Telangana SCERT Solution Class X (10) Maths Chapter 6 Progressions Exercise 6.1
(6) Progressions
Exercise – 6.1
(Q1) In which of the following situations, the list of number involved forms an arithmetic progression why?
Ans: (i) = The minimum taxi fare is RS 20 for the first km and there after RS 8 for each addition km.
=> Solution: The fare for the first km = RS 20
Fare for each additional km = RS 8
For ‘2’ km fare = RS 20 + 8 = RS 28
For ‘3’ km fare = RS 20 + RS 8 + RS 8 = RS 36
∴ The list of fare = 20, 28, 36
a2 – a1 = 28 – 20 = 8
a3 – a2 = 36 – 28 = 8
Since the common is same. It forms on arithmetic progression
(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 quantity of air be ‘x’ units.
First time = x
Second time = x – 1/4
Third time = x – 1/4 – 1/4
The list of amount of air is
x, x – 1/4, x – 4,
a2 – a1 = x – 1/4, – x = – 1/4
a3 – a2 = (x – 2/4) – (x – 1/4)
= x – 2/4 – x + 1/4
= -2/4 + 1/4
= -2+1/4
a3 – a2 = – 1/4
Since the common difference is same it forms an A.P.
(iii) The cost digging a well, after every metre of digging, when it cost RS 150 for the first metre and rises by RS 50 for each subsequent metre.
=> Solution:
For digging the well,
The cost first metre = RS 150
Rise in cost after first metre for every metre = RS 50
The cost for first two metre = RS 150 + RS 50
= RS 200
The cost for first three metres = RS 150 + RS 50 + RS 50
= RS 250
The list of costs in RS 150, 200, 250,
a2 – a1 = 200 – 150 = 50
a3 – a2 = 250 – 200 = 50
Since the common difference is same it forms an A.P.
(iv) The amount of money in the account at the end of each year, when RS 10000 is deposited at compound interest at 8% per annum.
=> Solution:
P = RS 10000 R = 8%
Amount, A = p (1 + R/100)n
Amount after one year (A1) = 10,000 (1 + 8/100)1
= 10000 (108/100)
= 10800
Amount after two years (A2) = 10800 (1 + 8/100)
= 10800 (108/100)
= 11664
Amount after three years (A3) = 11664 (1+8/100)
= 11664 (108/100)
= 12597.12/100 = 12597.12
List of amounts, 10800, 11664, 12597.12
Since the common difference is not equal.
It does not form an A.P.
(Q2) Write first four terms of the A.P, when the first term a and the common difference d are given as follows.
(i) a = 10, d = 10
=> Solution:
a1 = a = 10
a2 = a + d = 10 + 10 = 20
a3 = a + 2d = 10 + 2 (10) = 10+20 = 30
a4 = a + 3d = 10 + 3(10) = 10+30 = 40
or
a3 = a2 + d = 20+10 = 30
a4 = a3 + d = 30 + 10 = 40
a5 = a4 + d = 40+10 = 50
∴ The first four terms are, 10, 20, 30 & 40
(ii) a = -2, d = 0
=> Solution:
a1 = a = -2
a2 = a1 + d = – 2 + 0 = – 2
a3 = a2 + d = – 2 + 0 = – 2
a4 = a3 + d = – 2 + 0 = – 2
∴ The first four terms are, – 2, – 2, – 2 and – 2
(iii) a = 4, d = -3
=> Solution:
a1 = a = 4
a2 = a1 + d = 4 + (-3) = 4 – 3 = 1
a3 = a2 + d = 1 + (-3) = 1 – 3 = -2
a4 = a3 + d = – 2 + (-3) = – 2 – 3 = – 5
The first four terms are,
4, 1, – 2 & – 5
(iv) a = – 1, d = 1/2
= Solution:
a1 = a = – 1
a2 = a1 + d = – 1 + 1/2 = -2+1/2 = -1/2
a3 = a2 + d = – 1/2 + 1/2 = 0
a4 = a3 + d = 0 + 1/2 = 1/2
The first four terms are, -1, – 1/2, 0 and 1/2
(v) a = 1.25, d = 0.25
=> Solution:
a1 = a = – 1.25
a2 = a1 + d = – 1.25 + (-0, 25)
= 1.25 – 0.25
= – 1.5
a3 = a2 + d = – 1.5 + (-0.25)
= – 1.5 – 0.25
= – 1.75
a4 = a3 + d = – 1.75 + (-0.25)
= – 1.75 – 0.25
= -2
The first four terms are, -1.25, – 1.5, -1.75 and – 2
(Q3) For the following A P.S, write the first term and the common difference
(i) 3, 1, – 1. – 3
=> Solution:
The given A.P. is
3, 1, -1, -3
a1 = a = 3
d = a2 – a1
= 1 – 3
d = – 2
The first term = 3 and common difference = -2
(ii) – 5, – 1, 3, 7
=> Solution:
The given A.P. is
-5, -1, 3, 7,
a1 = a = -5, a2 = -1, a3 = 3
a2 – a1 = -1 + 5 = – 4
a3 – a2 = 3 – (-1) = 3 + 1 = 4
d = 4
The first term = -5 and common difference = 4
(iii) 1/3, 5/3, 9/3, 13/3
=> Solution:
The given A.P. is, 1/3, 5/3, 9/3, 13/3,
a1 = a = 1/3
a2 = 5/3,
a2 – a1 = 5/3 – 1/3 = 4/3
a3 – a2 = 9/3 – 5/3 = 4/3
∴ The first term = 1/3 and
The common difference = 4/3
(iv) 0.6, 1.7, 2.8, 3.9,
=> Solution:
The given A.P. is,
0.6, 1.7, 2.8, 3.9,
a1 = a = 0.6, a2 = 1.7, a3 = 2.8,
a2 – a1 = 1.7 – 0.6 = 1.1
a3 – a2 = 2.8 – 1.7 = 1.1
d = 1.1
The first term = 0.6 and
The common difference = 1.1
(Q4) Which of the following are A.P.S? If they form an A.P, find the common difference d and write the next three terms
(i) 2, 4, 8, 16
=> Solution:
a1 = a = 2, a2 = 4, a3 = 8,
a2 – a1 = 4 – 2 = 2
a3 – a2 = 8 – 4 = 4
a2 – a1 ≠ a3 – a2
∴ It doesn’t form an A.P.
(ii) 2, 5/2, 3, 7/2. —–
= Solution:
a1 = a = 2, a2 = 5/2, a3 = 3, —–
a2 – a1 = 5/2 – 2 = 5-4/2 = 1/2
a3 – a2 = 3 – 5/2 = 6-5/2 = 1/2
∴ It forms an A.P.
Common difference (d) = 1/2
a4 – a3 = 7/2 – 3 = 7-6/2 = 1/2
a5 = a4 + d = 7/2 + 1/2 = 8/2 = 4
a6 = a5 + d = 4 + 1/2 = 8+1/2 = 9/2
a7 = a6 + d = 9/2 + 1/2 = 10/2 = 5
∴ Three more terms are,
4, 9/2 and 5
(iii) -1.2, -3.2, -5.2, -7.2
=> Solution:
a1 = a = – 1.2, a2 = -3.2, a3 = 5.2
a2 – a1 = – 3.2 – (- 1.2) = – 3.2 + 1.2 = – 2
a3 – a2 = – 5.2 – ( – 3.2) = – 5.2 + 3.2 = – 2
a4 – a3 = – 7.2 – ( – 5.2) = – 7.2 + 5.2 = – 2
It forms an A.P.
Common difference (d) = – 2
a5 = a4 + d = – 7.2 + (-2) = – 7.2 – 2 = – 9.2
a6 = a5 + d = – 9.2 + (-2) = – 9.2 – 2 = – 11.2
a7 = a6 + d = – 11.2 + (-2) = – 11.2 – 2 = – 13.2
∴ Three more terms are,
– 9.2, – 11.2 and – 13.2
(iv) – 10, – 6, – 2, 2,
=> Solution:
a1 = a = – 10, a2 = – 6, a3 = – 2, a4 = 2
a2 – a1 = – 6 – (-10) = – 6 + 10 = 4
a3 – a2 = -2 – (-6) = – 2 + 6 = 4
a4 – a3 = 2 – (-2) = 2+2 = 4
∴ It forms an A.P.
Common difference (d) = 4
a5 = a4 + d = 2 + 4 = 6
a6 = a5 + d = 6 + 4 = 10
a7 = a6 + d = 10 + 4 = 14
Three more terms are, 6, 10 and 14
(v) 3, 3 + √2, 3 + 2√2, 3 + 3√2,
=> Solution:
a1 = a = 3, a2 = 3 + √2, a3 = 3 + 2√2
a2 – a1 = 3 + √2 – 3 = √2
a3 – a2 = 3 + 2√2 – (3 + √2) = 3+ 2√2 – 3 – √2
= 2√2 – √2
= √2 (2-1)
= √2
a4 – a3 = 3 + 3√2 – (3 + 2√2)
= 3 + 3√2 – 3 – 2√2
= 3√2 – 2√2
= √2 (3 – 2)
= √2
It forms an A.P
Common difference (d) = √2
a5 = a4 + d = 3 + 3√2 + √2 = 3 + 4√2
a6 = a5 + d = 3 + 4√2 + √2 = 3 + 5√2
a7 = a6 + d = 3 + 5√2 + √2 = 3 + 6√2
∴ Three more terms are,
3 + 4√2, 3 + 5√2 and 3 + 6√2
(vi) 0.2, 0.22, 0.222, 0.2222,
=> Solution:
a1 = a = 0.2, a2 = 0.22,
a2 – a1 = 0.22 – 0.2 = 0.02
a3 – a2 = 0.222 – 0.22 = 0.002
a2 – a1 ≠ a3 – a2
∴ It doesn’t forms an A.P.
(vii) 0, -4, -8, -12,
=> Solution:
a1 = a = 0, a2 = 1 – 4, a3 = -8,
a2 – a1 = -4 – 0 = – 4
a3 – a2 = – 8 – (-4) = – 8 + 4 = – 4
a4 – a3 = – 12 – (-8) = – 12 + 8 = – 4
∴ It forms an A.P.
Common difference (d) = – 4
a5 = a4 + d = – 12 + (-4) = – 12 – 4 = – 16
a6 = a5 + d = – 16 + (-4) = – 16 – 4 = – 20
a7 = a6 + d = – 20 + (-4) = – 20 – 4 = – 24
∴ Three more terms are, – 16, – 20 and – 24
(viii) –1/2, -1/2, -1/2, -1/2,
=> Solution:
a1 = a = -1/2
a2 – a1 = -1/2 – (-1/2) = -1/2 + 1/2 =0
a3 – a2 = -1/2 – (-1/2) = -1/2 +1/2 = 0
a4 – a3 = -1/2 – (-1/2) = -1/2 + 1/2 = 0
∴ It forms an A.P.
Common difference (d) = 0
a5 = a4 + d = -1/2 + 0 = -1/2
a6 = a5 + d = -1/2 + 0 = -1/2
a7 = a6 + d = -1/2 + 0 = -1/2
∴ Three more terms are, -1/2, -1/2 and -1/2
(ix) 1, 3, 9, 27, —
=> Solution:
a1 = a = 1, a2 = 3, a3 = 9
a2 – a1 = 3 – 1 = 2
a3 – a3 = 9 – 3 = 6
∴ a3 – a2 ≠ a2 – a1
∴ It doesn’t forms an A.P.
(x) a, 2a, 3a, 4a
=> Solution:
a1 = a, a2 = 2a
a2 – a1 = 2a – a = a
a3 – a2 = 3a – 2a = a (3 – 2) = a
a4 – a3 = 4a – 3a = a (4 – 3) = a
∴ It forms an A.P.
Common difference (d) = a
a5 = a4 + d = 4a + a = 5a
a6 = a5 + d = 5a + a = 6a
a7 = a6 + d = 6a + a = 7a
∴ Three more terms are,
5a, 6a, and 7a
(xi) a, a2, a3, a4
=> Solution:
a1 = a = a, a2 = a2,
a2 – a1 = a2 – a = a (a – 1)
a3 – a2= a3 – a2 = a2 (a2 – 1)
a2 – a1 ≠ a3 – a2
It doesn’t forms an A.P.
(xiii) √3, √6, √9, √12
=> Solution:
a1 = a = √3, a2 = √6,
a2 – a1 = √6 – √3
a3 – a2 = √9 – √6 = 3 – √6
∴ a2 – a1 ≠ a3 – a2
∴ It doesn’t forms an A.P.
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