Question

maths 10th arithmetic progression

Answer 1

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.

Answer

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.

(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.

Answer

Let the initial volume of air in a cylinder be V litres. In each stroke, the vacuum pump removes 1/4 of air remaining in the cylinder at a time. In other words, after every stroke, only 1 - 1/4 = 3/4th part of air will remain.
Therefore, volumes will be V, 3V/4 , (3V/4)2 , (3V/4)3...
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.

(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.

Answer

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.

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

Answer

We know that if Rs P is deposited at r% compound interest per annum for n years, our money will be



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

Answer

(i) a = 10, d = 10
Let the 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
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) a = - 2, d = 0
Let the series be a1, a2, a3, a4 …
a1 = a = -2
a2 = a1 + d = - 2 + 0 = - 2
a3 = a2 + d = - 2 + 0 = - 2
a4 = a3 + d = - 2 + 0 = - 2
Therefore, the series will be - 2, - 2, - 2, - 2 …
First four terms of this A.P. will be - 2, - 2, - 2 and - 2.

(iii) a = 4, d = - 3
Let the series be a1, a2, a3, a4 …
a1 = a = 4
a2 = a1 + d = 4 - 3 = 1
a3 = a2 + d = 1 - 3 = - 2
a4 = a3 + d = - 2 - 3 = - 5
Therefore, the series will be 4, 1, - 2 - 5 …
First four terms of this A.P. will be 4, 1, - 2 and - 5.

(iv) a = - 1, d = 1/2
Let the series be a1, a2, a3, a4 …a1 = a= -1
a2 = a1 + d = -1 + 1/2 = -1/2
a3 = a2 + d = -1/2 + 1/2 = 0
a4 = a3 + d = 0 + 1/2 = 1/2
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) a = - 1.25, d = - 0.25
Let the series be a1, a2, a3, a4 …
a1 = a = - 1.25
a2 = a1 + d = - 1.25 - 0.25 = - 1.50
a3 = a2 + d = - 1.50 - 0.25 = - 1.75
a4 = a3 + d = - 1.75 - 0.25 = - 2.00
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.

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