Exercise 5.1
1In which of the following situations, does the list of numbers involved make an arithmetic progression, and why?
(i) The taxi fare after each km when the fare is ₹ 15 for the first km and ₹ 8 for each additional km.
Fares (in ₹): 15, 23, 31, 39, ...
Each term is obtained by adding 8 to the previous term.
Yes, it is an AP with d = 8
(ii) The amount of air present in a cylinder when a vacuum pump removes 14 of the air remaining in the cylinder at a time.
Let the air at the start be V. After each removal, 34 of the air remains.
Amounts: V, 3V4, 9V16, 27V64, ...
a2 – a1 = –V4 ; a3 – a2 = –3V16
The differences are not equal.
No, it is not an AP.
(iii) The cost of digging a well after every metre of digging, when it costs ₹ 150 for the first metre and rises by ₹ 50 for each subsequent metre.
Costs (in ₹): 150, 200, 250, 300, ...
Each term is obtained by adding 50 to the previous term.
Yes, it is an AP with d = 50
(iv) The amount of money in the account every year, when ₹ 10000 is deposited at compound interest at 8 % per annum.
Amounts (in ₹): 10000(1.08), 10000(1.08)2, 10000(1.08)3, ...
i.e., 10800, 11664, 12597.12, ...
a2 – a1 = 864 ; a3 – a2 = 933.12
The differences are not equal.
No, it is not an AP.
2Write first four terms of the AP, when the first term a and the common difference d are given as follows: (i) a = 10, d = 10 (ii) a = –2, d = 0 (iii) a = 4, d = – 3 (iv) a = – 1, d = 12 (v) a = – 1.25, d = – 0.25
The terms are a, a + d, a + 2d, a + 3d.
(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 = 12
–1, –12, 0, 12
(v) a = –1.25, d = –0.25
–1.25, –1.50, –1.75, –2.00
3For the following APs, write the first term and the common difference: (i) 3, 1, – 1, – 3, . . . (ii) – 5, – 1, 3, 7, . . . (iii) 13, 53, 93, 133, . . . (iv) 0.6, 1.7, 2.8, 3.9, . . .
(i) 3, 1, –1, –3, ...
a = 3, d = –2
(ii) –5, –1, 3, 7, ...
a = –5 ; d = –1 – (–5) = 4
a = –5, d = 4
(iii) 13, 53, 93, 133, ...
a = 13 ; d = 53 – 13 = 43
a = 13, d = 43
(iv) 0.6, 1.7, 2.8, 3.9, ...
a = 0.6 ; d = 1.7 – 0.6 = 1.1
a = 0.6, d = 1.1
4Which of the following are APs? If they form an AP, find the common difference d and write three more terms.
A list is an AP if ak+1 – ak is the same for all k.
(i) 2, 4, 8, 16, ...
4 – 2 = 2 ; 8 – 4 = 4 ; 16 – 8 = 8
Not an AP
(ii) 2, 52, 3, 72, ...
AP; d = 12; next terms: 4, 92, 5
(iii) –1.2, –3.2, –5.2, –7.2, ...
AP; d = –2; next terms: –9.2, –11.2, –13.2
(iv) –10, –6, –2, 2, ...
AP; d = 4; next terms: 6, 10, 14
(v) 3, 3 + √2, 3 + 2√2, 3 + 3√2, ...
AP; d = √2; next terms: 3 + 4√2, 3 + 5√2, 3 + 6√2
(vi) 0.2, 0.22, 0.222, 0.2222, ...
0.22 – 0.2 = 0.02 ; 0.222 – 0.22 = 0.002
Not an AP
(vii) 0, –4, –8, –12, ...
AP; d = –4; next terms: –16, –20, –24
(viii) –12, –12, –12, –12, ...
AP; d = 0; next terms: –12, –12, –12
(ix) 1, 3, 9, 27, ...
Not an AP
(x) a, 2a, 3a, 4a, ...
AP; d = a; next terms: 5a, 6a, 7a
(xi) a, a2, a3, a4, ...
a2 – a = a(a – 1) ; a3 – a2 = a2(a – 1)
The differences are not equal.
Not an AP
(xii) √2, √8, √18, √32, ...
The list is √2, 2√2, 3√2, 4√2, ... ; each difference = √2
AP; d = √2; next terms: 5√2 (= √50), 6√2 (= √72), 7√2 (= √98)
(xiii) √3, √6, √9, √12, ...
√6 – √3 = √3(√2 – 1) ; √9 – √6 = √3(√3 – √2)
The differences are not equal.
Not an AP
(xiv) 12, 32, 52, 72, ...
The list is 1, 9, 25, 49, ... ; differences are 8, 16, 24
Not an AP
(xv) 12, 52, 72, 73, ...
The list is 1, 25, 49, 73, ... ; each difference = 24
AP; d = 24; next terms: 97, 121, 145
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Exercise 5.2
1Fill 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:
We use an = a + (n – 1)d.
(i) a = 7, d = 3, n = 8
an = 28
(ii) a = –18, n = 10, an = 0
d = 2
(iii) d = –3, n = 18, an = –5
–5 = a + 17 × (–3) ⇒ a = –5 + 51 = 46
a = 46
(iv) a = –18.9, d = 2.5, an = 3.6
3.6 = –18.9 + (n – 1) × 2.5 ⇒ (n – 1) = 22.52.5 = 9 ⇒ n = 10
n = 10
(v) a = 3.5, d = 0, n = 105
a105 = 3.5 + 104 × 0 = 3.5
an = 3.5
2Choose 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, –12, 2, . . ., is (A) 28 (B) 22 (C) –38 (D) –4812
(i)
a = 10, d = –3 ; a30 = 10 + 29 × (–3) = 10 – 87 = –77
(C) –77
(ii)
a = –3, d = –12 – (–3) = 52 ; a11 = –3 + 10 × 52 = –3 + 25 = 22
(B) 22
3In the following APs, find the missing terms in the boxes: (i) 2, ☐, 26 (ii) ☐, 13, ☐, 3 (iii) 5, ☐, ☐, 912 (iv) – 4, ☐, ☐, ☐, ☐, 6 (v) ☐, 38, ☐, ☐, ☐, – 22
(i) 2, ☐, 26
a = 2, a3 = 26 ⇒ 2 + 2d = 26 ⇒ d = 12 ; a2 = 2 + 12 = 14
2, 14, 26
(ii) ☐, 13, ☐, 3
a2 = a + d = 13 ; a4 = a + 3d = 3 ⇒ 2d = –10 ⇒ d = –5 ; a = 18
18, 13, 8, 3
(iii) 5, ☐, ☐, 912
a = 5, a4 = 5 + 3d = 192 ⇒ 3d = 92 ⇒ d = 32
5, 132, 8, 912
(iv) –4, ☐, ☐, ☐, ☐, 6
a = –4, a6 = –4 + 5d = 6 ⇒ d = 2
–4, –2, 0, 2, 4, 6
(v) ☐, 38, ☐, ☐, ☐, –22
a2 = a + d = 38 ; a6 = a + 5d = –22 ⇒ 4d = –60 ⇒ d = –15 ; a = 53
53, 38, 23, 8, –7, –22
4Which term of the AP : 3, 8, 13, 18, . . . ,is 78?
a = 3, d = 5, an = 78
78 = 3 + (n – 1) × 5 ⇒ (n – 1) = 15 ⇒ n = 16
78 is the 16th term.
5Find the number of terms in each of the following APs : (i) 7, 13, 19, . . . , 205 (ii) 18, 1512, 13, . . . , – 47
(i) 7, 13, 19, ..., 205
a = 7, d = 6 ; 205 = 7 + (n – 1) × 6 ⇒ (n – 1) = 33 ⇒ n = 34
Number of terms 34
(ii) 18, 1512, 13, ..., –47
a = 18, d = –52 ; –47 = 18 + (n – 1) × –52 ⇒ (n – 1) × 52 = 65 ⇒ n – 1 = 26 ⇒ n = 27
Number of terms 27
6Check whether – 150 is a term of the AP : 11, 8, 5, 2 . . .
a = 11, d = –3 ; –150 = 11 + (n – 1)(–3) ⇒ 3(n – 1) = 161 ⇒ n = 1643
n is not a positive integer.
No, –150 is not a term of this AP.
7Find the 31st term of an AP whose 11th term is 38 and the 16th term is 73.
a + 10d = 38 ; a + 15d = 73 ⇒ 5d = 35 ⇒ d = 7 ; a = 38 – 70 = –32
a31 = –32 + 30 × 7 = 178
31st term 178
8An AP consists of 50 terms of which 3rd term is 12 and the last term is 106. Find the 29th term.
a + 2d = 12 ; a + 49d = 106 ⇒ 47d = 94 ⇒ d = 2 ; a = 8
a29 = 8 + 28 × 2 = 64
29th term 64
9If the 3rd and the 9th terms of an AP are 4 and – 8 respectively, which term of this AP is zero?
a + 2d = 4 ; a + 8d = –8 ⇒ 6d = –12 ⇒ d = –2 ; a = 8
an = 0 ⇒ 8 + (n – 1)(–2) = 0 ⇒ n – 1 = 4 ⇒ n = 5
The 5th term is zero.
10The 17th term of an AP exceeds its 10th term by 7. Find the common difference.
a17 – a10 = 7 ⇒ (a + 16d) – (a + 9d) = 7 ⇒ 7d = 7
Common difference d = 1
11Which term of the AP : 3, 15, 27, 39, . . . will be 132 more than its 54th term?
a = 3, d = 12 ; a54 = 3 + 53 × 12 = 639
Required term = 639 + 132 = 771
771 = 3 + (n – 1) × 12 ⇒ n – 1 = 64 ⇒ n = 65
The 65th term
12Two APs have the same common difference. The difference between their 100th terms is 100, what is the difference between their 1000th terms?
Let the first terms be a and b, and the common difference be d.
(a + 99d) – (b + 99d) = 100 ⇒ a – b = 100
Difference of 1000th terms = (a + 999d) – (b + 999d) = a – b = 100
Difference between their 1000th terms 100
13How many three-digit numbers are divisible by 7?
The smallest three-digit number divisible by 7 is 105 and the largest is 994.
AP: 105, 112, 119, ..., 994 ; a = 105, d = 7
994 = 105 + (n – 1) × 7 ⇒ n – 1 = 127 ⇒ n = 128
Three-digit numbers divisible by 7 128
14How many multiples of 4 lie between 10 and 250?
The first multiple of 4 after 10 is 12 and the last before 250 is 248.
AP: 12, 16, 20, ..., 248 ; a = 12, d = 4
248 = 12 + (n – 1) × 4 ⇒ n – 1 = 59 ⇒ n = 60
Multiples of 4 between 10 and 250 60
15For what value of n, are the nth terms of two APs: 63, 65, 67, . . . and 3, 10, 17, . . . equal?
nth term of first AP = 63 + (n – 1) × 2 ; nth term of second AP = 3 + (n – 1) × 7
63 + 2(n – 1) = 3 + 7(n – 1) ⇒ 5(n – 1) = 60 ⇒ n = 13
n = 13
16Determine the AP whose third term is 16 and the 7th term exceeds the 5th term by 12.
a7 – a5 = 12 ⇒ 2d = 12 ⇒ d = 6
a3 = a + 2d = 16 ⇒ a = 4
The AP is 4, 10, 16, 22, ...
17Find the 20th term from the last term of the AP : 3, 8, 13, . . ., 253.
Writing the AP in reverse order: 253, 248, 243, ..., 3. Here a = 253, d = –5.
a20 = 253 + 19 × (–5) = 253 – 95 = 158
20th term from the last 158
18The 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.
(a + 3d) + (a + 7d) = 24 ⇒ a + 5d = 12 ... (1)
(a + 5d) + (a + 9d) = 44 ⇒ a + 7d = 22 ... (2)
(2) – (1): 2d = 10 ⇒ d = 5 ; a = 12 – 25 = –13
First three terms –13, –8, –3
19Subba 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?
Salaries: 5000, 5200, 5400, ... ; a = 5000, d = 200
7000 = 5000 + (n – 1) × 200 ⇒ n – 1 = 10 ⇒ n = 11
The 11th year from 1995 is 1995 + 10 = 2005.
His income reached ₹ 7000 in 2005
20Ramkali 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.
a = 5, d = 1.75, an = 20.75
20.75 = 5 + (n – 1) × 1.75 ⇒ n – 1 = 15.751.75 = 9 ⇒ n = 10
n = 10
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Exercise 5.3
1Find the sum of the following APs: (i) 2, 7, 12, . . ., to 10 terms. (ii) –37, –33, –29, . . ., to 12 terms. (iii) 0.6, 1.7, 2.8, . . ., to 100 terms. (iv) 115, 112, 110, . . ., to 11 terms.
We use Sn = n2[2a + (n – 1)d].
(i) 2, 7, 12, ..., to 10 terms
a = 2, d = 5 ; S10 = 5[4 + 9 × 5] = 5 × 49 = 245
245
(ii) –37, –33, –29, ..., to 12 terms
a = –37, d = 4 ; S12 = 6[–74 + 11 × 4] = 6 × (–30) = –180
–180
(iii) 0.6, 1.7, 2.8, ..., to 100 terms
a = 0.6, d = 1.1 ; S100 = 50[1.2 + 99 × 1.1] = 50 × 110.1 = 5505
5505
(iv) 115, 112, 110, ..., to 11 terms
a = 115, d = 112 – 115 = 160
S11 = 112[ 215 + 10 × 160] = 112 × 930 = 3320
3320
2Find the sums given below: (i) 7 + 1012 + 14 + . . . + 84 (ii) 34 + 32 + 30 + . . . + 10 (iii) –5 + (–8) + (–11) + . . . + (–230)
(i) 7 + 1012 + 14 + ... + 84
a = 7, d = 72, l = 84 ; 84 = 7 + (n – 1) × 72 ⇒ n – 1 = 22 ⇒ n = 23
S = n2(a + l) = 232(7 + 84) = 20932
Sum 20932 = 104612
(ii) 34 + 32 + 30 + ... + 10
a = 34, d = –2, l = 10 ; 10 = 34 + (n – 1)(–2) ⇒ n = 13
S = 132(34 + 10) = 286
Sum 286
(iii) –5 + (–8) + (–11) + ... + (–230)
a = –5, d = –3, l = –230 ; –230 = –5 + (n – 1)(–3) ⇒ n – 1 = 75 ⇒ n = 76
S = 762(–5 – 230) = 38 × (–235) = –8930
Sum –8930
3In an AP:
(i) given a = 5, d = 3, an = 50, find n and Sn.
50 = 5 + (n – 1) × 3 ⇒ n = 16 ; S16 = 162(5 + 50) = 440
n = 16, Sn = 440
(ii) given a = 7, a13 = 35, find d and S13.
35 = 7 + 12d ⇒ d = 73 ; S13 = 132(7 + 35) = 273
d = 73, S13 = 273
(iii) given a12 = 37, d = 3, find a and S12.
37 = a + 11 × 3 ⇒ a = 4 ; S12 = 122(4 + 37) = 246
a = 4, S12 = 246
(iv) given a3 = 15, S10 = 125, find d and a10.
a + 2d = 15 ; S10 = 5(2a + 9d) = 125 ⇒ 2a + 9d = 25
Putting a = 15 – 2d: 30 – 4d + 9d = 25 ⇒ d = –1 ; a = 17
a10 = 17 + 9 × (–1) = 8
d = –1, a10 = 8
(v) given d = 5, S9 = 75, find a and a9.
S9 = 92(2a + 8 × 5) = 75 ⇒ 2a + 40 = 503 ⇒ a = –353
a9 = –353 + 8 × 5 = 853
a = –353, a9 = 853
(vi) given a = 2, d = 8, Sn = 90, find n and an.
n2[4 + (n – 1) × 8] = 90 ⇒ n(4n – 2) = 90 ⇒ 2n2 – n – 45 = 0
(n – 5)(2n + 9) = 0 ⇒ n = 5 (n cannot be negative)
a5 = 2 + 4 × 8 = 34
n = 5, an = 34
(vii) given a = 8, an = 62, Sn = 210, find n and d.
Sn = n2(8 + 62) = 35n = 210 ⇒ n = 6
62 = 8 + 5d ⇒ d = 545
n = 6, d = 545
(viii) given an = 4, d = 2, Sn = –14, find n and a.
4 = a + (n – 1) × 2 ⇒ a = 6 – 2n
Sn = n2(a + 4) = n2(10 – 2n) = n(5 – n) = –14 ⇒ n2 – 5n – 14 = 0
(n – 7)(n + 2) = 0 ⇒ n = 7 ; a = 6 – 14 = –8
n = 7, a = –8
(ix) given a = 3, n = 8, S = 192, find d.
82[6 + 7d] = 192 ⇒ 6 + 7d = 48 ⇒ d = 6
d = 6
(x) given l = 28, S = 144, and there are total 9 terms. Find a.
92(a + 28) = 144 ⇒ a + 28 = 32 ⇒ a = 4
a = 4
4How many terms of the AP : 9, 17, 25, . . . must be taken to give a sum of 636?
a = 9, d = 8 ; n2[18 + (n – 1) × 8] = 636 ⇒ n(4n + 5) = 636
4n2 + 5n – 636 = 0 ⇒ (n – 12)(4n + 53) = 0 ⇒ n = 12
Number of terms 12
5The first term of an AP is 5, the last term is 45 and the sum is 400. Find the number of terms and the common difference.
S = n2(5 + 45) = 25n = 400 ⇒ n = 16
45 = 5 + 15d ⇒ d = 83
n = 16, d = 83
6The first and the last terms of an AP are 17 and 350 respectively. If the common difference is 9, how many terms are there and what is their sum?
350 = 17 + (n – 1) × 9 ⇒ n – 1 = 37 ⇒ n = 38
S = 382(17 + 350) = 19 × 367 = 6973
38 terms; sum = 6973
7Find the sum of first 22 terms of an AP in which d = 7 and 22nd term is 149.
149 = a + 21 × 7 ⇒ a = 2
S22 = 222(2 + 149) = 11 × 151 = 1661
S22 = 1661
8Find the sum of first 51 terms of an AP whose second and third terms are 14 and 18 respectively.
d = 18 – 14 = 4 ; a = 14 – 4 = 10
S51 = 512[20 + 50 × 4] = 512 × 220 = 5610
S51 = 5610
9If the sum of first 7 terms of an AP is 49 and that of 17 terms is 289, find the sum of first n terms.
S7 = 72(2a + 6d) = 49 ⇒ a + 3d = 7
S17 = 172(2a + 16d) = 289 ⇒ a + 8d = 17
Subtracting: 5d = 10 ⇒ d = 2 ; a = 1
Sn = n2[2 + (n – 1) × 2] = n2
Sn = n2
10Show that a1, a2, . . ., an, . . . form an AP where an is defined as below: (i) an = 3 + 4n (ii) an = 9 – 5n. Also find the sum of the first 15 terms in each case.
(i) an = 3 + 4n
a1 = 7, a2 = 11, a3 = 15, ... ; an+1 – an = 4 (a constant)
So, it is an AP with a = 7, d = 4.
S15 = 152[14 + 14 × 4] = 152 × 70 = 525
S15 = 525
(ii) an = 9 – 5n
a1 = 4, a2 = –1, a3 = –6, ... ; an+1 – an = –5 (a constant)
So, it is an AP with a = 4, d = –5.
S15 = 152[8 + 14 × (–5)] = 152 × (–62) = –465
S15 = –465
11If the sum of the first n terms of an AP is 4n – n2, what is the first term (that is S1)? What is the sum of first two terms? What is the second term? Similarly, find the 3rd, the 10th and the nth terms.
S1 = 4 – 1 = 3 ; so a1 = 3
S2 = 8 – 4 = 4 ; a2 = S2 – S1 = 1
S3 = 12 – 9 = 3 ; a3 = S3 – S2 = –1
S10 = 40 – 100 = –60 ; S9 = 36 – 81 = –45 ; a10 = –60 – (–45) = –15
an = Sn – Sn–1 = (4n – n2) – [4(n – 1) – (n – 1)2] = 5 – 2n
S1 = a1 = 3, S2 = 4, a2 = 1, a3 = –1, a10 = –15, an = 5 – 2n
12Find the sum of the first 40 positive integers divisible by 6.
AP: 6, 12, 18, ..., 240 ; a = 6, l = 240, n = 40
S40 = 402(6 + 240) = 20 × 246 = 4920
Sum 4920
13Find the sum of the first 15 multiples of 8.
AP: 8, 16, 24, ..., 120 ; a = 8, l = 120, n = 15
S15 = 152(8 + 120) = 15 × 64 = 960
Sum 960
14Find the sum of the odd numbers between 0 and 50.
AP: 1, 3, 5, ..., 49 ; a = 1, d = 2 ; 49 = 1 + (n – 1) × 2 ⇒ n = 25
S25 = 252(1 + 49) = 625
Sum 625
15A contract on construction job specifies a penalty for delay of completion beyond a certain date as follows: ₹ 200 for the first day, ₹ 250 for the second day, ₹ 300 for the third day, etc., the penalty for each succeeding day being ₹ 50 more than for the preceding day. How much money the contractor has to pay as penalty, if he has delayed the work by 30 days?
AP: 200, 250, 300, ... ; a = 200, d = 50, n = 30
S30 = 302[400 + 29 × 50] = 15 × 1850 = 27750
Penalty ₹ 27750
16A sum of ₹ 700 is to be used to give seven cash prizes to students of a school for their overall academic performance. If each prize is ₹ 20 less than its preceding prize, find the value of each of the prizes.
Let the first prize be ₹ a. Here d = –20, n = 7, S7 = 700.
72[2a + 6 × (–20)] = 700 ⇒ 2a – 120 = 200 ⇒ a = 160
The prizes are (in ₹) 160, 140, 120, 100, 80, 60 and 40
17In a school, students thought of planting trees in and around the school to reduce air pollution. It was decided that the number of trees, that each section of each class will plant, will be the same as the class, in which they are studying, e.g., a section of Class I will plant 1 tree, a section of Class II will plant 2 trees and so on till Class XII. There are three sections of each class. How many trees will be planted by the students?
Each class has 3 sections. So, trees planted by Classes I to XII are 3, 6, 9, ..., 36.
a = 3, d = 3, n = 12 ; S12 = 122(3 + 36) = 6 × 39 = 234
Total trees planted 234
18A spiral is made up of successive semicircles, with centres alternately at A and B, starting with centre at A, of radii 0.5 cm, 1.0 cm, 1.5 cm, 2.0 cm, . . . as shown in Fig. 5.4. What is the total length of such a spiral made up of thirteen consecutive semicircles? (Take π = 227)
Length of a semicircle of radius r = πr.
l1 = 0.5π, l2 = 1.0π, l3 = 1.5π, ... (an AP with a = 0.5π, d = 0.5π)
Total length = 132[2 × 0.5π + 12 × 0.5π] = 132 × 7π = 45.5π
= 45.5 × 227 = 143
Total length of the spiral 143 cm
19200 logs are stacked in the following manner: 20 logs in the bottom row, 19 in the next row, 18 in the row next to it and so on (see Fig. 5.5). In how many rows are the 200 logs placed and how many logs are in the top row?
AP: 20, 19, 18, ... ; a = 20, d = –1, Sn = 200
n2[40 + (n – 1)(–1)] = 200 ⇒ n(41 – n) = 400 ⇒ n2 – 41n + 400 = 0
(n – 16)(n – 25) = 0 ⇒ n = 16 or 25
For n = 25, a25 = 20 – 24 = –4, which is not possible. So, n = 16.
Top row = a16 = 20 + 15 × (–1) = 5
16 rows; logs in the top row 5
20In a potato race, a bucket is placed at the starting point, which is 5 m from the first potato, and the other potatoes are placed 3 m apart in a straight line. There are ten potatoes in the line (see Fig. 5.6). A competitor starts from the bucket, picks up the nearest potato, runs back with it, drops it in the bucket, runs back to pick up the next potato, runs to the bucket to drop it in, and she continues in the same way until all the potatoes are in the bucket. What is the total distance the competitor has to run?
Distances of the potatoes from the bucket are 5, 8, 11, ... m. For each potato she runs there and back.
Distances run: 2 × 5, 2 × 8, 2 × 11, ... = 10, 16, 22, ... (a = 10, d = 6, n = 10)
S10 = 102[20 + 9 × 6] = 5 × 74 = 370
Total distance 370 m
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Exercise 5.4 (Optional)
1Which term of the AP : 121, 117, 113, . . ., is its first negative term?
a = 121, d = –4 ; an = 121 + (n – 1)(–4) = 125 – 4n
For an < 0: 125 – 4n < 0 ⇒ n > 3114
The least integer value is n = 32. (Check: a32 = 125 – 128 = –3)
First negative term 32nd term
2The sum of the third and the seventh terms of an AP is 6 and their product is 8. Find the sum of first sixteen terms of the AP.
(a + 2d) + (a + 6d) = 6 ⇒ a + 4d = 3 ⇒ a = 3 – 4d
(a + 2d)(a + 6d) = 8 ⇒ (3 – 2d)(3 + 2d) = 8 ⇒ 9 – 4d2 = 8 ⇒ d = ±12
When d = 12: a = 1.
S16 = 8[2 + 15 × 12] = 8 × 192 = 76
When d = –12: a = 5.
S16 = 8[10 – 15 × 12] = 8 × 52 = 20
S16 = 76 or 20
3A ladder has rungs 25 cm apart (see Fig. 5.7). The rungs decrease uniformly in length from 45 cm at the bottom to 25 cm at the top. If the top and the bottom rungs are 212 m apart, what is the length of the wood required for the rungs?
Number of rungs = 25025 + 1 = 11
The lengths form an AP with a = 45, l = 25 and n = 11.
Total length = 112(45 + 25) = 112 × 70 = 385 cm
Length of wood required 385 cm (= 3.85 m)
4The houses of a row are numbered consecutively from 1 to 49. Show that there is a value of x such that the sum of the numbers of the houses preceding the house numbered x is equal to the sum of the numbers of the houses following it. Find this value of x.
Sum of houses before x = Sx–1 = (x – 1)x2
Sum of houses after x = S49 – Sx = 49 × 502 – x(x + 1)2
(x – 1)x2 = 1225 – x(x + 1)2 ⇒ x2 – x = 2450 – x2 – x ⇒ 2x2 = 2450
x2 = 1225 ⇒ x = 35 (x is positive)
Such a value exists: x = 35
5A small terrace at a football ground comprises of 15 steps each of which is 50 m long and built of solid concrete. Each step has a rise of 14 m and a tread of 12 m (see Fig. 5.8). Calculate the total volume of concrete required to build the terrace.
Volume of 1st step = 14 × 12 × 50 = 6.25 m3
Volume of 2nd step = 24 × 12 × 50 = 12.5 m3, and so on.
The volumes form an AP with a = 6.25, d = 6.25 and n = 15.
S15 = 152[12.5 + 14 × 6.25] = 152 × 100 = 750
Total volume of concrete 750 m3
