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NCERT New Syllabus Mathematics
Class: 10
Exercise 5.2
Topic: Arithmetic Progressions
Introduction to Arithmetic Progressions
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.
1) The nth term an of the AP with first term a and common difference d is given by
an = a + (n – 1) d.
2) an is also called the general term of the AP. If there are m terms in the AP, then
am represents the last term which is sometimes also denoted by l.
EXERCISE 5.2
Q1. Fill in the blanks in the following table, given that a is the first term, d is the common difference, and an the nth term of the AP:
| 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:
1) The nth term an of an Arithmetic Progression (AP) with the first term 'a' and
common difference 'd' is given by: an = a + (n – 1) d.
2) In general, an is called the nth term of an AP, and ar is referred to as the general
term or the rth term of an AP.
3) Sometimes 'an' represents the last term of an AP and is denoted by 'l'.
Solution:
i) a = 7, d = 3, n = 8, find an.
1) According to the problem,
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
2) So, here an = 28.
ii) a = – 18, an = 0, n = 10, find d.
1) According to the problem,
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.
2) So, here d = 2.
iii) d = – 3, an = – 5, n = 18, find a.
1) According to the problem,
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.
2) So, here a = 46.
iv) a = – 18.9, d = 2.5, an = 3.6, find n.
1) According to the problem,
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.
2) So, here n = 10.
v) a = 3.5, d = 0, n = 105, find an.
1) According to the problem,
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
2) So, here an = 3.5.
Q2. Choose the correct choice in the following and justify :
(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:
1) The nth term an of an Arithmetic Progression (AP) with the first term 'a' and
common difference 'd' is given by: an = a + (n – 1) d.
2) In general, an is called the nth term of an AP, and ar is referred to as the general
term or the rth term of an AP.
3) Sometimes 'an' represents the last term of an AP and is denoted by 'l'.
Solution:
(i) 30th term of the AP: 10, 7, 4, . . . , is
(A) 97 (B) 77 (C) –77 (D) – 87
1) According to the problem,
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
3) So, the answer is C i.e. a30 = – 77.
(ii) 11th term of the AP: – 3, – 1/2, 2, . . ., is
(A) 28 (B) 22 (C) –38 (D) – 48 and 1/2
1) According to the problem,
d = a2 – a1
d = (– 1/2) – (– 3)
d = – 1/2 + 3
d = (6 – 1)/2d = 5/2
2) So here,
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
3) So, the answer is B i.e. a11 = 22.
Q3. In the following APs, find the missing terms in the boxes :
(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:
1) The nth term an of an Arithmetic Progression (AP) with the first term 'a' and
common difference 'd' is given by: an = a + (n – 1) d.
2) In general, an is called the nth term of an AP, and ar is referred to as the general
term or the rth term of an AP.
3) Sometimes 'an' represents the last term of an AP and is denoted by 'l'.
Solution:
(i) 2, 囗, 26
1) Here,
a1 = 2, a2 = ?, a3 = 26
2) Using the formula an = a + (n – 1) d, we can find the value of d and all other terms.
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
3) From equation 1 and equation 2, we have a = 2, d = 12.
4) Using the formula an = a + (n – 1) d, we can find the value of a2.
an = a + (n – 1) d
a2 = 2 + (2 – 1) (12)
a2 = 2 + 12(1)
a2 = 14
5) So, the missing term a2 is 14.
(ii) 囗, 13, 囗, 3
1) Here,
a1 = ? (say a), a2 = 13, a3 = ?, a4 = 3
2) Using the formula an = a + (n – 1) d, we can find the value of d and all other terms.
a1 = a --------- equation 1
a2 = a + (2 – 1)(d)
a2 = a + d
13 = a + d
a + d = 13 --------- equation 2
3) Using formula an = a + (n – 1) d, and a4 = 3 we have,
an = a + (n – 1) d
a4 = a + (4 – 1)(d)
a4 = a + 3d
3 = a + 3d
a + 3d = 3 --------- equation 3
4) Subtract equation 2 from equation 3, and we get,
a + 3d = 3
a + d = 13
( – ) ( – ) ( – )
---------------------------
2d = – 10
d = – 10/2
d = – 5 --------- equation 4
5) Put d = – 5 from equation 4 in equation 2, we get,
a + d = 13
a + (– 5) = 13
a – 5 = 13
a = 13 + 5
a = 18 --------- equation 5
6) Now we will find a3.
an = a + (n – 1) da3 = a + (3 – 1)(d)a3 = 18 + 2 (– 5)a3 = 18 – 10
a3 = 8
7) So, the missing terms a1 = 18, and a3 = 8.
(iii) 5, 囗, 囗, 9½
1) Here,
a1 = a = 5, a2 = ?, a3 = ?, a4 = 9½
2) Using the formula an = a + (n – 1) d, we can find the value of d and all other terms.
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
3) Using formula an = a + (n – 1) d, a = 5, d = 3/2 we have,
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
4) Now we will find a2,
an = a + (n – 1) da3 = a + (3 – 1)(d)
a3 = a + 2d
a3 = 5 + 2(3/2)
a3 = 5 + 3
a3 = 8 --------- equation 4
5) So, the missing term a2 = 13/2, and a3 = 8.
(iv) – 4, 囗, 囗, 囗, 囗, 6
1) Here,
a1 = a = – 4, a2 = ?, a3 = ?, a4 = ?, a5 = ?, a6 = 6.
2) Using the formula an = a + (n – 1) d, we can find the value of d and all other
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
3) Using formula an = a + (n – 1) d, a = – 4, d = 2 we have,
an = a + (n – 1) d
a2 = a + (2 – 1)(d)
a2 = a + d
a2 = – 4 + 2
a2 = – 2 --------- equation 3
4) Now we will find a3,
an = a + (n – 1) da3 = a + (3 – 1)(d)
a3 = a + 2d
a3 = – 4 + 2(2)
a3 = – 4 + 4
a3 = 0 --------- equation 4
5) Now we will find a4,
an = a + (n – 1) da4 = a + (4 – 1)(d)
a4 = a + 3d
a4 = – 4 + 3(2)
a4 = – 4 + 6
a4 = 2 --------- equation 5
6) Now we will find a5,
an = a + (n – 1) da5 = a + 4da5 = a + (5 – 1)(d)
a5 = – 4 + 4(2)
a5 = – 4 + 8a5 = 4 --------- equation 6
7) So, the missing term a2 = – 2, a3 = 0, a4 = 2, a5 = 4.
(v) 囗, 38, 囗, 囗, 囗, – 22
1) Here,
a1 = a = ?, a2 = 38, a3 = ?, a4 = ?, a5 = ?, a6 = – 22.
2) Using the formula an = a + (n – 1) d, we can find the value of d and all other
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
3) Using formula an = a + (n – 1) d, a6 = – 22 we have,
an = a + (n – 1) d
a6 = a + (6 – 1)(d)
a6 = a + 5d
– 22 = a + 5d
a + 5d = – 22 --------- equation 3
4) Subtract equation 2 from equation 3, and we get,
a + 5d = – 22
a + d = 38
( – ) ( – ) ( – )
---------------------------
4d = – 60
d = – 60/4
d = – 15 --------- equation 4
5) Put d = – 15 from equation 4 in equation 2, we get,
a + d = 38
a + (– 15) = 38
a – 15 = 38
a = 38 + 15
a = 53 --------- equation 5
6) Now we will find a3,
an = a + (n – 1) da3 = a + (3 – 1)(d)
a3 = a + 2d
a3 = 53 + 2(– 15)
a3 = 53 – 30
a3 = 23 --------- equation 6
7) Now we will find a4,
an = a + (n – 1) da4 = a + (4 – 1)(d)
a4 = a + 3d
a4 = 53 + 3(– 15)
a4 = 53 – 45
a4 = 8 --------- equation 7
8) Now we will find a5,
an = a + (n – 1) da5 = a + 4da5 = a + (5 – 1)(d)
a5 = 53 + 4(– 15)
a5 = 53 – 60a5 = – 7 --------- equation 6
7) So, the missing term a1 = 53, a3 = 23, a4 = 8, a5 = – 7.
Q4. Which term of the AP : 3, 8, 13, 18, . . . , is 78?
Solution:
1) Here,
a1 = a = 3, a2 = 8, a3 = 13, a4 = 18.
2) Here.
d = a2 – a1
d = 8 – 3
d = 5 --------- equation 1
3) Here.
d = a3 – a2
d = 13 – 8
d = 5 --------- equation 2
4) From equation 1 and equation 2 as d = a2 – a1 = a3 – a2 = 5.
5) Using the formula an = a + (n – 1) d,
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
6) So, the 16th term of the given AP is 78.
Q5. Find the number of terms in each of the following APs :
(i) 7, 13, 19, . . . , 205 (ii) 18, 15½, 13, . . . , – 47
Explanation:
1) The nth term an of an Arithmetic Progression (AP) with the first term 'a' and
common difference 'd' is given by: an = a + (n – 1) d.
2) In general, an is called the nth term of an AP, and ar is referred to as the general
term or the rth term of an AP.
3) Sometimes 'an' represents the last term of an AP and is denoted by 'l'.
Solution:
(i) 7, 13, 19, . . . , 205
1) Here,
a1 = a = 7, a2 = 13, a3 = 19, an = 205.
2) Here.
d = a2 – a1
d = 13 – 7
d = 6 --------- equation 1
3) Here.
d = a3 – a2
d = 19 – 13
d = 6 --------- equation 2
4) From equation 1 and equation 2 as d = a2 – a1 = a3 – a2 = 6.
5) Using the formula an = a + (n – 1) d,
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
6) The number of terms in the given AP is 34.
(ii) 18, 15½, 13, . . . , – 47
1) Here,
a1 = a = 18, a2 = 15½, a3 = 13, an = – 47.
2) Here.
d = a2 – a1
d = 15½ – 18
d = 31/2 – 18
d = (31 – 36)/2d = – 5/2 --------- equation 1
3) Here.
d = a3 – a2
d = 13 – 15½
d = 13 – 31/2
d = (26 – 31)/2d = – 5/2 --------- equation 2
4) From equation 1 and equation 2 as d = a2 – a1 = a3 – a2 = – 5/2.
5) Using the formula an = a + (n – 1) d,
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
6) The number of terms in the given AP is 27.
Q6. Check whether – 150 is a term of the AP: 11, 8, 5, 2 . . .
1) Here,
a1 = a = 11, a2 = 8, a3 = 5, a4 = 2.
2) Here.
d = a2 – a1
d = 8 – 11
d = – 3
3) So d = a2 – a1 = – 3.
4) Using the formula an = a + (n – 1) d,
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
5) As, n = 164/3 is not an integer, (– 150) is not a term of this AP.
Q7. Find the 31st term of an AP whose 11th term is 38 and the 16th term is 73.
Explanation:
1) The nth term an of an Arithmetic Progression (AP) with the first term 'a' and
common difference 'd' is given by: an = a + (n – 1) d.
2) In general, an is called the nth term of an AP, and ar is referred to as the general
term or the rth term of an AP.
3) Sometimes 'an' represents the last term of an AP and is denoted by 'l'.
Solution:
1) Here,
a11 = 38, a16 = 73, find a31.
2) Let the first term be a and the common difference be d.
an = a + (n – 1) d
a11 = a + (11 – 1)(d)
a11 = a + 10d
38 = a + 10d
a + 10d = 38 --------- equation 1
3) Using formula an = a + (n – 1) d, a16 = 73 we have,
an = a + (n – 1) d
a16 = a + (16 – 1)(d)
a16 = a + 15d
73 = a + 15d
a + 15d = 73 --------- equation 2
4) Subtract equation 1 from equation 2, and we get,
a + 15d = 73
a + 10d = 38
( – ) ( – ) ( – )
---------------------------
5d = 35
d = 35/5
d = 7 --------- equation 3
5) Put d = 7 from equation 3 in equation 1, we get,
a + 10d = 38
a + 10(7) = 38
a + 70 = 38
a = 38 – 70
a = – 32 --------- equation 4
6) Now we will find a31,
an = a + (n – 1) da31 = a + (31 – 1)(d)
a31 = a + 30d
a31 = – 32 + 30(7)
a31 = – 32 + 210
a31 = 178 --------- equation 5
7) So, here a31 = 178.
Q8. An AP consists of 50 terms of which 3rd term is 12 and the last term is
106. Find the 29th term.
Solution:
1) Here,
n = 50, a3 = 12, a50 = 106, find a29.
2) Let the first term be "a" and the common difference be "d".
an = a + (n – 1) d
a3 = a + (3 – 1)(d)
a3 = a + 2d
12 = a + 2d
a + 2d = 12 --------- equation 1
3) Using formula an = a + (n – 1) d, a50 = 106 we have,
an = a + (n – 1) d
a50 = a + (50 – 1)(d)
a50 = a + 49d
106 = a + 49d
a + 49d = 106 --------- equation 2
4) Subtract equation 1 from equation 2, and we get,
a + 49d = 106
a + 2d = 12
( – ) ( – ) ( – )
---------------------------
47d = 94
d = 94/47
d = 2 --------- equation 3
5) Put d = 2 from equation 3 in equation 1, we get,
a + 2d = 12
a + 2(2) = 12
a + 4 = 12
a = 12 – 4
a = 8 --------- equation 4
6) Now we will find a29,
an = a + (n – 1) da29 = a + (29 – 1)(d)
a29 = a + 28d
a29 = 8 + 28(2)
a29 = 8 + 56
a29 = 64--------- equation 5
7) So, here a29 = 64.
Q9. If the 3rd and the 9th terms of an AP are 4 and – 8 respectively, which
term of this AP is zero?
Solution:
1) Here,
a3 = 4, a9 = – 8, find which term is 0.
2) Let the first term be "a" and the common difference be "d".
an = a + (n – 1) d
a3 = a + (3 – 1)(d)
a3 = a + 2d
4 = a + 2d
a + 2d = 4 --------- equation 1
3) Using formula an = a + (n – 1) d, a9 = – 8 we have,
an = a + (n – 1) d
a9 = a + (9 – 1)(d)
a9 = a + 8d
– 8 = a + 8d
a + 8d = – 8 --------- equation 2
4) Subtract equation 1 from equation 2, and we get,
a + 8d = – 8
a + 2d = 4
( – ) ( – ) ( – )
---------------------------
6d = – 12
d = – 12/6
d = – 2 --------- equation 3
5) Put d = – 2 from equation 3 in equation 1, we get,
a + 2d = 4
a + 2(– 2) = 4
a – 4 = 4
a = 4 + 4
a = 8 --------- equation 4
6) Now we will find n when an = 0.
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
7) So, here the 5th term is 0.
Q10. The 17th term of an AP exceeds its 10th term by 7. Find the common
difference.
Solution:
1) Here,
a17 = a10 + 7 --------- equation 1.
2) Let the first term be "a" and the common difference be "d".
an = a + (n – 1) d
a17 = a + (17 – 1)(d)
a17 = a + 16d --------- equation 2
3) Using formula an = a + (n – 1) d, a9 = – 8 we have,
an = a + (n – 1) d
a10 = a + (10 – 1)(d)
a10 = a + 9d --------- equation 3
4) From equations 1, 2, and 3 we have,
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
5) So, here the common difference is 1.
Q11. Which term of the AP : 3, 15, 27, 39, . . . will be 132 more than its 54th
term?
Solution:
1) Here,
a1 = a = 3, a2 = 15, a3 = 27, a4 = 39.
2) Here.
d = a2 – a1
d = 15 – 3
d = 12 --------- equation 1
3) Using the formula an = a + (n – 1) d,
an = a + (n – 1) d
a54 = 3 + 12 (54 – 1)
a54 = 3 + 12 (53)
a54 = 3 + 636
a54 = 639 --------- equation 2
4) Now we will find n for the term more by 132 than the 54th term.
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
6) So, 65th term is 132 more than 54th term.
7) Second method: our term is 132 more than the 54th term. As the difference is 12,
132 will give us 132/12 = 11. So our term is 54 + 11 = 65. So, 65th term is 132 more than 54th term.
Q12. Two APs have the same common difference. The difference between
their 100th terms is 100, what is the difference between their 1000th terms?
Solution:
1) Let the first terms of two APs be "x" and "y", and the common difference be "d".
2) The 100th term first AP of which the first term is x:
an = a + (n – 1) d
a100 = x + (100 – 1)(d)
a100 = x + 99d --------- equation 1
3) The 100th term of the second AP of which the first term is y:
an = a + (n – 1) d
a100 = y + (100 – 1)(d)
a100 = y + 99d --------- equation 2
4) As the difference between their 100th term 100, we have,
(x + 99d) – (y + 99d) = 100
(x – y) = 100 --------- equation 3
5) The 1000th term first AP of which the first term is x:
an = a + (n – 1) d
a1000 = x + (1000 – 1)(d)
a1000 = x + 999d --------- equation 4
6) The 1000th term second AP of which the first term is y:
an = a + (n – 1) d
a1000 = y + (1000 – 1)(d)
a1000 = y + 999d --------- equation 5
7) Now we find the difference between their 1000th terms,
The difference = (x + 999d) – (y + 999d)
The difference = (x – y) --------- equation 6
8) From equations 3 and 6, we have,
The difference = 100
9) So, the difference between their 1000th terms is 100.
Q13. How many three-digit numbers are divisible by 7?
Solution:
1) Here the first 3-digit number which is divisible by 7 is a = 105 and the common
difference d = 7.
2) The greatest 3-digit number is 999. When 7 divides 999, the remainder is 5.
So 999 – 5 divisible by 7. i.e 994 is divisible by 7. So here an = 994.
3) Using the formula an = a + (n – 1) d, we have,
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
Q14. How many multiples of 4 lie between 10 and 250?
Solution:
1) Here the first number between 10 and 250 which is divisible by 4 is a = 12 and
the common difference d = 4.
2) When 4 divides 250, the remainder 2. So 250 – 2 is divisible by 4. i.e. 248 is
divisible by 4. So here an = 248.
3) Using the formula an = a + (n – 1) d, we have,
(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.
Q15. For what value of n, are the nth terms of two APs: 63, 65, 67, . . . and 3, 10, 17, . . . equal?
Solution:
1) For first AP, a1 = a = 63, a2 = 65, a3 = 67.
2) Here.
d = a2 – a1
d = 65 – 63
d = 2
2) The nth term first AP will be,
an = a + (n – 1) d
an = 63 + (n – 1)(2)
an = 63 + 2(n – 1) --------- equation 1
3) For first AP, a1 = a = 3, a2 = 10, a3 = 17.
4) Here.
d = a2 – a1
d = 10 – 3
d = 7
5) The nth term second AP will be,
an = a + (n – 1) d
an = 3 + (n – 1)(7)
an = 3 + 7(n – 1) --------- equation 2
6) As the nth term of both the APs are the same, from equations 1 and 2 we have,
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
Q16. Determine the AP whose third term is 16 and the 7th term exceeds the 5th term by 12.
Solution:
1) Let the first term of an AP be "a" and the common difference be "d".
2) Using the formula an = a + (n – 1) d,
an = a + (n – 1) d
a3 = a + (3 – 1) d
16 = a + 2d
a + 2d = 16 --------- equation 1
3) Now we will find the 5th term.
an = a + (n – 1) d
a5 = a + (5 – 1) d
a5 = a + 4d --------- equation 2
4) Now we will find the 7th term.
an = a + (n – 1) d
a7 = a + (7 – 1) d
a7 = a + 6d --------- equation 3
5) As the 7th term exceeds the 5th term by 12, we have,
a7 = a5 + 12 --------- equation 4
6) So, from equations 2, 3, and 4, we have,
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
7) Put d = 6 from equation 5 in equation 1, and we get.
a + 2d = 16
a + 2(6) = 16
a + 12 = 16
a = 16 – 12
a = 4 --------- equation 6
8) The AP will be 4, 10, 16, 22...
Q17. Find the 20th term from the last term of the AP : 3, 8, 13, . . ., 253.
Solution:
1) To find the 20th term from the last term of AP 3, 8, 13, ....., 253, we first rewrite
the AP in reverse order.
2) The reversed AP becomes 253, ..., 13, 8, 3.
3) Here,
a1 = a = 253, a2 = ... , an-2 = 13, an-1 = 8, an = 3.
4) The common difference d is :
d = an – an-1
d = 3 – 8
d = – 5 --------- equation 1
5) Using the formula for the nth term of an AP: an = a + (n – 1) d,
an = a + (n – 1) d
a20 = 253 + (20 – 1)(– 5)
a20 = 253 – 5(19)
a20 = 253 – 95
a20 = 158
6) Therefore, the 20th term from the last term of the AP : 3, 8, 13, . . ., 253, is
158.
Q18. The sum of the 4th and 8th terms of an AP is 24 and the sum of the 6th
and 10th terms is 44. Find the first three terms of the AP.
Solution:
1) Let the first term be "a" and the common difference be "d".
2) Here a4 + a8 = 24 ---------- equation 1
3) Here a6 + a10 = 44 ---------- equation 2
4) Now we will find a4 , a8 , a6 , and a10,
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
5) From equations 1, 3, and 4, we have,
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
9) Now we will find the first 3 terms of an AP.
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
10) From equations 11, 12, and 13, the first 3 terms are – 13, – 8, and – 3.
Q19. Subba Rao started work in 1995 at an annual salary of Rs 5000 and
received an increment of Rs 200 each year. In which year did his income reach Rs 7000?
Solution:
1) Subba Rao started work in 1995 with an annual salary of ₹5000.
Therefore, the first term (a) of the arithmetic progression (AP) is:
a = 5000.
2) The increment is ₹200 each year, so the common difference () is:
d = 200.
3) We need to find the year when his income reached ₹7000, which means:
an = 7000.
4) Using the formula for the th term of an AP: an = a + (n – 1) d:
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
5) Therefore, Subba Rao’s income reached ₹7000 in the 11th year of his service.
Since he started in 1995, after 10 years, his income would reach ₹7000 in:
1995 + 10 = 2005
6) Hence, in 2005,(11th year) Subba Rao’s income reached ₹7000.
Q20. Ramkali saved ₹ 5 in the first week of a year and then increased her
weekly savings by ₹ 1.75. If in the nth week, her weekly savings become ₹ 20.75, find n.
Solution:
1) Ramkali saved ₹ 5 in the first week.
Therefore, the first term (a) of the arithmetic progression (AP) is:
a = 5.
2) The weekly increase in her savings is ₹ 1.75, so the common difference
() is: d = 1.75.
3) We need to find the week when her savings reached ₹ 20.75, which means:
an = 20.75.
4) Using the formula for the th term of an AP: an = a + (n – 1) d:
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
5) Therefore, in the 10th week, Ramkali’s savings will be ₹20.75.
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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