Arithmetic Progressions Solutions
Sum of the first n terms of an AP (Sn) –
AP = a, a2, a3, _ _ _ _
Que. 1. Find the sum of the following arithmetic progressions:
1. 2, 7, 12, _ _ _ _ up to 10 terms
a = 2 , n = 10
d = a2 – a1 = 7 – 2 = 5
= 5 [ 4 + 9 X 5 ]
= 5 [ 4 + 45 ]
= 5 X 49
S10 = 245
2. -37, -33, -29, _ _ _ _, up to 12 terms
a = -37 , n = 12
d = a2 – a1 = -38 – (-37) = -33 + 37
= 4
S12 = 6 [ -74 + 11 X 4 ]
= 6 [ -74 + 44 ]
= 6 X – 30
S12 = -180
3. 0.6, 1.7, 2.8, _ _ _ _, up to 100 terms
a = 0.6 , n = 100
d = a2 – a1 = 1.7 – 0.6 = 1.1
= 50 [ 1.2 + 99 X 1.1 ]
= 50 [ 1.2 + 108.9 ]
= 50 [ 110.1 ]
S100 = 5505
Que. 2. Find the sums given below:
2. 34 + 32 + 30 + _ _ _ _ + 10
AP = 34, 32, 30, _ _ _ _, 10
a = 34 , d = a2 – a1 = 32 – 34
an = 10 = -2
a + (n – 1) d = 10
34 + (n – 1) – 2 = 10
-2 (n – 1) = 10 – 34
-2 (n – 1) = -24
n – 1 = 12
n = 12 + 1
n = 13
S13 = 286
3. -5 + (-8) + (-11) + _ _ _+ (-230)
AP = -5, -8, -11, _ _ _ , -230
a = -5 , d = a2 – a1 = -8 – (-5) = -8 + 5
an = -230, = -3
a + (n – 1) d = -230
-5 + (n – 1) – 3 = -230
-3 (n – 1) = -230 + 5
-3 (n – 1) = -225
n – 1 = 75
n = 75 + 1
n = 76
= 38 [ -5 – 230 ]
= 38 [ -235 ]
S76 = -8930
Que. 3. In an A.P.
(i) Given a = 5, d = 3, and aₙ = 50. Find n and Sₙ.
a = 5, d = 3, an = 50
n = ? , Sn = ?
a + (n – 1) d = an
5 + (n – 1) 3 = 50
3 (n – 1) = 50 – 5
3 (n – 1) = 45
n – 1 = 15
n = 16
= 8 [ 10 + 45 ]
= 8 X 45
S16 = 440
(ii) Given a = 7 and a₁₃ = 35. Find d and S₁₃.
a = 7 , a13 = 35 , n = 13
d = ? , s13 = ?
a13 = a + (n – 1) d
35 = 7 + (13 – 1) d
35 – 7 = 12d
= 13 X 21
= 273
(iii) Given a = 37 and d = 3. Find a and S₁₂.
a12 = 37 , d = 3
a = ? , S12 = ?
a12 = a + (n – 1) d
37 = a + (12 – 1 ) 3
37 = a + 11 X 3
37 = a + 33
37 – 33 = a
a = 4
= 6 [ 8 + 11 X 3 ]
= 6 [ 8 + 33 ]
= 6 X 41
S12 = 246
(iv) Given a₃ = 15 and S₁₀ = 125. Find d and a₁₀.
a3 = 15 , s10 = 125
a10 = ? , d = ?
a3 = 15
a + 2d = 15 – (1)
S10 = 125
5 (2a + 9d) = 125
2a + 9d = 25 – (2)
Multiply equation (1) by 2,
2a + 4d = 30 – (3)
Equation (2) – (3)
2a + 9d = 25
-2a + -4d = -30
5d = -5
d = -1
Substitute the value of d in equation (1),
a + 2d = 15
a + 2 (-1) = 15
a – 2 = 15
a = 15 + 2
a = 17
a10 = a + 9d
= 17 + 9 (-1)
= 17 – 9
= 8
a10 = 8
(v) Given d = 5 and S₉ = 75. Find a and a₉.
d = 5 , S9 = 75
a = ? , a9 = ?
S9 = 75
(vi) Given a = 2, d = 8, and Sₙ = 90. Find n and aₙ.
a = 2 , d = 8 , Sn = 90
n = ? , an = ?
Sn = 90
2n (2n – 1) = 90
4n2 – 2n = 90
4n2 – 2n – 90 = 0
2 (2n2 – n – 45) = 0
2n2 – n – 45 = 0
2n2 – 10n + 9n – 45 = 0
2n (n – 5) + 9 (n – 5) = 0
(n – 5) (2n + 9) = 0
n – 5 = 0 , 2n + 9 = 0
n = 5 , 2n = -9
Discarding this value,
an = a + (n – 1) d
a5 = 2 + (5 – 1) X 8
a5 = 34
(vii) Given a = 8, aₙ = 62, and Sₙ = 210. Find n and d.
a = 8 , an = 62 , Sn = 210
n = ? , d = ?
an = 62
a + (n – 1) d = 62
8 + (n – 1) d = 62
(n – 1) d = 62 – 8
(n – 1) d = 54 – (1)
Sn = 210
n [ 16 + (n – 1) d ] = 210 X 2
From equation (1),
(n – 1) d = 54
(6 – 1) d = 54
(viii) Given aₙ = 4, d = 2, and Sₙ = -14. Find n and a.
an = 4 , d = 2 , Sn = -14
n = ? , a = ?
an = 4
a + (n – 1) d = 4
a + (n – 1) 2 = 4
a + 2n – 2 = 4
a – 2n = 4 + 2
a + 2n = 6
a = 6 – 2n – (1)
Sn = -14
n (a + n – 1) = -14
n (a + n – 1) = -14 – (2)
From equations (1) and (2),
n [ 6 – 2n + n – 1 ] = -14
n [ -n + 5 ] = -14
-n2+ 5n = -14
n2 – 5n – 14 = 0
n2 – 7n + 2n – 14 = 0
n (n – 7) + 2 (n – 7) = 0
(n – 7) (n + 2) = 0
n – 7 = 0 , n + 2 = 0
n = 7 n = -2
The value of n in equation (1),
a = 6 – 2n
a = 6 – 2 (7)
a = 6 – 14
a = -8
(ix) Given a = 3, n = 8 and S = 192, find d.
a = 3 , n = 8 , S = 192
d = ?
48 = 6 + 7d
48 – 6 = 7d
42 = 7d
d = 6
(x) Given l = 28, S = 144 and total term is 9, find a?
l = 28 , S = 144 , n = 9
a = ?
16 X 2 = a + 28
32 – 28 = a
4 = a
a = 4
Que. 4. How many terms of the A.P. 9, 17, 25… should be taken to obtain a sum of 636?
AP = 9, 17, 25, _ _ _ _
Sn = 636 , n = ?
a = 9 , d = a1 – a2 = 17 – 9 = 8
636 = n [ 9 + 4n – 4 ]
636 = n [ 5 + 4n ]
636 = 5n + 4n2
0 = 4n2 + 5n – 636
quadratic formula,
Que. 5. For an A.P., the first term is 5, the last term is 45, and the sum is 400. Find the number of terms and the common difference.
a = 5 , an = 45 , Sn = 400
n = ? , d = ?
an = 45
a + (n – 1) d = 45
5 + (n – 1) d = 45
(n – 1) d = 45 – 5
(n – 1) d = 40 – (1)
Sn = 400
n [ 10 + (n – 1) d ] = 400 X 2
From equation (1),
n [ 10 + 40 ] = 800
n X 50 = 800
n = 16
Substituting the value of n in equation (1),
(16 – 1) d = 40
15d = 40
Que. 6. The first and last terms of an A.P. are 17 and 350, respectively. If the common difference is 9, how many terms are there, and what is their sum?
a = 17 , an = l = 350 , d = 9
n = ? , Sn = ?
an = 350
an = a + (n – 1) d
350 = 17 + (n – 1) 9
350 – 17 = 9 (n – 1)
333 = 9 (n – 1)
n – 1 = 37
n = 37 + 1
n = 38
= 19 X 367
S38 = 6973
Que. 7. Find the sum of the first 22 terms of an A.P. in which d = 7 and the 22nd term is 149.
S22 = ? , d = 7 , a22 = 149
a22 = 149
a + 21d = 149
a + 21 X 7 = 149
a + 147 = 149
a = 149 – 147
a = 2
= 11 X 151
S22 = 1661
Que. 8. Find the sum of the first 51 terms of an A.P. whose second and third terms are 14 and 18, respectively.
S51 = ? , a2 = 14 , a3 = 18
d = a3 – a2 = 18 – 14
d = 4
a2 = a + d
14 = a + 4
14 – 4 = a
a = 10
= 51 X 110
S51 = 5610
Que. 9. If the sum of the first 7 terms of an A.P. is 49 and the sum of the first 17 terms is 289, find the sum of its first n terms.
S7 = 49 , S17 = 289
Sn = ?
S7 = 49
By doing equations (1)-(2),
a + 3d = 7
-a + -8d = -17
-5d = -10
d = 2
From equation (1),
a + 3d = 7
a + 3 (2) = 7
a = 7 – 6
a = 1
Que. 10. Show that a₁, a₂, …, aₙ, … forms an A.P. if aₙ is defined as follows:
AP = a1, a2, _ _ _ _, an
- an = 3 + 4n
= n = 1
a1 = 3 + 4 (1)
a = 7
n = 15
a15 = 3 + 4 (15)
= 3 + 60
a15 = 63
= 15 X 35
S15 = 525
2. an = 9 – 5n
n = 1
a1 = 9 – 5 (1)
a = 9 – 5
a = 4
n = 15
a15 = 9 – 5 (15)
= 9 – 75
a15 = -66
S15 = -465
Que. 12. Find the sum of the first 40 positive integers that are divisible by 6.
1, 2, 3, 4, _ _ _ _
AP = 6, 12, 18, _ _ _ _ a40 (240)
a = 6 , d = a2 – a1 = 12 – 6 = 6
n = 40
an = a + (n – 1) d
a40 = 6 + (40 – 1) 6
a40 = 6 + 39 X 6
= 6 + 234
= 240
= 20 X 246
S40 = 4920
Que. 13. Find the sum of the first 15 multiples of 8.
AP = 8, 16, 24, _ _ _ _
a = 8 , d = 8 , n = 15
= 15 X 64
S15 = 960
Que. 14. Find the sum of odd numbers between 0 and 50
AP = 1, 3, 5, _ _ _ _ 49
a = 1 , d = 3 – 1 = 2 , an = 49
an = 49
a + (n – 1) d = 49
1 + (n – 1) 2 = 49
2 (n – 1) = 48
n – 1 = 24
n = 24 + 1
n = 25
= 25 X 25
S25 = 625
Que. 15. In a contract for construction work, the provision for a penalty for delaying the work beyond a certain date is as follows: ₹200 for the first day, ₹250 for the second day, ₹300 for the third day, and so on—meaning the penalty for each succeeding day is ₹50 more than that of the preceding day. How much money will a contractor have to pay as a penalty if he delays the work by 30 days?
AP = 200, 250, 300, _ _ _ _
n = 30 , a = 200 , d = 50
= 15 [ 400 + 29 X 50 ]
= 15 [ 400 + 1450 ]
= 15 X 1850
S30 = 27750
Que. 16. A sum of ₹ 700 has been earmarked to give 7 cash prizes to the students of a school for their overall academic performance. If each prize is ₹ 20 less than the prize immediately preceding it, find the value of each prize.
S7 = 700 , n = 7
AP = x, x – 20, x – 40, _ _ _ _
a = x , d = x – 20 – x = -20
100 = x – 60
100 + 60 = x
X = 160 = a
Que. 17. Students of a school decided to plant trees inside and outside the school premises to reduce air pollution. It was decided that each section of every class would plant a number of trees equal to its class number. For example, a section of Class I would plant 1 tree, a section of Class II would plant 2 trees, a section of Class III would plant 3 trees, and so on, up to Class XII. There are three sections in each class. What will be the total number of trees planted by the students of this school?
1, 2, 3, _ _ _ _ 12
AP = 3 X 1, 3 X 2, 3 X 3, _ _ _ _ 3 X 12
= 3, 6, 9, _ _ _ 36
a = 3 , d = 6 – 3 = 3 , a12 = 36
n = 12 , S12 = ?
= 6 X 39
S12 = 234
Que. 18. A spiral is made by drawing successive semicircles with centres alternately at A and B, starting with centre A, and having radii of 0.5 cm, 1.0 cm, 1.5 cm, 2.0 cm, … . What is the total length of this spiral made up of thirteen consecutive semicircles?
n = 13 , a = 0.5
AP = 0.5, 1.0, 1.5, 2, _ _ _
d = a2 – a1 = 1.0 – 0.5 = 0.5 cm
Que. 19. 200 logs are stacked such that there are 20 logs in the bottom row, 19 in the next row, 18 in the row above that, and so on. In how many rows are these 200 logs placed, and how many logs are there in the top row?
AP = 20, 19, 18, 17, _ _ _ _
Sn = 200 , n = ? , an = ?
a = 20 , d = 19 – 20 = -1
200 X 2 = n [ 40 – 1n + 1 ]
400 = n [ -n + 41 ]
400 = -n2 + 41n
n2 – 41n + 400 = 0
n2 – 16n – 25n + 400 = 0
n (n – 16) -25 (n – 16) = 0
(n – 16) (n – 25) = 0
n – 16 = 0 n – 25 = 0
n = 16 n = 25
an = a + (n – 1) d
a16 = 20 + (16 – 1) (-1)
= 20 – 15
a16 = 5
an = a + (n – 1) d
a25 = 20 + (25 – 1) (-1)
= 20 – 24
a25 = -4
Therefore, upon rejecting this value,
