Punjab State Board PSEB 10th Class Maths Book Solutions Chapter 5 Arithmetic Progressions Ex 5.1 Textbook Exercise Questions and Answers.

## PSEB Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.1

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 15 for the first km and 8 for each additional km.

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

(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 msubsequent metre.

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

Solution:

(i) Let T_{n} denotes the taxi fare in n^{th} km.

According to question,

T_{1} = 15 km;

T_{2} = 15 + 8 = 23;

T_{3} = 23 + 8 = 31

Now, T_{3} – T_{2} = 31 – 23 = 8

T_{2} – T_{1} = 23 – 15 = 8

Here, T_{3} – T_{2} = T_{2} – T_{1} = 8

∴ given situation form an AP.

(ii) Let T_{n} denotes amount of air present in a cylinder.

According to question,

T_{1} = x;

T_{2} = x – \(\frac{1}{4}\)x

= \(\frac{4-1}{4}\)x = \(\frac{3}{4}\)x

T_{3} = \(\frac{3}{4} x-\frac{1}{4}\left[\frac{3}{4} x\right]=\frac{3}{4} x-\frac{3}{16} x\)

= \(\left(\frac{12-3}{16}\right) x=\frac{9}{16}\)x and so on

Now, T_{3} – T_{2} = \(\frac{9}{16}\)x – \(\frac{3}{4}\)x

= \(\left(\frac{9-12}{16}\right) x=-\frac{3}{16}\)x

T_{2} – T_{1} = \(\frac{3}{4}\)x – x

= \(\left(\frac{3-4}{4}\right) x=-\frac{1}{4}\)x

Here, T_{3} – T_{2} ≠ T_{2} – T_{1}

∴ given situation donot form an AP.

(iii) Let T_{n} denotes cost of digging a well for the nth metre,

According to question,

T_{1} = ₹ 150; T_{2} = (150 + 50) = ₹ 200;

T_{3} = ₹ (200 + 5o) = 250 and so on

Now, T_{3} – T_{2} = ₹ (250 – 200) = 50

T_{2} – T_{1} = ₹ (200 – 150) = 50

Here, T_{3} – T_{2} = T_{2}– T_{1} = 50

∴ given situation form an A.P.

(iv) Let T_{n} denotes amount of money in the nth year.

According to question

T_{1} = ₹ 10,000

T_{2} = ₹ 10,000 + ₹ \(\left[\frac{10,000 \times 8 \times 1}{100}\right]\)

= ₹ 10,000 + ₹ 800 = ₹ 10,800

T_{3} = ₹ 10,800 + ₹ \(\left[\frac{10,800 \times 8 \times 1}{100}\right]\)

= ₹ 10,800 + ₹ 864

= ₹ 11,640 and so on.

Now, T_{3} – T_{2} = ₹ (11,640 – 10,800) = ₹ 840

T_{2} – T_{1} = ₹ (10,800 – 10,000) = ₹ 800

Here, T_{3} – T_{2} ≠ T_{2} – T_{1}

∴ given situation do not form an A.P.

Question 2.

Write first four terms of the AP, when the first term a and the common difference d are given as follows:

(1) a = 10, d = 10

(ii) a = -2, d = 0

(iii) a = 4, d = -3

(iv) a = -1, d = \(\frac{1}{2}\)

(w) a = -1.25, d = -0.25

Solution:

(i) Given that first term = a = 10

and common difference = d = 10

∴ T_{1} = a = 10;

T_{2} = a + d = 10 + 10 = 20;

T_{3} = a + 2d

= 10 + 2 × 10 = 10 + 20 = 30;

T_{4} = a + 3d = 10 + 3 × 10

= 10 + 30 = 40

Hence, first four terms of an A.P. are 10, 20, 30, 40………….

(ii) Given that first term = a = -2

and common iifference = d = 0

∴ T_{1} = a = -2;

T_{2} = a + d = -2 + 0 = -2

T_{3} = a + 2d = -2 + 2 × 0 = -2

T_{4} = a + 3d = -2 + 3 × 0 = -2

Hence, first four terms of an A.P. are -2, -2, -2, -2,…………….

(iii) Given that first term = a = 4

and common difference = d = -3

∴ T_{1} = a = 4;

T_{2}= a + d = 4 – 3 = 1

T_{3} = a + 2d = 4 + 2(-3) = 4 – 6 = -2

T_{4} = a + 3d = 4 + 3(-3) = 4 – 9 = -5

Hence, first four terms of an A.P. are 4, 1, -2, -5, ……….

(iv) Given that first term = a = -1

and common difference = d = \(\frac{1}{2}\)

∴ T_{1} = a = -1; T_{2} = a + d

= -1 + \(\frac{1}{2}\) = \(-\frac{1}{2}\)

T_{3} = a + 2d = -1 + 2(\(\frac{1}{2}\))

= -1 + 1 = 0

T_{4} = a + 3d = -1 + 3(\(\frac{1}{2}\))

= \(\frac{-2+3}{2}=\frac{1}{2}\)

Hence, first four terms of an AP are -1, –\(\frac{1}{2}\), 0, \(\frac{1}{2}\), …………..

(v) Given that first term = a = – 1.25

and common difference = d = – 0.25

∴ T_{1} = a = – 1.25;

T_{2} = a + d = – 1.25 – 0.25 = -1.50

T_{3} = a + 2d = – 1.25 + 2(- 0.25)

= – 1.25 – 0.50 = – 1.75

T_{4} = a + 3d = – 1.25 + 3(- 0.25)

= – 1.25 – 0.75 = – 2

Hence, first four terms of an A.P. are – 1.25, – 1.50, – 1.75, – 2, ……………..

Question 3.

For the following APs, wilte 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, ………..

Solution:

(i) Given A.P., is 3, 1, -1, -3, ………

Here T_{1} = 3, T_{2} = 1,

T_{3} = -1, T_{4} = -3

First term = T_{1} = 3

Now, T_{2} – T_{1} = 1 – 3 = – 2

T_{3} – T_{2} = – 1 – 1 = -2

T_{4} – T_{3} = -3 + 1 = -2

∴ T_{2} – T_{1} = T_{3} – T_{2} = T_{4} – T_{3} = – 2

Hence, common difference = – 2 and first term = 3.

(ii) Given A.P. is – 5, – 1, 3, 7, ………….

Here T_{1} = – 5, T_{2} = – 1,

T_{3} = 3, T_{4} = 7

First term T_{1} = -5

Now, T_{2} – T_{1} = -1 + 5 = 4

T_{3}– T_{2} = 3 + 1 = 4

T_{4} – T_{3} = 7 – 3 = 4

∴ T_{2} – T_{1} = T_{3} – T_{2} = T_{4} – T_{3} = 4

Hence, common difference = 4 and first term = – 5.

(iii) Given AP. is:

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

Here T_{1} = \(\frac{1}{3}\), T_{2} = \(\frac{5}{3}\),

T_{3} = \(\frac{9}{3}\), T_{4} = \(\frac{13}{3}\)

First term = T_{1} = \(\frac{1}{3}\)

Now, T_{2} – T_{1} = \(\frac{5}{3}-\frac{1}{3}=\frac{5-1}{3}=\frac{4}{3}\)

T_{3} – T_{2} = \(\frac{9}{3}-\frac{5}{3}=\frac{9-5}{3}=\frac{4}{3}\)

T_{4} – T_{3} = \(\frac{13}{3}-\frac{9}{3}=\frac{13-9}{3}=\frac{4}{3}\)

∴ T_{2} – T_{1} = T_{3} – T_{2} = T_{4} – T_{3} = \(\frac{4}{3}\)

Hence, common difference = \(\frac{4}{3}\) and first term = \(\frac{1}{3}\).

(iv) Given A.P. is 0.6, 1.7, 2.8, 3.9,…

Here, T_{1} = 0.6, T_{2} = 1.7, T_{3} = 2.8, T_{4} = 3.9

First term = T_{1} = 0.6

Now, T_{2} – T_{1} = 1.7 – 0.6 = 1.1

T_{3} – T_{2} = 2.8 – 1.7 = 1.1

T_{4} – T_{3} = 3.9 – 2.8 = 1.1

Hence, common difference = 1.1 and first term = 0.6.

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 + √2, 3 + 2√2, 3 + 3√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) √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}, 73, ………….

Solution:

(i) Given terms are 2, 4, 8, 16 ………………

Here T_{1} = 2, T_{2} = 4, T_{3} = 8, T_{4} = 16

T_{2} – T_{1} = 4 – 2 = 2

T_{3} – T_{2} = 8 – 4 = 4

∵ T_{2} – T_{1} ≠ T_{3} – T_{2}

Hence, given terms do not form an A.P.

(ii) Given terms are 2, \(\frac{5}{2}\), 3, \(\frac{7}{2}\), ………

Here T_{1} = 2, T_{2} = 4, T_{3} = 3, T_{4} = 16

T_{2} – T_{1} = \(\frac{4}{4}\) – 2 = \(\frac{5-4}{2}\) = \(\frac{1}{2}\)

T_{3} – T_{2} = 3 – \(\frac{5}{2}\) = \(\frac{6-5}{2}=\frac{1}{2}\)

T_{4} – T_{3} = \(\frac{7}{2}-3=\frac{7-6}{2}=\frac{1}{2}\)

∵ T_{2} – T_{1} = T_{3} – T_{2} = T_{4} – T_{3} = \(\frac{1}{2}\)

∴ Common difference = d = \(\frac{1}{2}\)

Now, T_{5} = a + 4d = 2 + 4\(\frac{1}{2}\) = 4

T_{6} = a + 5d = 2 + 5(\(\frac{1}{2}\)) = \(\frac{4+5}{2}=\frac{9}{2}\)

T_{7} = a + 6d = 2 + 6(\(\frac{1}{2}\)) = 2 + 3 = 5.

(iii) Given terms are – 1.2, – 3.2, – 5.2, – 7.2, …………

Here T_{1} = – 1.2, T_{2} = – 3.2,

T_{3} = – 5.2, T_{4} = – 7.2

T_{2} – T_{1} = – 3.2 + 1.2 = – 2

T_{3} – T_{2} = – 5.2 + 3.2 = – 2

T_{ 4} – T_{3} = – 7.2 + 5.2 = – 2

∵ T_{2} – T_{1} = T_{3} – T_{2} = T_{4} – T_{3} = – 2

∴ Common difference = d = – 2

Now, T_{5} = a + 4d = – 1.2 + 4(-2) = – 1.2 – 8 = – 9.2

T_{6} = a + 5d = – 1.2 + 5(-2) = – 1.2 – 10 = – 11.2

T_{7} = a + 6d = – 1.2 + 6(-2) = -1.2 – 12 = – 13.2

(iv) Given terms are – 10, – 6, – 2, 2, ………..

Here T_{1} = – 10,T_{2} = – 6

T_{3} = – 2, T_{4}=2 .

T_{2} – T_{1} = – 6 + 10 = 4

T_{3} – T_{2} = – 2 + 6 =4

T_{4} – T_{3} = 2 + 2 = 4

∵ T_{2} – T_{1}=T_{3} – T_{2} = T_{4} – T_{3} = 4 .

∴ Common difference = d = 4

Now, T_{5} = a + 4d = – 10 + 4(4) = – 10 + 16 = 6

T_{6} = a + 5d = – 10 + 5(4) = – 10 + 20 = 10

T_{7} = a + 6d = – 10 + 6(4) = – 10 + 24 = 14.

(v) Given terms are 3, 3 + √2, 3 + 2√2, 3 + 3√2, …………

Here T_{1} = 3, T_{2} = 3 + √2,

T_{3} = 3 + 2√2, T_{4}= 3 + 3√2

T_{2} – T_{1} = 3 + √2 – 3 = √2

T_{3} – T2 = 3 + 2√2 – (3 + √2)

= 3 + 2√2 – 3 – √2 = √2

T_{4} – T_{3} = 3 + 3√2 – (3 + 2√2)

= 3 + 3√2 – 3 – 2√2 = √2

∵ T_{2} -T_{1} = T_{3} – T_{2} = T_{4} – T_{3} = √2

∴ Common difference = d = √2

Now, T_{5} = a + 4d = 3 + 4(√2) = 3 + 4√2

T_{6} = a + 5d = 3 + 5√2

T_{7} = a + 6d = 3 + 6√2

(vi) Given terms are 0.2, 0.22, 0.222, 0.2222, …………..

Here Here T_{1} = 0.2, T_{2} = 0.22,

T_{3} = 0.222, T_{4} = 0.2222.

T_{2} – T_{1} = 0.22 – 0.2 = 0.02

T_{3} – T_{2} = 0.222 – 0.22 = 0.002

∵ T_{2} – T_{1} ≠ T_{3} – T_{2}

∴ given terms do not form an A.P.

(vii) Given terms are 0, -4, -8, -12

Here T_{1} = 0, T_{2} = -4,

T_{3} = -8, T_{4} = -12

T_{2} – T_{1} = – 4 – 0 = -4

T_{3} – T_{2}= – 8 + 4 = -4

T_{4} – T_{3}= – 12 + 8 = -4.

T_{2} – T_{1} = T_{3} – T_{2} = T_{4} – T_{3}

∴ Common difference = d = -4

Now, T_{5}= a + 4d = 0 + 4(-4) = -16

T_{6} = a + 5d = 0 + 5(-4) = -20

T_{7} = a + 6d = 0 + 6(-4) = -24.

(viii) Given terms are \(-\frac{1}{2}\), \(-\frac{1}{2}\), \(-\frac{1}{2}\), \(-\frac{1}{2}\), ……….

Here T_{1} = \(-\frac{1}{2}\), T_{2} = –\(\frac{1}{2}\)

T_{3} = \(-\frac{1}{2}\), T_{4} = \(-\frac{1}{2}\)

T_{2} – T_{1} = \(-\frac{1}{2}\) + \(\frac{1}{2}\) = 0

T_{3} – T_{2} = \(-\frac{1}{2}\) + \(\frac{1}{2}\) = 0

∵ T_{2} – T_{1} = T_{3} – T_{2} = 0

∴ Common difference = d = 0

Now, T_{5} = T_{6} = T_{7} = –\(\frac{1}{2}\)

[∵ a = –\(\frac{1}{2}\), d = 0]

(ix) Given terms are 1, 3, 9, 27

T_{1} = 1, T_{2} = 3, T_{3} = 9, T_{4} = 27

T_{2} – T_{1} = 3 1 = 2

T_{3} – T_{2} = 9 – 3 = 6.

∵ T_{2} – T_{1} ≠ T_{3} – T_{2}

∴ Given terms do not form an A.P.

(x) Given terms are a, 2a, 3a, 4a, …

T_{1} = a, T_{2} = 2a, T_{3} = 3a, T4 = 4a

T_{2} – T_{1} = 2a – a = a

T_{3} – T_{2} = 3a – 2a = a

T_{4} – T_{3} = 4a – 3a = a

∵ T_{2} – T_{1} = T_{3} – T_{2} = T_{4} – T_{3} = a

∴ Common difference = d = a

Now T_{5} = a + 4d = a + 4(a) = a + 4a = 5a

T_{6} = a + 5d = a + 5a = 6a

T_{7} = a + 6d = a + 6a = 7a

(xi) Given terms are a, a^{2}, a^{3}, a^{4}, …………

T_{1} = a, T_{2} = a^{2}, T3 = a^{3}, T_{4} = a^{4}

T_{2} – T_{1} = a^{2} – a

T_{3} – T_{2} = a^{3} – a^{2}

∵ T_{2} – T_{1} ≠ T_{3} – T_{2}

∴ Given terms do not form an A.P.

(xii) Given terms are √2, √8, √18, √32, …………

T_{1} = √2, T_{2} = √8, T_{3} = √18, T_{4} = √32

or T_{1} = √2, T_{2} = 2√2 T_{3} = 3√2, T_{4} = 4√2

T_{2} – T_{1} = 2√2 – √2 = √2

T_{3} – T = 3√2 – 2√2 = √2

T_{4} – T_{3} = 4√2 – 3√2 = √2

∵ T_{2} – T_{1} = T_{3} – T_{2} = T_{4} – T_{3}= √2

∴ Common difference = d = √2

Now, T_{5} = a + 4d = √2 + 4√2 = 5√2

T_{6} = a + 5d = √2 + 5√2 = 6√2

T_{7} = a + 6d = √2 + 6√2 = 7√2

(xiii) Given terms are √3, √6, √9, √12, ……………..

T_{1} = √3, T_{2}= √6, T_{3}= √9, T_{4}= √12

or T_{1} = √3, T_{2} = √6, T_{3} = 3, T_{4} = 2√3

T_{4} – T_{1} = √6 – √3

T_{3} – T_{2} = 3 – √6

∵ T_{2} – T_{1} ≠ T_{3} – T_{2}

∴Given terms do not form an A.P.

(xiv) Given terms are 1^{2}, 3^{2}, 5^{2}, 7^{2}, ………..

T_{1} = 1^{2}, T_{2} = 3^{2}, T_{3} = 5^{2}, T_{4} = 7^{2}

or T_{1} = 1, T_{2} = 9, T_{3} = 25, T_{4} = 49

T_{4} – T_{1} = 9 – 1 = 8

T_{3} – T_{2} = 25 – 9 = 16

∵ T_{2} – T_{1} ≠ T_{3} – T_{2}

∴ Given terms do not form an A.P.

(xv) Given terms are 12, 52, 72, 73

T_{1} = 12, T_{2} = 52, T_{3} = 72, T_{4} = 73

or T_{1} = 1, T_{2} = 25, T_{3} = 49, T_{4} = 73

T_{2} – T_{1} = 25 – 1 = 24

T_{3} – T_{2} =49 – 24= 24

T_{4} – T_{3} = 73 – 49 = 24

∵ T_{2} – T_{1} = T_{3} – T_{2} = T_{4} – T_{3} = 24

∴ Common difference = d = 24

T_{5} = a + 4d = 1 + 4(24) = 1 + 96 = 97

T_{6} = a + 5d = 1 + 5(24) = 1 + 120 = 121

T_{7} = a + 6d = 1 +6(24) = 1 + 144 = 145