Arithmetic Progressions class 10th
Que. 1. Fill in the blanks in the following table, where ‘a’ is the first term, ‘d’ is the common difference, and ‘an‘ is the nth term of the AP:
1. a = 7 , d = 3 , n = 8 , an = ?
an = a + (n – 1) d
a8 = 7 + (8 – 1) 3
= 7 + 7 X 3
= 7 + 21
a8 = 28
2. a = 18 , d = ? , n = 10 , an = 0
an = a + (n – 1) d
0 = -18 + (10 – 1) X d
0 = -18 + 9d
18 = 9d
d = 2
3. a = ? , d = -3 , n = 18 , an = -5
an = a + (n – 1) d
-5 = a + (18 – 1) (-3)
-5 = a + 17 X -3
-5 = a – 51
-5 + 51 = a
a = 46
4. a = -18.9 , d = 2.5 , n = ? , an = 3.6
an = a + (n – 1) d
3.6 = -18.9 + (n – 1) d
3.6 + 18.9 = 2.5n – 2.5
22.5 = 2.5n – 2.5
22.5 + 2.5 = 2.5n
25.0 = 2.5n
n = 10
5. a = 3.5 , d = 0 , n = 105 , an = ?
a105 = a + (n – 1) d
a105 = 3.5 + (105 – 1) X 0
a105 = 3.5 + 0
a105 = 3.5
Que. 2. Choose the correct answer in the following and justify your choice:
1. The 30th term of the A.P.: 10, 7, 4, … is:
AP = 10, 7, 4 _ _ _ _ _ a30
a = 10
d = a2 – a1
= 7 – 10 = -3
an = a + (n – 1) d
a30 = 10 + (30 – 1) X 0
= 10 + 29 X – 3
= 10 – 87
a30 = -77
Que. 3.Find the missing terms in the boxes for the following arithmetic progressions:
1. AP – 2 , _____ , 26
a = 2 , a2 = ? , a3 = 26
a = 2 _ _ _ _ – (1)
a3 = 26
a + (n – 1) d = 26
2 + (3 – 1) d = 26
2d = 26 – 2
2d = 24
d = 12
a2 = a + (n – 1) d
= 2 + (2 – 1) X 12
= 2 + 12
a2 = 14
2. a, 13, a3, 3
a2 = 13
a + (n – 1) d = 13
a + (2 – 1) d = 13
a + d = 13 _ _ _ _ – (1)
a4 = 3
a + (4 – 1) d = 3
a + 3d = 3 _ _ _ _ (2)
Subtracting equation (2) from (1),
a + d = 13
-a + -3d = -3
-2d = 10
d = -5
Substituting the value of d in equation (1),
a + (-5) = 13
a – 5 = 13
a = 13 + 5
a = 18
a3 = a + (n – 1) d
= 18 + (3 – 1) (-5)
= 18 – 10
a3 = 8
4. -4, a2, a3, a4, a5, 6
a = -4
[ a6 = 6 ]
a + 5d = 6
-4 + 5d = 6
5d = 6 + 4
5d = 10
[ d = 2 ]
a2 = a + d
= -4 + 2 = -2
a3 = a + 2d
= -4 + 2 X 2
= -4 + 4
[ a3 = 0 ]
a4 = a + 3d
= -4 + 3 X 2
= -4 + 6
[ a4 = 2 ]
a5 = a + 4d
= -4 + 4 X 2
= -4 + 8
[ a5 = 4 ]
5. a, 38, a3, a4, a5, -22
a2 = 38 , a6 = -22
a + (n – 1) d = 38
a + (2 – 1) d = 38
a + d = 38 – (1)
a6 = -22
a + 5d = -22 – (2)
From equations (1) and (2),
a + d = 38
a + -5d = – + 22
-4d = 60
-d = 15
[ d = -15 ]
From equation (1),
a + d = 38
a + (-15) = 38
a = 38 – 15
[ a = 23 ]
Que. 4. Which term of the A.P. 3, 8, 13, 18… is 78?
3, 8, 13, 18 _ _ _ _, 78
a = 3 d = a2 – a1
an = 78 d = 8 – 3
[ d = 5 ]
an = a + (n – 1) d
78 = 3 + (n – 1) 5
78 – 3 = (n – 1) 5
75 = 5 (n – 1)
n – 1 = 15
n = 15 + 1
n = 16
Que. 5. How many terms are there in each of the following arithmetic progressions?
(i) 7, 13, 19, …, 205
an = a + (n – 1) d
205 = 7 + (n – 1) 6
205 – 7 = 6 (n – 1)
198 = 6 (n – 1)
33 = n – 1
n = 33 + 1
n = 34
-65 X 2 = -5 (n – 1)
130 = 5 (n – 1)
n – 1 = 26
n = 26 + 1
n = 27
Que. 6. Is -150 a term of the A.P. 11, 8, 5, 2…? Why?
A.P = 11, 8, 5, 2 _ _ _ _, – 150
a = 11 d = a2 – a1
an = -150 d = 8 – 11
[ d = -3 ]
an = a + (n – 1) d
-150 = 11 + (n – 1) – 3
-150 – 11 = -3 (n – 1)
-161 = -3 (n – 1)
53.6 = n – 1
n = 53.6 + 1
n = 54.6
Que. 7. Find the 31st term of an A.P. whose 11th term is 38 and the 16th term is 73.
a31 = ?
a11 = 38
a16 = 73
a11 = 38
a + 10d = 38 – (1)
a16 = 73
a + 15d = 73 – (2)
Subtracting equation (2) from (1),
a + 10d = 38
a + 15d = 73
-5d = -35
d = 7
Substituting the value of d in equation (1),
a + 10d = 38
a + 10(7) = 38
a + 70 = 38
a = 38 – 72
a = -32
a31 = a + (n – 1) d
= -32 + (31 – 1) 7
= -32 + 30 X 7
= -32 + 210
a31 = 178
Que. 8. An A.P. consists of 50 terms of which the third term is 12 and the last term is 106. Find the 29th term.
a3 = 12
a50 = 106
a29 = ?
Total terms = 50
a3 = 12
a + 2d = 12 – (1)
Substituting the value of d in equation (1),
a50 = 106
a + 49d = 106 – (2)
Subtracting equation (2) from (1),
a + 2d = 12
a + 49d = 106
-47d = -94
d = 2
a + 2d = 12
a + 2 X 2 = 12
a = 12 – 4
a = 8
a29 = a + (n – 1) d
= 8 + (29 – 1) 2
= 8 + 28 X 2
= 8 + 56
a29 = 64
Que. 9. If the third and ninth terms of an A.P. are 4 and 8 respectively, which term of this A.P. is zero?
a3 = 4
a9 = -8
a3 = 4
a + 2d = 4 – (1)
a9 = -8
a + 8d = -8 – (2)
Subtracting equation (2) from (1),
a + 2d = 4
a + 8d = -8
-6d = 12
d = -2
Substituting the value of d in equation (1),
a + 2d = 4
a + 2 (-2) = 4
a – 4 = 4
a = 4 + 4
a = 8
an = 0
a + (n – 1) d = 0
8 + (n – 1) (-2) = 0
-2 (n – 1) = -8
n = 4 + 1
n = 5
Que. 10. The 17th term of an A.P. exceeds its 10th term by 7. Find the common difference.
a17 = a10 + 7
a + 16d = a + 9d + 7
a + 16d – a – 9d = 7
7d = 7
d = 1
Que. 11. Which term of the A.P. 3, 15, 27, 39… will be 132 more than its 54th term?
AP = 3, 15, 27, 39, _ _ _ _ a54 + 132
an = a54 + 132
a = 3 d = a2 – a1 = 15 – 3
d = 12
an = a54 + 132
a + (n – 1) d = a + (54 – 1) d + 132
3 + (n – 1) 12 = 3 + 53 X 12 + 132
3 + 12 (n – 1) = 135 + 636
12 (n – 1) = 135 + 636 – 3
12 (n – 1) = 768
n – 1 = 64
n = 65
Que. 12. Two arithmetic progressions have the same common difference. If the difference between their 100th terms is 100, what is the difference between their 1000th terms?
A.P = a, a2, a3, _ _ _ , a100, _ _ _ a1000
A.P2 = A, A2, A3, _ _ _ , A100,_ _ _ A1000
Common difference = d
According to the question,
-a100 – A100 = 100
a + 99d – (A + 99d) = 100
a + 99d – A – 99d = 100
a – A = 100 – (1)
a1000 – A1000
a + 999d – (A + 999d)
a + 999d – A – 999d
a – A
From equation (1),
[ a – A = 100 ]
Que.13. How many three-digit numbers are divisible by 7?
100, 101, 102, _ _ _ , 999
AP = 105, 112, 119, _ _ _ 994
a = 105
d = a2 – a1 = 112 – 105
d = 7
an = 994
a + (n – 1) d = 994
105 + (n – 1) 7 = 994
7 (n – 1) = 994 – 105
7 (n – 1) = 889
n – 1 = 127
n = 127 + 1
n = 128
Que. 14. How many multiples of 4 lie between 10 and 250?
10, 11, 12, …………………, 250
AP – 12, 16, 20, …………………..248
a = 12
d = 4
an = 248
a + (n -1) d = an
12 + (n -1) 4 = 248
4 (n -1) = 248 – 12
4 (n -1) = 236
n -1 = 59
n = 59 + 1
n = 60
Que. 15. For what value of n are the n terms of the two arithmetic progressions 63, 65, 67, … and 3, 10, 17, … equal?
AP1 = 63, 65, 67 ………………..an
AP2 = 3, 10, 17…………………an
AP1 = 63, 65, 67
a = 63, d = 65 – 63 = 2
AP2 = 3, 10, 17
A = 3, D = 10 – 3 =7
an = an
a + (n – 1) d = A + (n – 1) D
63 + (n – 1) 2 = 3 + (n – 1) 7
63 + 2n -2 = 3 + 7n -7
61 + 2n = 7n – 4
2n – 7n = -4 – 61
-5n = -65
n = 13
Que. 16. Find the A.P. whose third term is 16 and the 7th term exceeds the 5th term by 12 ?
a3 = 16
a7 = a5 +12
a3 = 16
a + (n -1) d = 16
a + (3 -1) d = 16
a + 2d = 16 ——–(1)
a7 = a5 + 12
a + 6d = a + 4d +12
a + 6d -a -4d = 12
2d = 12
d = 6 ————-(2)
Substituting the value of d in equation (1),
a + 2d = 16
a + 2(6) =16
a + 12 =16
a = 16 -12
a = 4
a2 = a + d
= 4 + 6
a2 = 10
a4 = a + 3d
= 4 + 3 X 6
= 4 + 18
a4 = 22
AP – 4, 10,16, 22………………..
Que. 17. Find the 20th term from the last term of the A.P.: 3, 8, 13, …, 253.
AP = 3, 8, 13, ………………..,253
AP = 253, ………………….. 13, 8, 3
a = 253
d = -(a2 – a1) = – (8 – 3)
= -5
a20 = a + (n -1) d
= 253 + (20 – 1) X -5
= 253 + 19 X -5
= 253 – 95
a20 = 158
Que. 18. The sum of the 4th and 8th terms of an A.P. is 24, and the sum of the 6th and 10th terms is 44. Find the first three terms of this A.P.
a4 + a8 = 24
a6 + a10 = 44
a4 + a8 = 24
a + 3d + a + 7d = 24
2a + 10d = 24
2 (a + 5d) = 24
a + 5d = 12 ————–(1)
a6 + a10 = 44
a + 5d + a + 9d = 44
2a + 14d = 44
2 (a + 7d) = 44
a + 7d = 22 —————–(2)
Subtracting equation (2) from (1),
a + 5d = 12
a + 7d = 22
-2d = -10
d = 5
Substituting the value of d in equation (1),
a + 5d =12
a + 5(5) = 12
a + 25 = 12
a = 12 – 25
a = -13
a2 = a + (n -1) d
= -13 + (2 -1) 5
= -13 + 5
a2 = -8
a3 = a + (n -1) d
= -13 + (3 -1) 5
= -13 + 2 X 5
= -13 + 10
a3 = -3
AP = -13, -8, -3………….
Que. 19. Subba Rao started work in 1995 at an annual salary of ₹5,000 and received an increment of ₹200 each year. In which year did his salary reach ₹7,000?
AP = 5000, 5200, 5400 ………..7000
d = 200, a = 5000, an = 7000
an = a + (n -1) d
7000 = 5000 + (n -1) X 200
7000 – 5000 = 200 (n -1)
2000 = 200 (n – 1)
n – 1 = 10
n = 10 +1
n = 11
His salary will be ₹7000 in the 11th year.
Que. 20. Ramkali saved ₹50 in the first week of a year and then increased her weekly savings by ₹17.5. If in the *n*th week her weekly savings become ₹207.50, find n.
AP – 50, 67.5, ………………..207.50
a = 50, d = 17.5, an = 207.50
a + (n -1) d = an
50 + (n – 1) 17.5 = 207.50
17.5 (n – 1) = 207.50 – 50
17.5 (n – 1) = 157.50
n – 1 = 9
n = 9 + 1
n = 10
