Students can practice TS Class 10 Maths Solutions Chapter 6 Progressions Ex 6.1 to get the best methods of solving problems.

## TS 10th Class Maths Solutions Chapter 6 Progressions Exercise 6.1

Question 1.

In which of the following situations, does the list of numbers involved make an arithmetic progression, and why ?

i) The taxi fare after each km when the fare is ₹ 20 for the first km and rises by ₹ 8 for each additional km.

Solution:

Fare for the first km = ₹ 20 = a

Fare for each km after the first = ₹ 8 = d

∴ The fares would be 20, 28, 36, 44, …….

The above list forms an A.P.

Since each term in the list, starting from the second can be obtained by adding ‘8’ to its preceding term.

ii) The amount of air present in a cylinder when a vacuum pump removes 1/4^{th} of the air remaining in the cylinder at a time.

Solution:

Let the amount of air initially present in the cylinder be 1024 lit.

First it removes \(\frac{1}{4}\)^{th} of the volume

i.e., \(\frac{1}{4}\) × 1024 = 256

Remaining air present in the cylinder = 768

At second time it removes \(\frac{1}{4}\)^{th} of 768

i.e., \(\frac{1}{4}\)^{th} × 768 = 192

Remaining air in the cylinder = 768 – 192 = 576

Again at third time it removes \(\frac{1}{4}\)^{th} of 576

i.e., \(\frac{1}{4}\) × 576 = 144

Remaining air in the cylinder

= 576 – 144 = 432

i.e., the volume of the air present in the cylinder after 1^{st}, 2^{nd}, 3^{rd} …. times is 1024, 768, 576, 432,…….

Here, a_{2} – a_{1} = 768 – 1024 = -256

a_{3} – a_{2} = 576 – 768 = – 192

a_{4} – a_{3} = 432 – 576 = – 144

Thus the difference between any two successive terms is not equal to a fixed number.

∴ The given situation doesn’t show an A.R

iii) 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.

Solution:

Cost for digging the first metre = ₹ 150

Cost for digging subsequent metres = ₹ 50 each i.e.,

The list is 150, 200, 250, 300, 350, …..

Here d = a_{2} – a_{1} = a_{3} – a_{2}

= a_{4} – a_{3} = ……… = 50

∴ The given situation represents an A.P

iv) The amount of money in the account every year, when ₹ 10,000 is deposited at compound interest at 8% per annum.

Solution:

Amount deposited initially = P = ₹ 10,000

Rate of interest = R = 8% p.a [at C.I.]

∴ A = P(1 + \(\frac{\mathrm{R}}{100}\))^{n}

The terms 10800, 11664, 12597.12, ….

a_{2} – a_{2} = 800

a_{3} – a_{2} = 864

a_{4} – a_{3} = 933.12

…………………….

Here a = 10,000

But a_{2} – a_{1} ≠ a_{3}

– a_{2} ≠ a_{4} – a_{3}

∴ The given situation doesn’t represent an A.P.

Question 2.

Write first four of the AP, when first term a and the common difference d 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

Solution:

Question 3.

For the following A.Ps, write the first term and the common difference :

i) 3, 1, -1, – 3, …..

ii) -5,-1, 3, 7, …….

iii) \(\frac{1}{3}\), \(\frac{5}{3}\), \(\frac{9}{3}\), \(\frac{13}{3}\),………

iv) 0.6, 1.7, 2.8, 3.9,…….. (A.P. Mar ’15)

Solution:

Question 4.

Which of the following are APs ? If they form an AP, find the common difference d and write three more terms.

i) 2, 4, 8, 16, ……..

ii) 2, \(\frac{5}{2}\), 3, \(\frac{7}{2}\),………..

iii) – 1.2, -3.2, -5.2, -7.2, …….

iv) -10, -6, -2, 2,……….

v) 3, 3 + \(\sqrt{2}\), 3 + 2\(\sqrt{2}\), 3 + 3\(\sqrt{2}\),……….

vi) 0.2, 0.22, 0.222, 0.2222,……….

vii) 0, -4, -8, -12,……….

viii) –\(\frac{1}{2}\), –\(\frac{1}{2}\), –\(\frac{1}{2}\), –\(\frac{1}{2}\),…………

ix) 1, 3, 9, 27,……………….

x) a, 2a, 3a, 4a,………

xi) a, a^{2}, a^{3}, a^{4},………..

xii) \(\sqrt{2}\), \(\sqrt{8}\), \(\sqrt{18}\), \(\sqrt{32}\),…………

xiii) \(\sqrt{3}\), \(\sqrt{6}\), \(\sqrt{9}\), \(\sqrt{12}\),…………

Solution: