1In which of the following situations, does the list of numbers involved make an arithmetic progression, and why?
Answer & Reference:
(i) Taxi fare: First km = ₹15, each additional km = ₹8
Sequence: 15, 23, 31, 39,...
Common difference (d) = 8 (constant) Yes, it forms an AP. (Page 49-50)
(ii) Air in cylinder: Vacuum pump removes 1/4 of remaining air
If initial air = 1 unit, then sequence: 1, 3/4, (3/4)², (3/4)³,...
Common ratio = 3/4 (not common difference) No, it does not form an AP. (Page 49)
(iii) Cost of digging: First metre = ₹150, each subsequent metre = +₹50
Sequence: 150, 200, 250, 300,...
Common difference (d) = 50 (constant) Yes, it forms an AP. (Page 49)
(iv) Compound interest: Principal = ₹10000, rate = 8% p.a.
Amount after 1 year = 10000(1.08), after 2 years = 10000(1.08)², etc.
This is a geometric progression, not arithmetic. No, it does not form an AP. (Page 49)
2Write first four terms of the AP, when the first term a and the common difference d are given as follows:
Answer & Reference:
(i) a = 10, d = 10 → 10, 20, 30, 40
(ii) a = -2, d = 0 → -2, -2, -2, -2
(iii) a = 4, d = -3 → 4, 1, -2, -5
(iv) a = -1, d = 1/2 → -1, -1/2, 0, 1/2
(v) a = -1.25, d = -0.25 → -1.25, -1.50, -1.75, -2.00
(Page 52)
3For the following APs, write the first term and the common difference:
Answer & Reference:
(i) 3, 1, -1, -3,... → a = 3, d = 1-3 = -2
(ii) -5, -1, 3, 7,... → a = -5, d = -1-(-5) = 4
(iii) 1/3, 5/3, 9/3, 13/3,... → a = 1/3, d = 5/3 - 1/3 = 4/3
(iv) 0.6, 1.7, 2.8, 3.9,... → a = 0.6, d = 1.7-0.6 = 1.1
(Page 52)
4Which of the following are APs? If they form an AP, find the common difference d and write three more terms.
Answer & Reference:
(i) 2, 4, 8, 16,... → d₁ = 2, d₂ = 4 → Not constant → Not AP
(ii) 5/2, 7/2, 9/2, 11/2,... → d = 1 → Next terms: 13/2, 15/2, 17/2
(iii) -1.2, -3.2, -5.2, -7.2,... → d = -2 → Next terms: -9.2, -11.2, -13.2
(iv) -10, -6, -2, 2,... → d = 4 → Next terms: 6, 10, 14
(v) 3, 3+√2, 3+2√2, 3+3√2,... → d = √2 → Next terms: 3+4√2, 3+5√2, 3+6√2
(vi) 0.2, 0.22, 0.222, 0.2222,... → d₁ = 0.02, d₂ = 0.002 → Not constant → Not AP
(vii) 0, -4, -8, -12,... → d = -4 → Next terms: -16, -20, -24
(viii) -1/2, -1/2, -1/2, -1/2,... → d = 0 → Next terms: -1/2, -1/2, -1/2
(ix) 1, 3, 9, 27,... → d₁ = 2, d₂ = 6 → Not constant → Not AP
(x) a, 2a, 3a, 4a,... → d = a → Next terms: 5a, 6a, 7a
(xi) a, a², a³, a⁴,... → d₁ = a²-a, d₂ = a³-a² → Not constant → Not AP
(xii) √2, √8, √18, √32,... → Simplify: √2, 2√2, 3√2, 4√2 → d = √2 → Next terms: 5√2, 6√2, 7√2
(xiii) √3, √6, √9, √12,... → d₁ = √6-√3, d₂ = √9-√6 → Not constant → Not AP
(xiv) 1², 3², 5², 7²,... → 1, 9, 25, 49 → d₁ = 8, d₂ = 16 → Not constant → Not AP
(xv) 1², 5², 7², 73 → 1, 25, 49, 73 → d = 24 → Next terms: 97, 121, 145
(Page 54-55)
5Fill in the blanks in the following table, given that a is the first term, d the common difference and an the nth term of the AP:
Answer & Reference:
a
d
n
an
7
3
8
28
-18
2
10
0
46
-3
18
-5
-18.9
2.5
10
3.6
3.5
0
105
3.5
Formula used: an = a + (n-1)d (Page 57)
6Choose the correct choice in the following and justify:
Answer & Reference:
(i) 30th term of AP: 10, 7, 4,...
a = 10, d = -3, n = 30
a30 = 10 + (30-1)(-3) = 10 - 87 = -77 → Option (C)
(ii) 11th term of AP: -3, -1/2, 2,...
a = -3, d = 5/2, n = 11
a11 = -3 + (11-1)(5/2) = -3 + 25 = 22 → Option (B)
(Page 58)
7In the following APs, find the missing terms in the boxes:
Answer & Reference:
(i) 2, __, 26 → a = 2, a3 = 26
26 = 2 + 2d → d = 12 → Missing term = 2 + 12 = 14
(ii) __, 13, __, 3 → Let terms be a, a+d, a+2d, a+3d
a+d = 13, a+3d = 3 → Solving: d = -5, a = 18 → Missing terms: 18, 8
(iii) 5, __, __, 9½ → a = 5, a4 = 19/2
19/2 = 5 + 3d → d = 3/2 → Missing terms: 13/2, 8
(iv) -4, __, __, __, __, 6 → a = -4, a6 = 6
6 = -4 + 5d → d = 2 → Missing terms: -2, 0, 2, 4
(v) __, 38, __, __, __, -22 → a2 = 38, a6 = -22
a+d = 38, a+5d = -22 → Solving: d = -15, a = 53 → Missing terms: 53, 23, 8, -7
(Page 58)
8Which term of the AP: 3, 8, 13, 18,..., is 78?
Answer & Reference:
a = 3, d = 5, an = 78
78 = 3 + (n-1)5 → 75 = 5(n-1) → n-1 = 15 → n = 16
So, 78 is the 16th term. (Page 58)
9Find the number of terms in each of the following APs:
Answer & Reference:
(i) 7, 13, 19,..., 205
a = 7, d = 6, an = 205
205 = 7 + (n-1)6 → 198 = 6(n-1) → n-1 = 33 → n = 34
(ii) 18, 15½, 13,..., -47
a = 18, d = -5/2, an = -47
-47 = 18 + (n-1)(-5/2) → -65 = (n-1)(-5/2) → n-1 = 26 → n = 27
(Page 59)
10Check whether -150 is a term of the AP: 11, 8, 5, 2...
Answer & Reference:
a = 11, d = -3, an = -150
-150 = 11 + (n-1)(-3) → -161 = -3(n-1) → n-1 = 161/3 ≈ 53.67
Since n must be a positive integer, -150 is not a term of this AP. (Page 59)
11Find the 31st term of an AP whose 11th term is 38 and the 16th term is 73.
Answer & Reference:
a11 = a + 10d = 38
a16 = a + 15d = 73
Subtracting: 5d = 35 → d = 7
Then a + 70 = 38 → a = -32
a31 = -32 + 30×7 = -32 + 210 = 178
(Page 59)
12An AP consists of 50 terms of which 3rd term is 12 and the last term is 106. Find the 29th term.
Answer & Reference:
a3 = a + 2d = 12
a50 = a + 49d = 106
Subtracting: 47d = 94 → d = 2
Then a + 4 = 12 → a = 8
a29 = 8 + 28×2 = 8 + 56 = 64
(Page 59)
13If the 3rd and the 9th terms of an AP are 4 and -8 respectively, which term of this AP is zero?
Answer & Reference:
a3 = a + 2d = 4
a9 = a + 8d = -8
Subtracting: 6d = -12 → d = -2
Then a - 4 = 4 → a = 8
Let an = 0 → 8 + (n-1)(-2) = 0 → 8 = 2(n-1) → n-1 = 4 → n = 5
So, the 5th term is zero. (Page 59)
14The 17th term of an AP exceeds its 10th term by 7. Find the common difference.
15Which term of the AP: 3, 15, 27, 39,... will be 132 more than its 54th term?
Answer & Reference:
a = 3, d = 12
a54 = 3 + 53×12 = 3 + 636 = 639
Required term = 639 + 132 = 771
Let an = 771 → 3 + (n-1)12 = 771 → 12(n-1) = 768 → n-1 = 64 → n = 65
(Page 59)
16Two APs have the same common difference. The difference between their 100th terms is 100, what is the difference between their 1000th terms?
Answer & Reference:
Let first AP: a, a+d, a+2d,...
Second AP: b, b+d, b+2d,...
Given: (a + 99d) - (b + 99d) = 100 → a - b = 100
Difference between 1000th terms: (a + 999d) - (b + 999d) = a - b = 100
(Page 59)
17How many three-digit numbers are divisible by 7?
Answer & Reference:
First three-digit number divisible by 7: 105
Last three-digit number divisible by 7: 994
AP: 105, 112, 119,..., 994
a = 105, d = 7, an = 994
994 = 105 + (n-1)7 → 889 = 7(n-1) → n-1 = 127 → n = 128
(Page 59)
18How many multiples of 4 lie between 10 and 250?
Answer & Reference:
First multiple of 4 after 10: 12
Last multiple of 4 before 250: 248
AP: 12, 16, 20,..., 248
a = 12, d = 4, an = 248
248 = 12 + (n-1)4 → 236 = 4(n-1) → n-1 = 59 → n = 60
(Page 59)
19For what value of n, are the nth terms of two APs: 63, 65, 67,... and 3, 10, 17,... equal?
Answer & Reference:
First AP: a1 = 63, d1 = 2 → an = 63 + (n-1)2 = 61 + 2n
Second AP: a2 = 3, d2 = 7 → an = 3 + (n-1)7 = 7n - 4
Set equal: 61 + 2n = 7n - 4 → 65 = 5n → n = 13
(Page 59)
20Determine the AP whose third term is 16 and the 7th term exceeds the 5th term by 12.
Answer & Reference:
a3 = a + 2d = 16
a7 - a5 = 12 → (a+6d) - (a+4d) = 12 → 2d = 12 → d = 6
Then a + 12 = 16 → a = 4
AP: 4, 10, 16, 22,...
(Page 59)
21Find the 20th term from the last term of the AP: 3, 8, 13,..., 253.
Answer & Reference:
Total terms: a = 3, d = 5, an = 253
253 = 3 + (n-1)5 → 250 = 5(n-1) → n = 51
20th term from last = (51-20+1)th = 32nd term from start
a32 = 3 + 31×5 = 3 + 155 = 158
Alternative: Reverse AP: 253, 248,... with d = -5
a20 = 253 + 19×(-5) = 253 - 95 = 158
(Page 60)
22The 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.
Answer & Reference:
(a+3d) + (a+7d) = 24 → 2a + 10d = 24 → a + 5d = 12 ...(1)
(a+5d) + (a+9d) = 44 → 2a + 14d = 44 → a + 7d = 22 ...(2)
Subtract (1) from (2): 2d = 10 → d = 5
From (1): a + 25 = 12 → a = -13
First three terms: -13, -8, -3
(Page 60)
23Subba Rao started work in 1995 at an annual salary of ₹5000 and received an increment of ₹200 each year. In which year did his income reach ₹7000?
Answer & Reference:
a = 5000, d = 200, an = 7000
7000 = 5000 + (n-1)200 → 2000 = 200(n-1) → n-1 = 10 → n = 11
Year = 1995 + 10 = 2005
(Page 60)
24Ramkali 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.
Answer & Reference:
a = 5, d = 1.75, an = 20.75
20.75 = 5 + (n-1)1.75 → 15.75 = 1.75(n-1) → n-1 = 9 → n = 10
(Page 60)
25Find the sum of the following APs:
Answer & Reference:
(i) 2, 7, 12,..., to 10 terms
a = 2, d = 5, n = 10
S10 = 10/2 [2×2 + (10-1)5] = 5[4 + 45] = 5×49 = 245
(ii) -37, -33, -29,..., to 12 terms
a = -37, d = 4, n = 12
S12 = 12/2 [2×(-37) + 11×4] = 6[-74 + 44] = 6×(-30) = -180
(iii) 0.6, 1.7, 2.8,..., to 100 terms
a = 0.6, d = 1.1, n = 100
S100 = 100/2 [2×0.6 + 99×1.1] = 50[1.2 + 108.9] = 50×110.1 = 5505
(iv) 1/15, 1/12, 1/10,..., to 11 terms
a = 1/15, d = 1/60, n = 11
S11 = 11/2 [2/15 + 10/60] = 11/2 [2/15 + 1/6] = 11/2 [9/30] = 11/2 × 3/10 = 33/20
(Page 68)
26Find the sums given below:
Answer & Reference:
(i) 7 + 10½ + 14 + ... + 84
a = 7, d = 7/2, l = 84
n = ? → 84 = 7 + (n-1)(7/2) → 77 = (n-1)(7/2) → n-1 = 22 → n = 23
S23 = 23/2 (7 + 84) = 23/2 × 91 = 1046.5
(ii) 34 + 32 + 30 + ... + 10
a = 34, d = -2, l = 10
n = ? → 10 = 34 + (n-1)(-2) → -24 = -2(n-1) → n = 13
S13 = 13/2 (34 + 10) = 13/2 × 44 = 286
(iii) -5 + (-8) + (-11) + ... + (-230)
a = -5, d = -3, l = -230
n = ? → -230 = -5 + (n-1)(-3) → -225 = -3(n-1) → n = 76
S76 = 76/2 (-5 - 230) = 38 × (-235) = -8930
(Page 68)
27Fig. 5.1: Ladder with decreasing rung lengths
Answer & Reference:
Fig. 5.1: Ladder with rungs decreasing uniformly by 2 cm from bottom to top (Page 49)
28Fig. 5.2: Squares with increasing side lengths
Answer & Reference:
Fig. 5.2: Number of unit squares in squares with side 1, 2, 3,... units (Page 50)