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 Tn denotes the taxi fare in nth km.
According to question,
T1 = 15 km;
T2 = 15 + 8 = 23;
T3 = 23 + 8 = 31
Now, T3 – T2 = 31 – 23 = 8
T2 – T1 = 23 – 15 = 8
Here, T3 – T2 = T2 – T1 = 8
∴ given situation form an AP.
(ii) Let Tn denotes amount of air present in a cylinder.
According to question,
T1 = x;
T2 = x – \(\frac{1}{4}\)x
= \(\frac{4-1}{4}\)x = \(\frac{3}{4}\)x
T3 = \(\frac{3}{4} x-\frac{1}{4}\left[\frac{3}{4} x\right]=\frac{3}{4} x-\frac{3}{16} x\)
= \(\left(\frac{12-3}{16}\right) x=\frac{9}{16}\)x and so on
Now, T3 – T2 = \(\frac{9}{16}\)x – \(\frac{3}{4}\)x
= \(\left(\frac{9-12}{16}\right) x=-\frac{3}{16}\)x
T2 – T1 = \(\frac{3}{4}\)x – x
= \(\left(\frac{3-4}{4}\right) x=-\frac{1}{4}\)x
Here, T3 – T2 ≠ T2 – T1
∴ given situation donot form an AP.
(iii) Let Tn denotes cost of digging a well for the nth metre,
According to question,
T1 = ₹ 150; T2 = (150 + 50) = ₹ 200;
T3 = ₹ (200 + 5o) = 250 and so on
Now, T3 – T2 = ₹ (250 – 200) = 50
T2 – T1 = ₹ (200 – 150) = 50
Here, T3 – T2 = T2– T1 = 50
∴ given situation form an A.P.
(iv) Let Tn denotes amount of money in the nth year.
According to question
T1 = ₹ 10,000
T2 = ₹ 10,000 + ₹ \(\left[\frac{10,000 \times 8 \times 1}{100}\right]\)
= ₹ 10,000 + ₹ 800 = ₹ 10,800
T3 = ₹ 10,800 + ₹ \(\left[\frac{10,800 \times 8 \times 1}{100}\right]\)
= ₹ 10,800 + ₹ 864
= ₹ 11,640 and so on.
Now, T3 – T2 = ₹ (11,640 – 10,800) = ₹ 840
T2 – T1 = ₹ (10,800 – 10,000) = ₹ 800
Here, T3 – T2 ≠ T2 – T1
∴ 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
∴ T1 = a = 10;
T2 = a + d = 10 + 10 = 20;
T3 = a + 2d
= 10 + 2 × 10 = 10 + 20 = 30;
T4 = 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
∴ T1 = a = -2;
T2 = a + d = -2 + 0 = -2
T3 = a + 2d = -2 + 2 × 0 = -2
T4 = 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
∴ T1 = a = 4;
T2= a + d = 4 – 3 = 1
T3 = a + 2d = 4 + 2(-3) = 4 – 6 = -2
T4 = 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 = \(\frac{1}{2}\)
∴ T1 = a = -1; T2 = a + d
= -1 + \(\frac{1}{2}\) = \(-\frac{1}{2}\)
T3 = a + 2d = -1 + 2(\(\frac{1}{2}\))
= -1 + 1 = 0
T4 = 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
∴ T1 = a = – 1.25;
T2 = a + d = – 1.25 – 0.25 = -1.50
T3 = a + 2d = – 1.25 + 2(- 0.25)
= – 1.25 – 0.50 = – 1.75
T4 = 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 T1 = 3, T2 = 1,
T3 = -1, T4 = -3
First term = T1 = 3
Now, T2 – T1 = 1 – 3 = – 2
T3 – T2 = – 1 – 1 = -2
T4 – T3 = -3 + 1 = -2
∴ T2 – T1 = T3 – T2 = T4 – T3 = – 2
Hence, common difference = – 2 and first term = 3.
(ii) Given A.P. is – 5, – 1, 3, 7, ………….
Here T1 = – 5, T2 = – 1,
T3 = 3, T4 = 7
First term T1 = -5
Now, T2 – T1 = -1 + 5 = 4
T3– T2 = 3 + 1 = 4
T4 – T3 = 7 – 3 = 4
∴ T2 – T1 = T3 – T2 = T4 – T3 = 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 T1 = \(\frac{1}{3}\), T2 = \(\frac{5}{3}\),
T3 = \(\frac{9}{3}\), T4 = \(\frac{13}{3}\)
First term = T1 = \(\frac{1}{3}\)
Now, T2 – T1 = \(\frac{5}{3}-\frac{1}{3}=\frac{5-1}{3}=\frac{4}{3}\)
T3 – T2 = \(\frac{9}{3}-\frac{5}{3}=\frac{9-5}{3}=\frac{4}{3}\)
T4 – T3 = \(\frac{13}{3}-\frac{9}{3}=\frac{13-9}{3}=\frac{4}{3}\)
∴ T2 – T1 = T3 – T2 = T4 – T3 = \(\frac{4}{3}\)
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, T1 = 0.6, T2 = 1.7, T3 = 2.8, T4 = 3.9
First term = T1 = 0.6
Now, T2 – T1 = 1.7 – 0.6 = 1.1
T3 – T2 = 2.8 – 1.7 = 1.1
T4 – T3 = 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, a2, a3, a4, ……………….
(xii) √2, √8, √18, √32, …………
(xiii) √3, √6, √9, √12, ……………..
(xiv) 12, 32, 52, 72, ………..
(xv) 12, 52, 72, 73, ………….
Solution:
(i) Given terms are 2, 4, 8, 16 ………………
Here T1 = 2, T2 = 4, T3 = 8, T4 = 16
T2 – T1 = 4 – 2 = 2
T3 – T2 = 8 – 4 = 4
∵ T2 – T1 ≠ T3 – T2
Hence, given terms do not form an A.P.
(ii) Given terms are 2, \(\frac{5}{2}\), 3, \(\frac{7}{2}\), ………
Here T1 = 2, T2 = 4, T3 = 3, T4 = 16
T2 – T1 = \(\frac{4}{4}\) – 2 = \(\frac{5-4}{2}\) = \(\frac{1}{2}\)
T3 – T2 = 3 – \(\frac{5}{2}\) = \(\frac{6-5}{2}=\frac{1}{2}\)
T4 – T3 = \(\frac{7}{2}-3=\frac{7-6}{2}=\frac{1}{2}\)
∵ T2 – T1 = T3 – T2 = T4 – T3 = \(\frac{1}{2}\)
∴ Common difference = d = \(\frac{1}{2}\)
Now, T5 = a + 4d = 2 + 4\(\frac{1}{2}\) = 4
T6 = a + 5d = 2 + 5(\(\frac{1}{2}\)) = \(\frac{4+5}{2}=\frac{9}{2}\)
T7 = a + 6d = 2 + 6(\(\frac{1}{2}\)) = 2 + 3 = 5.
(iii) Given terms are – 1.2, – 3.2, – 5.2, – 7.2, …………
Here T1 = – 1.2, T2 = – 3.2,
T3 = – 5.2, T4 = – 7.2
T2 – T1 = – 3.2 + 1.2 = – 2
T3 – T2 = – 5.2 + 3.2 = – 2
T 4 – T3 = – 7.2 + 5.2 = – 2
∵ T2 – T1 = T3 – T2 = T4 – T3 = – 2
∴ Common difference = d = – 2
Now, T5 = a + 4d = – 1.2 + 4(-2) = – 1.2 – 8 = – 9.2
T6 = a + 5d = – 1.2 + 5(-2) = – 1.2 – 10 = – 11.2
T7 = a + 6d = – 1.2 + 6(-2) = -1.2 – 12 = – 13.2
(iv) Given terms are – 10, – 6, – 2, 2, ………..
Here T1 = – 10,T2 = – 6
T3 = – 2, T4=2 .
T2 – T1 = – 6 + 10 = 4
T3 – T2 = – 2 + 6 =4
T4 – T3 = 2 + 2 = 4
∵ T2 – T1=T3 – T2 = T4 – T3 = 4 .
∴ Common difference = d = 4
Now, T5 = a + 4d = – 10 + 4(4) = – 10 + 16 = 6
T6 = a + 5d = – 10 + 5(4) = – 10 + 20 = 10
T7 = a + 6d = – 10 + 6(4) = – 10 + 24 = 14.
(v) Given terms are 3, 3 + √2, 3 + 2√2, 3 + 3√2, …………
Here T1 = 3, T2 = 3 + √2,
T3 = 3 + 2√2, T4= 3 + 3√2
T2 – T1 = 3 + √2 – 3 = √2
T3 – T2 = 3 + 2√2 – (3 + √2)
= 3 + 2√2 – 3 – √2 = √2
T4 – T3 = 3 + 3√2 – (3 + 2√2)
= 3 + 3√2 – 3 – 2√2 = √2
∵ T2 -T1 = T3 – T2 = T4 – T3 = √2
∴ Common difference = d = √2
Now, T5 = a + 4d = 3 + 4(√2) = 3 + 4√2
T6 = a + 5d = 3 + 5√2
T7 = a + 6d = 3 + 6√2
(vi) Given terms are 0.2, 0.22, 0.222, 0.2222, …………..
Here Here T1 = 0.2, T2 = 0.22,
T3 = 0.222, T4 = 0.2222.
T2 – T1 = 0.22 – 0.2 = 0.02
T3 – T2 = 0.222 – 0.22 = 0.002
∵ T2 – T1 ≠ T3 – T2
∴ given terms do not form an A.P.
(vii) Given terms are 0, -4, -8, -12
Here T1 = 0, T2 = -4,
T3 = -8, T4 = -12
T2 – T1 = – 4 – 0 = -4
T3 – T2= – 8 + 4 = -4
T4 – T3= – 12 + 8 = -4.
T2 – T1 = T3 – T2 = T4 – T3
∴ Common difference = d = -4
Now, T5= a + 4d = 0 + 4(-4) = -16
T6 = a + 5d = 0 + 5(-4) = -20
T7 = a + 6d = 0 + 6(-4) = -24.
(viii) Given terms are \(-\frac{1}{2}\), \(-\frac{1}{2}\), \(-\frac{1}{2}\), \(-\frac{1}{2}\), ……….
Here T1 = \(-\frac{1}{2}\), T2 = –\(\frac{1}{2}\)
T3 = \(-\frac{1}{2}\), T4 = \(-\frac{1}{2}\)
T2 – T1 = \(-\frac{1}{2}\) + \(\frac{1}{2}\) = 0
T3 – T2 = \(-\frac{1}{2}\) + \(\frac{1}{2}\) = 0
∵ T2 – T1 = T3 – T2 = 0
∴ Common difference = d = 0
Now, T5 = T6 = T7 = –\(\frac{1}{2}\)
[∵ a = –\(\frac{1}{2}\), d = 0]
(ix) Given terms are 1, 3, 9, 27
T1 = 1, T2 = 3, T3 = 9, T4 = 27
T2 – T1 = 3 1 = 2
T3 – T2 = 9 – 3 = 6.
∵ T2 – T1 ≠ T3 – T2
∴ Given terms do not form an A.P.
(x) Given terms are a, 2a, 3a, 4a, …
T1 = a, T2 = 2a, T3 = 3a, T4 = 4a
T2 – T1 = 2a – a = a
T3 – T2 = 3a – 2a = a
T4 – T3 = 4a – 3a = a
∵ T2 – T1 = T3 – T2 = T4 – T3 = a
∴ Common difference = d = a
Now T5 = a + 4d = a + 4(a) = a + 4a = 5a
T6 = a + 5d = a + 5a = 6a
T7 = a + 6d = a + 6a = 7a
(xi) Given terms are a, a2, a3, a4, …………
T1 = a, T2 = a2, T3 = a3, T4 = a4
T2 – T1 = a2 – a
T3 – T2 = a3 – a2
∵ T2 – T1 ≠ T3 – T2
∴ Given terms do not form an A.P.
(xii) Given terms are √2, √8, √18, √32, …………
T1 = √2, T2 = √8, T3 = √18, T4 = √32
or T1 = √2, T2 = 2√2 T3 = 3√2, T4 = 4√2
T2 – T1 = 2√2 – √2 = √2
T3 – T = 3√2 – 2√2 = √2
T4 – T3 = 4√2 – 3√2 = √2
∵ T2 – T1 = T3 – T2 = T4 – T3= √2
∴ Common difference = d = √2
Now, T5 = a + 4d = √2 + 4√2 = 5√2
T6 = a + 5d = √2 + 5√2 = 6√2
T7 = a + 6d = √2 + 6√2 = 7√2
(xiii) Given terms are √3, √6, √9, √12, ……………..
T1 = √3, T2= √6, T3= √9, T4= √12
or T1 = √3, T2 = √6, T3 = 3, T4 = 2√3
T4 – T1 = √6 – √3
T3 – T2 = 3 – √6
∵ T2 – T1 ≠ T3 – T2
∴Given terms do not form an A.P.
(xiv) Given terms are 12, 32, 52, 72, ………..
T1 = 12, T2 = 32, T3 = 52, T4 = 72
or T1 = 1, T2 = 9, T3 = 25, T4 = 49
T4 – T1 = 9 – 1 = 8
T3 – T2 = 25 – 9 = 16
∵ T2 – T1 ≠ T3 – T2
∴ Given terms do not form an A.P.
(xv) Given terms are 12, 52, 72, 73
T1 = 12, T2 = 52, T3 = 72, T4 = 73
or T1 = 1, T2 = 25, T3 = 49, T4 = 73
T2 – T1 = 25 – 1 = 24
T3 – T2 =49 – 24= 24
T4 – T3 = 73 – 49 = 24
∵ T2 – T1 = T3 – T2 = T4 – T3 = 24
∴ Common difference = d = 24
T5 = a + 4d = 1 + 4(24) = 1 + 96 = 97
T6 = a + 5d = 1 + 5(24) = 1 + 120 = 121
T7 = a + 6d = 1 +6(24) = 1 + 144 = 145
