Math, asked by pratipatel313, 7 months ago

EXERCISE 5.1
1. In which of the following situations, does the list of numbers involved make an arithmetic
progression, and why?
The taxi fare after each km when the fare is * 15 for the first km and 3 8 for each
additional km.
() The amount of air present in a cylinder when a vacuum pump removes a
of the
4
air remaining in the cylinder at a time.
(i) The cost of digging a well after every metre of digging, when it costs * 150 for the
first metre and rises by 50 for each subsequent metre.
(iv) The amount of money in the account every year, when 10000 is deposited at
compound interest at 8 % per annum.
Write first four terms of the AP, when the first term a and the common difference d are
given as follows:
(i) a= 10, d=10
(ii) a=-2, d=0
(ii) a = 4 d=-3
1
(iv) a=-1, d=t​

Answers

Answered by muskaan12358
1

Answer:

(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, 3V/4 , (3V/4)2 , (3V/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.

2. 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

Solutions:

(i) a = 10, d = 10

Let us consider, the Arithmetic Progression series be a1, a2, a3, a4, a5 …

a1 = a = 10

a2 = a1+d = 10+10 = 20

a3 = a2+d = 20+10 = 30

a4 = a3+d = 30+10 = 40

a5 = a4+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 a1, a2, a3, a4, a5 …

a1 = a = -2

a2 = a1+d = – 2+0 = – 2

a3 = a2+d = – 2+0 = – 2

a4 = a3+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 a1, a2, a3, a4, a5 …

a1 = a = 4

a2 = a1+d = 4-3 = 1

a3 = a2+d = 1-3 = – 2

a4 = a3+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 a1, a2, a3, a4, a5 …

a2 = a1+d = -1+1/2 = -1/2

a3 = a2+d = -1/2+1/2 = 0

a4 = a3+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.

I hope it will help u.....

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