn’t forms a 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

Since, *a _{n}*

_{+1}–

*a*or the common difference is same every time.

_{n}Therefore, *d* = √2 and the given series forms a A.P.

Hence, next three 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)

Since, *a _{n}*

_{+1}–

*a*or the common difference is not same every time.

_{n}Therefore, the given series doesn’t form a 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

Since, *a _{n}*

_{+1}–

*a*or the common difference is not same every time.

_{n}Therefore, the given series doesn’t form a 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

Since, *a _{n}*

_{+1}–

*a*or the common difference is same every time.

_{n}Therefore, *d* = 24 and the given series forms a A.P.

Hence, next three terms are;

*a*_{5} = 73+24 = 97

*a*_{6} = 97+24 = 121

*a*_{7 }= 121+24 = 145