Arithmetic Progressions (AP) are a fundamental mathematical concept that forms the basis for understanding various real-world applications, ranging from financial planning to engineering design. In this chapter of the 10th-grade NCERT syllabus, we delve into sequences where the difference between consecutive terms remains constant. This simple yet powerful concept helps students recognize patterns and solve problems involving future predictions, geometric designs, and even natural phenomena.
An arithmetic progression represented as a sequence of numbers like a, a+d, a+2d, and so on, is defined by the first term (a) and the common difference (d). This progression opens up a range of possibilities for solving complex problems by understanding the behavior of numbers in a linear format.
In this blog, we’ll explore the key concepts of AP, including its general form, the formula for the nth term, and the sum of n terms. We’ll also solve various problems from the new NCERT syllabus, making it easier for students to grasp the concept and excel in their exams.
an = a + (n – 1) d.
am represents the last term which is sometimes also denoted by l.
EXERCISE 5.2
| a | d | n | an |
i | 7 | 3 | 8 | ----- |
ii | -18 | ----- | 10 | 0 |
iii | ----- | -3 | 18 | -5 |
iv | -18.9 | 2.5 | ----- | 3.6 |
v | 3.5 | 0 | 105 | ----- |
Explanation:
common difference 'd' is given by: an = a + (n – 1) d.
term or the rth term of an AP.
Solution:
a = 7, d = 3, n = 8,
an = a + (n – 1) d
an = 7 + 3(8 – 1)
an = 7 + 3(8 – 1)
an = 7 + 3(7)
an = 7 + 21an = 28
a = – 18, an = 0, n = 10,
an = a + (n – 1) d
0 = – 18 + d(10 – 1)
0 = – 18 + 9d
9d = 18
d = 18/9d = 2.
d = – 3, an = – 5, n = 18,
an = a + (n – 1) d
– 5 = a + (– 3)(18 – 1)
– 5 = a + (– 3)(17)
– 5 = a – 51
a = 51 – 5a = 46.
a = – 18.9, d = 2.5, an = 3.6,
an = a + (n – 1) d
3.6 = – 18.9 + (2.5)(n – 1)
(2.5)(n – 1) = 3.6 + 18.9
(2.5)(n – 1) = 22.5(n – 1) = 22.5/2.5
(n – 1) = 225/25
(n – 1) = 9
n = 9 + 1n = 10.
a = 3.5, d = 0, n = 105,
an = a + (n – 1) d
an = 3.5 + 0(105 – 1)
an = 3.5 + 0(104)
an = 3.5 + 0
an = 3.5
(i) 30th term of the AP: 10, 7, 4, . . . , is
(A) 97 (B) 77 (C) –77 (D) – 87
(ii) 11th term of the AP: – 3, – 1/2, 2, . . ., is
(A) 28 (B) 22 (C) –38 (D) – 48 and 1/2
Explanation:
common difference 'd' is given by: an = a + (n – 1) d.
term or the rth term of an AP.
Solution:
(A) 97 (B) 77 (C) –77 (D) – 87
d = a2 – a1
d = 7 – 102) So here,
d = – 3
a = 10, d = – 3, n = 30,
an = a + (n – 1) d
a30 = 10 + (30 – 1) (– 3)
a30 = 10 – 3(29)
a30 = 10 – 87
a30 = – 77
(ii) 11th term of the AP: – 3, – 1/2, 2, . . ., is
(A) 28 (B) 22 (C) –38 (D) – 48 and 1/2
d = a2 – a1
d = (– 1/2) – (– 3)
d = – 1/2 + 3
d = (6 – 1)/2d = 5/2
a = – 3, d = 5/2, n = 11,
an = a + (n – 1) d
a11 = – 3 + (11 – 1) (5/2)
a11 = – 3 + (5/2)(10)
a11 = – 3 + 25
a11 = 22
(i) 2, 囗, 26(ii) 囗, 13, 囗, 3(iii) 5, 囗, 囗, 9½(iv) – 4, 囗, 囗, 囗, 囗, 6(v) 囗, 38, 囗, 囗, 囗, – 22
Explanation:
common difference 'd' is given by: an = a + (n – 1) d.
term or the rth term of an AP.
Solution:
a1 = 2, a2 = ?, a3 = 26
a1 = a = 2 --------- equation 1
a3 = a + (3 – 1)(d)
a3 = 2 + d(2)
26 = 2 + 2d
2d = 26 – 22d = 24
d = 24/2
d = 12 --------- equation 2
an = a + (n – 1) d
a2 = 2 + (2 – 1) (12)
a2 = 2 + 12(1)
a2 = 14
a1 = ? (say a), a2 = 13, a3 = ?, a4 = 3
a1 = a --------- equation 1
a2 = a + (2 – 1)(d)
a2 = a + d
13 = a + d
a + d = 13 --------- equation 2
an = a + (n – 1) d
a4 = a + (4 – 1)(d)
a4 = a + 3d
3 = a + 3d
a + 3d = 3 --------- equation 3
a + 3d = 3
a + d = 13
2d = – 10
d = – 10/2
d = – 5 --------- equation 4
a + d = 13
a + (– 5) = 13
a – 5 = 13
a = 13 + 5
a = 18 --------- equation 5
an = a + (n – 1) da3 = a + (3 – 1)(d)a3 = 18 + 2 (– 5)a3 = 18 – 10
a3 = 8
a1 = a = 5, a2 = ?, a3 = ?, a4 = 9½
a1 = a = 5 --------- equation 1
an = a + (n – 1) d
a4 = 5 + (4 – 1)(d)
a4 = 5 + 3d
9½ = 5 + 3d
3d = 9½ – 5
3d = 19/2 – 5
3d = (19 – 10)/2
3d = 9/2
d = 3/2 --------- equation 2
an = a + (n – 1) d
a2 = a + (2 – 1)(d)
a2 = a + d
a2 = 5 + 3/2
a2 = (10 + 3)/2
a2 = 13/2 --------- equation 3
an = a + (n – 1) da3 = a + (3 – 1)(d)
a3 = a + 2d
a3 = 5 + 2(3/2)
a3 = 5 + 3
a3 = 8 --------- equation 4
a1 = a = – 4, a2 = ?, a3 = ?, a4 = ?, a5 = ?, a6 = 6.
terms.
a1 = a = – 4 --------- equation 1
an = a + (n – 1) d
a6 = – 4 + (6 – 1)(d)
a6 = – 4 + 5d
6 = – 4 + 5d
5d = 6 + 4
5d = 10
d = 10/5
d = 2 --------- equation 2
an = a + (n – 1) d
a2 = a + (2 – 1)(d)
a2 = a + d
a2 = – 4 + 2
a2 = – 2 --------- equation 3
an = a + (n – 1) da3 = a + (3 – 1)(d)
a3 = a + 2d
a3 = – 4 + 2(2)
a3 = – 4 + 4
a3 = 0 --------- equation 4
an = a + (n – 1) da4 = a + (4 – 1)(d)
a4 = a + 3d
a4 = – 4 + 3(2)
a4 = – 4 + 6
a4 = 2 --------- equation 5
an = a + (n – 1) da5 = a + 4da5 = a + (5 – 1)(d)
a5 = – 4 + 4(2)
a5 = – 4 + 8a5 = 4 --------- equation 6
a1 = a = ?, a2 = 38, a3 = ?, a4 = ?, a5 = ?, a6 = – 22.
terms.
a1 = a --------- equation 1
an = a + (n – 1) d
a2 = a + (2 – 1)(d)
a2 = a + d
38 = a + d
a + d = 38 --------- equation 2
an = a + (n – 1) d
a6 = a + (6 – 1)(d)
a6 = a + 5d
– 22 = a + 5d
a + 5d = – 22 --------- equation 3
a + 5d = – 22
a + d = 38
4d = – 60
d = – 60/4
d = – 15 --------- equation 4
a + d = 38
a + (– 15) = 38
a – 15 = 38
a = 38 + 15
a = 53 --------- equation 5
an = a + (n – 1) da3 = a + (3 – 1)(d)
a3 = a + 2d
a3 = 53 + 2(– 15)
a3 = 53 – 30
a3 = 23 --------- equation 6
an = a + (n – 1) da4 = a + (4 – 1)(d)
a4 = a + 3d
a4 = 53 + 3(– 15)
a4 = 53 – 45
a4 = 8 --------- equation 7
an = a + (n – 1) da5 = a + 4da5 = a + (5 – 1)(d)
a5 = 53 + 4(– 15)
a5 = 53 – 60a5 = – 7 --------- equation 6
Solution:
a1 = a = 3, a2 = 8, a3 = 13, a4 = 18.
d = a2 – a1
d = 8 – 3
d = 5 --------- equation 1
d = a3 – a2
d = 13 – 8
d = 5 --------- equation 2
an = a + (n – 1) d
78 = 3 + (n – 1)(d)
78 = 3 + 5(n – 1)
5(n – 1) = 78 – 3
5(n – 1) = 75
(n – 1) = 75/5
(n – 1) = 15
n = 15 + 1
n = 16 --------- equation 3
(i) 7, 13, 19, . . . , 205 (ii) 18, 15½, 13, . . . , – 47
Explanation:
common difference 'd' is given by: an = a + (n – 1) d.
term or the rth term of an AP.
Solution:
a1 = a = 7, a2 = 13, a3 = 19, an = 205.
d = a2 – a1
d = 13 – 7
d = 6 --------- equation 1
d = a3 – a2
d = 19 – 13
d = 6 --------- equation 2
an = a + (n – 1) d
205 = 7 + (n – 1)(6)
205 = 7 + 6(n v 1)
6(n – 1) = 205 – 7
6(n – 1) = 198
(n – 1) = 198/6
(n – 1) = 33
n = 33 + 1
n = 34
a1 = a = 18, a2 = 15½, a3 = 13, an = – 47.
d = a2 – a1
d = 15½ – 18
d = 31/2 – 18
d = (31 – 36)/2d = – 5/2 --------- equation 1
d = a3 – a2
d = 13 – 15½
d = 13 – 31/2
d = (26 – 31)/2d = – 5/2 --------- equation 2
an = a + (n – 1) d
– 47 = 18 + (n – 1)(– 5/2)
– 47 = 18 – 5(n – 1)/2
5(n – 1)/2 = 18 + 47
5(n – 1)/2 = 65
5(n – 1) = 2(65)
(n – 1) = 2(65)/5
(n – 1) = 2(13)
(n – 1) = 26
n = 26 + 1n = 27
a1 = a = 11, a2 = 8, a3 = 5, a4 = 2.
d = a2 – a1
d = 8 – 11
d = – 3
an = a + (n – 1) d
– 150 = 11 + (n – 1)(– 3)
– 150 = 11 – 3(n – 1)
3(n – 1) = 11 + 150
3(n – 1) = 161
(n – 1) = 161/3
n = (161/3) + 1
n = (161 + 3)/3
n = (164)/3
Explanation:
common difference 'd' is given by: an = a + (n – 1) d.
term or the rth term of an AP.
Solution:
a11 = 38, a16 = 73, find a31.
an = a + (n – 1) d
a11 = a + (11 – 1)(d)
a11 = a + 10d
38 = a + 10d
a + 10d = 38 --------- equation 1
an = a + (n – 1) d
a16 = a + (16 – 1)(d)
a16 = a + 15d
73 = a + 15d
a + 15d = 73 --------- equation 2
a + 15d = 73
a + 10d = 38
5d = 35
d = 35/5
d = 7 --------- equation 3
a + 10d = 38
a + 10(7) = 38
a + 70 = 38
a = 38 – 70
a = – 32 --------- equation 4
an = a + (n – 1) da31 = a + (31 – 1)(d)
a31 = a + 30d
a31 = – 32 + 30(7)
a31 = – 32 + 210
a31 = 178 --------- equation 5
106. Find the 29th term.
Solution:
n = 50, a3 = 12, a50 = 106, find a29.
an = a + (n – 1) d
a3 = a + (3 – 1)(d)
a3 = a + 2d
12 = a + 2d
a + 2d = 12 --------- equation 1
an = a + (n – 1) d
a50 = a + (50 – 1)(d)
a50 = a + 49d
106 = a + 49d
a + 49d = 106 --------- equation 2
a + 49d = 106
a + 2d = 12
47d = 94
d = 94/47
d = 2 --------- equation 3
a + 2d = 12
a + 2(2) = 12
a + 4 = 12
a = 12 – 4
a = 8 --------- equation 4
an = a + (n – 1) da29 = a + (29 – 1)(d)
a29 = a + 28d
a29 = 8 + 28(2)
a29 = 8 + 56
a29 = 64--------- equation 5
term of this AP is zero?
Solution:
a3 = 4, a9 = – 8, find which term is 0.
an = a + (n – 1) d
a3 = a + (3 – 1)(d)
a3 = a + 2d
4 = a + 2d
a + 2d = 4 --------- equation 1
an = a + (n – 1) d
a9 = a + (9 – 1)(d)
a9 = a + 8d
– 8 = a + 8d
a + 8d = – 8 --------- equation 2
a + 8d = – 8
a + 2d = 4
6d = – 12
d = – 12/6
d = – 2 --------- equation 3
a + 2d = 4
a + 2(– 2) = 4
a – 4 = 4
a = 4 + 4
a = 8 --------- equation 4
an = a + (n – 1) d0 = 8 + (n – 1)(– 2)
0 = 8 – 2(n – 1)
2(n – 1) = 8(n – 1) = 8/2
(n – 1) = 4
n = 5 --------- equation 5
difference.
Solution:
a17 = a10 + 7 --------- equation 1.
an = a + (n – 1) d
a17 = a + (17 – 1)(d)
a17 = a + 16d --------- equation 2
an = a + (n – 1) d
a10 = a + (10 – 1)(d)
a10 = a + 9d --------- equation 3
a17 = a10 + 7
a + 16d = a + 9d + 7
a + 16d – a – 9d = 7
16d – 9d = 7
7d = 7
d = 7/7
d = 1 --------- equation 4
term?
Solution:
a1 = a = 3, a2 = 15, a3 = 27, a4 = 39.
d = a2 – a1
d = 15 – 3
d = 12 --------- equation 1
an = a + (n – 1) d
a54 = 3 + 12 (54 – 1)
a54 = 3 + 12 (53)
a54 = 3 + 636
a54 = 639 --------- equation 2
an = a54 + 132
an = 639 + 1325) Now we will find the value of n.
an = 771 --------- equation 3
an = a + (n – 1) d
771 = 3 + 12 (n – 1)
12 (n – 1) = 771 – 3
12 (n – 1) = 768
(n – 1) = 768/12
(n – 1) = 64
n = 64 + 1
n = 65 --------- equation 4
132 will give us 132/12 = 11. So our term is 54 + 11 = 65. So, 65th term is 132 more than 54th term.
their 100th terms is 100, what is the difference between their 1000th terms?
Solution:
an = a + (n – 1) d
a100 = x + (100 – 1)(d)
a100 = x + 99d --------- equation 1
an = a + (n – 1) d
a100 = y + (100 – 1)(d)
a100 = y + 99d --------- equation 2
(x + 99d) – (y + 99d) = 100
(x – y) = 100 --------- equation 3
an = a + (n – 1) d
a1000 = x + (1000 – 1)(d)
a1000 = x + 999d --------- equation 4
an = a + (n – 1) d
a1000 = y + (1000 – 1)(d)
a1000 = y + 999d --------- equation 5
The difference = (x + 999d) – (y + 999d)
The difference = (x – y) --------- equation 6
The difference = 100
Solution:
difference d = 7.
So 999 – 5 divisible by 7. i.e 994 is divisible by 7. So here an = 994.
an = a + (n – 1) d
994 = 105 + 7 (n – 1)
7 (n – 1) = 994 – 105
7 (n – 1) = 889
(n – 1) = 889/7
(n – 1) = 127
4) Thus, the number of three-digit numbers divisible by 7 is 128.n = 127 + 1n = 128
Solution:
the common difference d = 4.
divisible by 4. So here an = 248.
(n – 1) = 59an = a + (n – 1) d248 = 12 + 4 (n – 1)4 (n – 1) = 248 – 12
4 (n – 1) = 236
(n – 1) = 236/4
n = 59 + 1
n = 604) Thus, the number of integers divisible by 4 between 10 and 250 is 60.
Solution:
d = a2 – a1
d = 65 – 63
d = 2
an = a + (n – 1) d
an = 63 + (n – 1)(2)
an = 63 + 2(n – 1) --------- equation 1
d = a2 – a1
d = 10 – 3
d = 7
an = a + (n – 1) d
an = 3 + (n – 1)(7)
an = 3 + 7(n – 1) --------- equation 2
63 + 2(n – 1) = 3 + 7(n – 1)
7(n – 1) – 2(n – 1) = 63 – 3
7n – 7 – 2n + 2 = 607) Therefore, the terms at the 13th position of these A.P.s are equal.
5n – 5 = 60
5n = 60 + 5
5n = 65
n = 65/5
n = 13
Solution:
an = a + (n – 1) d
a3 = a + (3 – 1) d
16 = a + 2d
a + 2d = 16 --------- equation 1
an = a + (n – 1) d
a5 = a + (5 – 1) d
a5 = a + 4d --------- equation 2
an = a + (n – 1) d
a7 = a + (7 – 1) d
a7 = a + 6d --------- equation 3
a7 = a5 + 12 --------- equation 4
a7 = a5 + 12
a + 6d = a + 4d + 12
a + 6d – a – 4d = 12
6d – 4d = 12
2d = 12
d = 12/2
d = 6 --------- equation 5
a + 2d = 16
a + 2(6) = 16
a + 12 = 16
a = 16 – 12
a = 4 --------- equation 6
Solution:
the AP in reverse order.
a1 = a = 253, a2 = ... , an-2 = 13, an-1 = 8, an = 3.
d = an – an-1
d = 3 – 8
d = – 5 --------- equation 1
an = a + (n – 1) d
a20 = 253 + (20 – 1)(– 5)
a20 = 253 – 5(19)
a20 = 253 – 95
a20 = 158
158.
and 10th terms is 44. Find the first three terms of the AP.
Solution:
a) First we will find a4
an = a + (n – 1) da4 = a + (4 – 1) d
a4 = a + 3d ---------- equation 3
b) Now we will find a8
an = a + (n – 1) da8 = a + (8 – 1) d
a8 = a + 7d ---------- equation 4
c) Now we will find a6
an = a + (n – 1) da6 = a + (6 – 1) d
a6 = a + 5d ---------- equation 5
d) Now we will find a10
an = a + (n – 1) da10 = a + (10 – 1) d
a10 = a + 9d ---------- equation 6
a4 + a8 = 24
a + 3d + a + 7d = 246) From equations 2, 5, and 6, we have,
2a + 10d = 24
2(a + 5d) = 24
(a + 5d) = 24/2
(a + 5d) = 12
a = 12 – 5d ---------- equation 7
a6 + a10 = 44(a + 5d) + (a + 9d) = 44
2a + 14d = 447) From equations 7 and 8, we have,
2(a + 7d) = 44
(a + 7d) = 44/2
(a + 7d) = 22
a = 22 – 7d ---------- equation 8
12 – 5d = 22 – 7d
7d – 5d = 22 – 128) Put d = 5 from equation 9 in equation 8, we get
2d = 10d = 5 ---------- equation 9
a = 22 – 7da = 22 – 7(5)a = 22 – 35
a = – 13 ---------- equation 10
a) First we will find a1
an = a + (n – 1) da1 = a + (1 – 1) da1 = – 13 ---------- equation 11
b) First we will find a2
an = a + (n – 1) da2 = a + (2 – 1) d
a2 = a + d
a2 = – 13 + 5(1)
a2 = – 8 ---------- equation 12
c) First we will find a3
an = a + (n – 1) d
a3 = a + (3 – 1) d
a3 = a + 2 d
a3 = – 13 + 5(2)
a3 = – 13 + 10
a3 = – 3 ---------- equation 13
received an increment of Rs 200 each year. In which year did his income reach Rs 7000?
Solution:
Therefore, the first term (a) of the arithmetic progression (AP) is:
a = 5000.
d = 200.
an = 7000.
Substituting the known values:
an = a + (n – 1) d7000 = 5000 + 200 (n – 1)
200 (n – 1) = 7000 – 5000
200 (n – 1) = 2000
(n – 1) = 2000/200
n – 1 = 10
n = 11 ---------- equation 1
Since he started in 1995, after 10 years, his income would reach ₹7000 in:
1995 + 10 = 2005
weekly savings by ₹ 1.75. If in the nth week, her weekly savings become ₹ 20.75, find n.
Solution:
Therefore, the first term (a) of the arithmetic progression (AP) is:
a = 5.
() is: d = 1.75.
an = 20.75.
Substituting the known values:
an = a + (n – 1) d20.75 = 5 + 1.75 (n – 1)
1.75 (n – 1) = 20.75 – 5
1.75 (n – 1) = 15.75
(n – 1) = 15.75/1.75
(n – 1) = (15.75 x 2)/(1.75 x 2)
(n – 1) = (31.5)/(3.5)
(n – 1) = (31.5 x 2)/(3.5 x 2)
(n – 1) = (63)/(7)
n – 1 = 9
n = 10 ---------- equation 1
Conclusion: The Power of Arithmetic Progressions
As we wrap up our journey through Arithmetic Progressions, we've seen how sequences unfold and reveal patterns in everyday life—from predicting future events to organizing data. The concept of common differences and the sum of terms gives us the ability to model various scenarios in both math and real-world situations. Whether you’re solving problems in exams or exploring the magic of number sequences, the knowledge of AP serves as a powerful tool. Keep practicing, and you’ll see how every sequence leads to new possibilities!
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