RS Aggarwal Solutions Chapter 11 Arithmetic Progression Exercise 11D Class 10 Maths
Chapter Name  RS Aggarwal Chapter 11 Arithmetic Progression 
Book Name  RS Aggarwal Mathematics for Class 10 
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Related Study  NCERT Solutions for Class 10 Maths 
Exercise 11D Solutions
1.Find the sum of each of the following Aps:
(i) 2, 7, 12, 17, ...... to 19 terms.
(ii) 9, 7, 5, 3..... to 14 terms
(iii) – 37, 33, 29, ..... to 12 terms.
(iv) 1/15, 1/12, 1/10, ...... to 11 terms.
(v) 0.6, 1.7, 2.8, ..... to 100 terms
Solution
(i) The given AP is 2, 7, 12, 17, ......
Here, a = 2 and d = 7 – 2 = 5
Using the formula, S_{n} = n/2[2a + (n – 1)d], we have
S_{n} = 19/2[2 × 2 + (19 – 1) × 5]
= 19/2 × (4 + 90)
= 19/2 × 94
= 893
(ii) The given AP is 9, 7, 5, 3, .....
Here, a = 9 and d = 7 – 9 =  2
Using the formula, Sn = n/2[2a + (n – 1)d], we have
S14 = 14/2 [2 × 9 + (14 – 1) × (2)]
= 7 × (18 – 26)
= 7 × (8)
=  56.
(iii) The given AP is 37, 33, 29, ......
Here, a =  37 and d =  33 – (37) =  33 + 37 = 4
Using the formula, S_{n} = n/2[2a + (n – 1)d], we have
S_{12 }= 12/2[2 × (37) + (12 – 1) × 4]
= 6 × (74 + 44)
= 6 × (30)
= 180
(iv) The given AP is 1/15, 1/12, 1/10, .......
Here, a = 1/15 and d = 1/12 – 1/15
= (5 – 4)/60
= 1/60
Using the formula, S_{n }= n/2[2a + (n – 1)d], we have
S_{11} = 11/2[2 × (1/15) + (11 – 1) × 1/60]
= 11/2 × (2/15 + 10/60)
= 11/2 × (18/60)
= 33/20
(v) The given AP is 0.6, 1.7, 2.8, ........
Here, a = 0.6 and d = 1.7 – 0.6 = 1.1
Using formula, S_{n} = n/2[2a + (n – 1)d], we have
S_{100 }= 100/2 [2 × 0.6 + (100 – 1) × 1.1]
= 50 × (1.2 + 108.9)
= 5505
2. Find the sum of each of the following arithmetic series:
(i) 7 + 10.1/2 + 14 + .......84
(ii) 34 + 32 + 30 + .....+ 10
(iii) (5) + (8) + (11) + ....+ (230)
Solution
(i) The given arithmetic series is 7 + 10.1/2 + 14 + .... + 84.
Here, a = 7, d = 10.1/2 – 7 = 21/2 – 7
= (21 – 4)/2
= 7/2 and l = 84.
Let the given series contains n terms. Then,
a_{n} = 84
⇒ 7 + (n – 1) × 7/2 = 84 [a_{n} = a + (n – 1)d]
⇒ (7/2).n + 7/2 = 84
⇒ (7/2).n = 84 – 7/2 = 161/2
⇒ n = 161/7 = 23
∴ Required sum = 23/2 × (7 + 84) [S_{n} = n/2(a + 1)]
= 23/3 × 91
= 2030/2
= 1046.1/2
(ii) The given arithmetic series is 34 + 32 + 30 + .... +10.
Here, a = 34, d = 32 – 34 = 2 and l = 10.
Let the given series contain n terms. Then,
a_{n} = 10
⇒ 34 + (n – 1) × (2) = 10 [a_{n} = a + (n – 1)d]
⇒  2n + 36 = 10
⇒  2n = 10 – 36 =  26
⇒ n = 13
∴ Required sum = 13/2 × (34 + 10) [S_{n }= n/2(a + 1)]
= 13/2 × 44
= 286
(iii) The given arithmetic series is (5) + (8) + (11) + ...... + (230).
Here, a =  5, d =  8 – (5)
=  8 + 5
=  3
and l = 230.
Let the given series contain n terms. Then,
a_{n} =  230
⇒  5 + (n – 1) × (3) =  230 [a_{n} = a + (n – 1)d]
⇒  3n – 2 =  230
⇒  3n =  230 + 2 =  228
⇒ n = 76
∴ Required sum = 76/2 × [(5) + (230)] [S_{n} = n/2(a + 1)]
= 76/2 × (235)
= 8930
3. Find the sum of first n terms of an AP whose nth term is (5 – 6n). Hence, find its sum of its first 20 terms.
Solution
Let a_{n} be the nth term of the AP.
∴ a_{n} = 5 – 6n
Putting n = 1, we get
First term, a = a_{1} = 5 – 6 × 1 =  1
Putting n = 2, we get
a_{2} = 5 – 6 × 2 =  7
Let d be the common difference of the AP.
∴ d = a_{2} – a_{1}
= 7 – (1)
=  7 + 1
=  6
Sum of first n terms of the AP, S_{n}
= n/2[2 × (1) + (n – 1) × (6)] {S_{n} = n/2[2a + (n – 1)d}
= n/2(2 – 6n + 6)
= n(2 – 3n)
= 2n – 3n^{2}
Putting n = 20, we get
S_{20} = 2 × 20 – 3 × 20^{2 }
= 40 – 1200
= 1160
4. The sum of the first n terms of an AP is (3n^{2} + 6n). Find the nth term and the 15^{th} term of this AP.
Solution
Let S_{n} denotes the sum of first n terms of the AP.
∴ S_{n} = 3n^{2} + 6n
⇒ S_{n1} = 3(n – 1)^{2} + 6(n – 1)
= 3(n^{2} – 2n + 1) + 6(n – 1)
= 3n^{2} – 3
∴ nth terms of the AP, a_{n }
S_{n} – S_{n 1}
= (3n^{2 }+ 6n) – (3n^{2} – 3)
= 6n + 3
Putting n = 15, we get
a_{15} = 6 × 15 + 3
= 90 + 3
= 93
Hence, the n^{th} term is (6n + 3) and 15^{th} term is 93.
5. The sum of the first n terms of an AP is given by S_{n} = (3n^{2} – n). Find its
(i) nth term,
(ii) first term and
(ii) common difference
Solution
Given:
S_{n} = (3n^{2 }– n) ...(i)
Replacing n by (n – 1) in (i), we get:
S_{n1} = 3(n – 1)^{2}  (n – 1)
= 3(n^{2} – 2n + 1) n + 1
= 3n^{2} – 7n + 4
(i) Now, T_{n} = (S_{n} – S_{n1})
= (3n^{2} – n) – (3n^{2} – 7n + 4) = 6n – 4
∴ n^{th} term, T_{n }= (6n – 4) .....(ii)
(ii) Putting n = 1 in (ii), we get:
T_{1} = (6 × 1) – 4 = 2
(iii) Putting n = 2 in (ii), we get:
T_{2} = (6 × 2) – 4 = 8
∴ Common difference, d = T_{2} – T_{1}
= 8  2
= 6
6. The sum of the first n terms of an AP is (5n^{2}/2 + 3n/2). Find its nth term and the 20^{th} term of this AP.
Solution
S_{n} = (5n^{2}/2 + 3n/2)
= 1/2 (5n^{2} + 3n) .....(i)
Replacing n by (n – 1) in (i), we get:
S_{n1} = 1/2 × [5(n – 1)^{2} + 3(n – 1)]
= 1/2 × [5n^{2} – 10n + 5 + 3n – 3]
= 1/2 × [5n^{2} – 7n + 2]
∴ T_{n} = S_{n} – S_{n1}
= 1/2(5n^{2} + 3n) – 1/2 × [5n^{2} – 7n + 2]
= 1/2(10n – 2)
= 5n – 1 .....(ii)
Putting n = 20 in (ii), we get
T_{20} = (5 × 20) – 1 = 99
Hence, the 20^{th} term is 99.
7. The sum of the first n term sofa an AP is (3n^{2}/2 + 5n/2). Find its nth term and the 25^{th} term.
Solution
Let S_{n} denotes the sum of first n terms of the AP.
∴ S_{n} = 3n^{2}/2 + 5n/2
⇒ S_{n1} = 3(n – 1)^{2}/2 + {5(n – 1)}/2
= 3(n^{2} – 2n + 1)/2 + 5(n – 1)/2
= (3n^{2 }– n – 2)/2
∴ n^{th} term of the AP, a_{n}
= S_{n} – S_{n1}
= (3n^{2} + 5n)/2 – (3n^{2} – n – 2)/2
= (6n + 2)/2
= 3n + 2
Putting n = 25, we get
a_{25} = 3 × 25 + 1 = 75 + 1 = 76
Hence, the nth term is (3n + 1) and 25^{th} term is 76.
8. How many terms of the AP 21, 18, 15, .... must be added to get the sum 0?
Solution
The given AP is 21, 18, 15, .....
Here, a = 21 and d = 18 – 21 =  3
Let the required number of terms be n. Then,
S_{n} = 0
⇒ n/2[2 × 21 + (n – 1) × (3)] = 0 {S_{n} = n/2 [2a + (n – 1)]}
⇒ n/2(42 – 3n + 3) = 0
⇒ n(45 – 3n) = 0
⇒ n = 0 or 45 – 3n = 0
⇒ n = 0 or n = 15
∴ n = 15 (Number of terms cannot be zero)
Hence, the required number of terms is 15.
9. How many terms of the AP 9, 17, 25, ..... must be taken so that their sum is 636?
Solution
The given AP is 9, 17, 25, ........
Here, a = 9 and d = 17 – 9 = 8
Let the required’ number of terms be n. Then,
S_{n }= 636
⇒ n/2[2 × 9 + (n – 1) × 8] = 636 {S_{n} = n/2[2a + (n – 1)d]}
⇒ n/2(18 + 8n – 8) = 636
⇒ n/2(10 + 8n) = 636
⇒ n(5 + 4n) = 636
⇒ 4n^{2} + 5n – 636 = 0
⇒ 4n^{2} – 48n + 53n – 636 = 0
⇒ 4n(n – 12) + 53(n – 12) = 0
⇒ (n – 12)(4n + 53) = 0
⇒ n – 12 = 0 or 4n + 53 = 0
⇒ n = 12 or n = (53/4)
∴ n = 12 (Number of terms cannot negative)
Hence, the required number of terms is 12.
10. How many terms of AP 63, 60, 57, 54, ...... must be taken so that their sum is 693? Explain the double answer:
Solution
The given AP is 63, 60, 57, 54, .......
Here, a = 63 and d = 60 – 63 =  3
Let the required number terms be n. Then,
S_{n} = 693
⇒ n/2[2 × 63 + (n – 1) × (3)] = 693 {S_{n} = n/2[2a + (n – 1)d]}
⇒ n/2(126 – 3n + 3) = 693
⇒ n(129 – 3n) = 1386
⇒ 3n^{2}  129n + 1386 = 0
⇒ 3n^{2} – 66n – 63n + 1386 = 0
⇒ 3n(n – 22) – 63(n – 22) = 0
⇒ (n – 22)(3n – 63) = 0
⇒ n – 22 = 0 or 3n – 63 = 0
⇒ n = 22 or n = 21
So, the sum of 21 terms as well as that of 22 terms is 693. This is because the 22^{nd} term of the AP is 0.
a_{22} = 63 + (22 – 1) × (3)
= 63 – 63
= 0
Hence, the required number of terms is 21 or 22.
11. How many terms of the AP 20, 19.1/3, 18.2/3, ..... must be taken so that their sum is 300? Explain the double answer.
Solution
The given AP is 20, 19.1/3, 18.2/3, ......
Here, a = 20 and d = 19.1/3 – 20 = 58/3 – 20
= (58 – 60)/3
= (2/3)
Let the required number of terms ne n. Then,
S_{n} = 300
⇒ n/2[2 × 20 + (n – 1) × (2/3)] = 300 {S_{n }= n/2[2a + (n – 1)d]}
⇒ n/2(40 – (2/3)n + 2/3) = 300
⇒ n/2 × (122 – 2n)/3 = 300
⇒ 122n – 2n^{2} = 1800
⇒ 2n^{2} – 122n + 1800 = 0
⇒ 2n^{2} – 50n – 72n + 1800 = 0
⇒ 2n(n – 25) – 72(n – 25) = 0
⇒ (n – 25)(2n – 72) = 0
⇒ n – 25 = 0 or 2n – 72 = 0
⇒ n = 25 or n = 36
So, the sum of first 25 terms as well as that of first 36 terms is 300. This because the sum of all terms from 26^{th} to 36^{th} is 0.
12. Find the sum of all odd numbers between 0 and 50.
Solution
All odd numbers between 0 and 50 are 1,
This is an in which a = 1, d = (3 – 1) = 2 and l = 49.
This is an AP in which a = 1, d = (3 – 1) = 2 and l = 49.
Let the number of terms be n.
Then, T_{n} = 49
⇒ a + (n – 1)d = 49
⇒ 1 + (n – 1) × 2 = 49
⇒ 2n = 50
⇒ n = 25
∴ Required sum = n/2(a + 1)
= 25/2[l + 49]
= 25 × 25
= 625
Hence, the required sum is 625.
13. Find the sum of all natural numbers between 200 and 400 which are divisible by 7.
Solution
Natural numbers between 200 and 400 which are divisible by 7 are 203, 210, .... 399.
This is an AP with a = 203, d = 7 and l = 399.
Suppose there are n terms in the AP. Then,
a_{n} = 399
⇒ 203 + (n – 1) × 7 = 399 [a_{n} = a + (n – 1)d]
⇒ 7n + 196 = 399
⇒ 7n = 399 – 196 = 203
⇒ 7n = 399 – 196 = 203
⇒ n = 29
∴ Required sum = 29/2(203 + 399) [S_{n} = n/2(a + l)]
= 29/2 × 602
= 8729
Hence, the required sum is 8729.
14. Find the sum of first positive integers divisible by 6.
Solution
The positive integers divisible by 6 are 6, 12, 18, ......
This an AP with a = 6 and d = 6.
Also, n = 40 (Given)
Using the formula, S_{n} = n/2[2a + (n – 1)d], we get
S_{40 }= 40/2 [2×6 + (40 – 1)×6]
= 20(12 + 234)
= 20 × 246
= 4920
Hence, the required sum is 4920.
15. Find the sum of first 15 multiples of 8.
Solution
The first 15 multiples of 8 are 8, 16, 24, 32, ......
This is an AP in which a = 8, d = (16 – 8) = 8 and n = 15.
Thus, we have:
l = a + (n – 1)d
= 8 + (15 – 1)8
= 120
∴ Required sum = n/2(a + l)
= 15/2[8 + 120]
= 15 × 64
= 960
Hence, the required sum is 960.
16. Find the sum of all multiples of 9 lying between 300 and 700.
Solution
The multiples of 9 lying between 300 and 700 are 306, 315, ......., 693.
This is AP with a = 306, d = 9 and l = 693.
Suppose these are n terms are n terms in the AP. Then,
a_{n} = 693
⇒ 306 + (n – 1) × 9 = 693 [a_{n} = a + (n – 1)d]
⇒ 9n + 297 = 693
⇒ 9n = 693 – 297 = 396
⇒ n = 44
∴ Required sum = 44/2(306 = 693) [S_{n} = n/2(a + l)]
= 22 × 999
= 21978
Hence, the required sum is 21978.
17. Find the sum of all threedigits natural numbers which are divisible by 13.
Solution
All threedigit numbers which are divisible by 13 are 104, 117, 130, 143, ...... 938.
This is an AP in which a = 104, d = (117 – 104) = 13 and l = 938
Let the number of terms be n.
Then, T_{n }= 938
⇒ a + (n – 1)d = 988
⇒ 104 + (n – 1) × 13 = 988
⇒ 13n = 897
⇒ n = 69
∴ Required sum = n/2(a + l)
= 69/2[104 + 988]
= 69 × 546
= 37674
Hence, the required sum is 37674.
18. Find the sum of first 100 even number which are divisible by 5.
Solution
The first few even natural numbers which are divisible by 5 are 10, 20, 30, 40, ...
This is an AP in which a = 10, d = (20 – 10) = 10 and n = 100
The sum of n terms of an AP is given by
S_{n} = n/2[2a + (n – 1)d]
(100/2) × [2 × 10 + (100 – 1) × 10] [∵ a = 10, d = 10 and n =100]
= 50 × [20 + 990]
= 50 × 1010
= 50500
Hence, the sum of the first hundred even natural numbers which are divisible by 5 is 50500.
19. Find the sum of the following:
(1 – 1/n) + (1 – 2/n) + (1 – 3/n) + ...... upto n terms.
Solution
On simplifying the given series, we get:
(1 – 1/n) + (1 – 2/n) + (1 – 3/n) + ..... n terms
= (1 + 1 + 1 + ..... n terms) – (1/n + 2/n + 3/n + ..... + n/n)
= n – (1/n + 2/n + 3/n + ..... + n/n)
Here, (1/n + 2/n + 3/n + ...... + n/n) is an AP whose first term is 1/n and the common difference is (2/n – 1/n)
= 1/n.
The sum of terms of an AP is given by
S_{n }= n/2[2a + (n – 1)d]
= n – [n/2{2 × (1/n) + (n – 1) × (1/n)}]
= n – [n/2 [(2/n) + (n – 1)/n]]
= n – {n/2(n + 1)/n}
= n – (n + 1)/2
= (n – 1)/2
20. In an AP. It is given that S_{5} + S_{7} = 167 and S_{10} = 235, then find the AP, where S_{n }denotes the sum of its first n terms.
Solution
Let a be the first term and d be the common difference of three AP. Then,
S_{5} + S_{7} = 167
⇒ 5/2(2a + 4d) + 7/2(2a + 6d) = 167 {S_{n} = n/2[2a + (n – 1)d]}
⇒ 5a + 10d + 7a + 21d = 167
⇒ 12a + 31d = 167 ...(1)
Also,
S_{n} = 235
⇒ 10/2(2a + 9d) = 235
⇒ 5(2a + 9d) = 235
⇒ 2a + 9d = 47
Multiplying both sides by 6, we get
12a + 54d = 282 ....(2)
Subtracting (1) from (2), we get
12a + 54d – 12a – 31d = 282 – 167
⇒ 23d = 115
⇒ d = 5
Putting d = 5 in (1), we get
12a + 31 × 5 = 167
⇒ 12a + 155 = 167
⇒ 12a = 167 – 155 = 12
⇒ a = 1
Hence, the AP is 1, 6, 11, 16, ....
21. In an AP, the first term is 2, the last term is 29 and the sum of all terms is 155. Find the common difference.
Solution
Here, a = 2, l = 29 and S_{n} = 155
Let d be the common difference of the given AP and n be the total number of terms.
Then, T_{n} = 29
⇒ a + (n – 1)d = 29
⇒ 2 + (n – 1)d = 29 ...(i)
The sum of n terms of an AP is given by
S_{n} = n/2[a + l] = 155
⇒ n/2[2 + 29] = (n/2) × 31 = 155
⇒ n = 10
Putting the value of n in (i), we get:
⇒ 2 + 9d = 29
⇒ 9d = 27
⇒ d = 3
Thus, the common difference of the given AP is 3.
22. In an AP, the first term is 4, the las term is 29 and the sum of all its terms is 150. Find its common difference.
Solution
Suppose there are n terms in the AP.
Here, a =  4, l = 29 and S_{n} = 150
S_{n} = 150
⇒ n/2 (4 + 29) = 150 [S_{n} = n/2(a + l)]
⇒ n = (150 × 2)/25 = 12
Thus, the AP contains 12 terms.
Let d be the common difference of the AP.
∴ a_{12} = 29
⇒ 4 + (12 – 1) × d = 29 [a_{n} = a + (n – 1)d]
⇒ 11d = 29 + 4 = 33
⇒ d = 3
Hence, the common difference of the AP is 3.
23. The 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?
Solution
Suppose there are n terms in the AP.
Here, a = 17, d = 9 and l = 350.
∴ a_{n} = 350
⇒ 17 + (n – 1) × 9 = 350 [a_{n} = a + (n – 1)d]
⇒ 9n + 8 = 350
⇒ 9n = 350 – 8 = 342
⇒ n = 38
Thus, there are 38 terms in the AP.
∴ S_{38} = 28/2 (17 + 350) [S_{n} = n/2(a + l)]
= 19 × 367
= 6973
Hence, the required sum is 6973.
24. The first and last terms of an AP are 5 and 45 respectively. If the sum of all its terms is 400, find the common difference and the number of terms.
Solution
Suppose there are n term in the AP.
Here, a = 5, l = 45 and S_{n} = 400
S_{n} = 400
⇒ n/2(5 + 45) = 400 [S_{n} = n/2(a + l)]
⇒ n/2 × 50 = 400
⇒ n = (400 × 2)/50
= 16
Thus, there are 16 terms in the AP.
Let d be the common difference of the AP.
∴ a_{16} = 45
⇒ 5 + (16 – 1) × d = 45 [a_{n} = a + (n – 1)d]
⇒ 15d = 45 – 5 = 40
⇒ d = 40/15 = 8/3
Hence, the common difference of the AP is 8/3.
25. In an AP, the first term is 22, nth terms is 11 and sum of first n terms is 66. Find the n and hence find the 4 common difference.
Solution
Here, a = 22, T_{n} = 11 and S_{n} = 66
Let d be the common difference of the given AP.
Then, T_{n} = 11
⇒ a + (n – 1)d = 22 + (n – 1)d = 11
⇒ (n – 1)d =  33 ....(i)
The sum of n terms of an AP is given by
S_{n} = n/2[2a + (n – 1)d] = 66 [Substituting the value of (n – 1)d from (i)]
⇒ n/2[2 × 22 + (33)]
= (n/2) × 11
= 66
⇒ n = 12
Putting the value of n in (i), we get:
11d = 33
⇒ d =  3
Thus, n = 12 and d = 3
26. The 12^{th} term of an AP is 13 and the sum of its first four terms is 24. Find the sum of its first 10 terms.
Solution
Let a be the first term and d be the common difference of the AP. Then,
a_{12} =  13
⇒ a + 11d =  13 ...(1) [a_{n} = a + (n – 1)d]
Also,
S_{4} = 24
⇒ 4/2(2a + 3d) = 24 {S_{n} = n/2[2a + (n – 1)d]}
⇒ 2a + 3d = 12 ...(2)
Solving (1) and (2), we get
2(13 – 11d) + 3d = 12
⇒  26 – 22d + 3d = 12
⇒  19d = 12 + 26 = 38
⇒ d =  2
Putting d =  2 in (1), we get
a + 11 × (2) =  13
⇒ a =  13 + 22 = 9
∴ Sum oof its first 20 terms, S_{10}
= 10/2[2 × 9 + (10 – 1) × (2)]
= 5 × (18 – 18)
= 5 × 0
= 0
Hence, the required sum s is 0.
27. The sum of the first 7 terms of an Ap is 182. If its 4^{th} and 17^{th} terms are in the ratio 1:5, find the AP.
Solution
Let a be the first term and d be the common difference of the AP.
∴ S_{7} = 182
⇒ 7/2(2a + 6d) = 182 {S_{n} = n/2[2a + (n – 1))d]}
⇒ a + 3d = 26 ....(1)
Also,
a_{4} : a_{17} = 1 : 5 (Given)
⇒ (a + 3d)/(a + 16d) = 1/5 [a_{n} = a + (n – 1)d]
⇒ 5a + 155d = a + 16d
⇒ d = 4a ....(2)
Solving (1) and (2), we get
a + 3 × 4a = 26
⇒ 13a = 26
⇒ a = 2
Putting a = 2 in in (2), we get
d = 4 × 2 = 8
Hence, the required AP is 2, 10, 18, 26, ......
28. The sum of first 9 terms of an AP is 81 and that of its first 20 terms is 400. Find the first term and common difference of the AP.
Solution
Here, a = 4, d =7 and l = 81
Let the nth term be 81.
Then T_{n} = 81
⇒ a + (n – 1)d = 4(n – 1)7 = 81
⇒ (n – 1)7 = 77
⇒ (n – 1) = 11
⇒ n = 12
Thus, there are 12 terms in AP.
The sum of n terms of an AP is given by
S_{n} = n/2[a + l]
∴ S_{12} = 12/2 [4 + 81]
= 6 × 85
= 510
Thus, the required sum is 510.
29. The sum of the first 7 terms of an AP is 49 and the sum of its first 17 term is 289. Find the sum of its first n terms.
Solution
Let a be the first term and d be the common difference of the given AP.
Then, we have:
S_{n} = n/2[2a + (n – 1)d]
S_{7} = 7/2 [2a + 6d] = 7[a + 3d]
S_{17} = 17/2 [2a + 16d] = 17[a + 8d]
However, S_{7} = 49 and S_{17} = 289
Now, 7(a + 3d) = 49
⇒ a + 3d = 7 ...(i)
Also, 17[a + 8d] = 289
⇒ a + 8d = 17 ...(ii)
Subtracting (i) from (ii), we get:
5d = 10
⇒ d = 2
Putting d = 2 in (i), we get
a + 6 = 7
⇒ a = 1
Thus, a = 1 and d = 2
∴ Sum of n terms of AP = n/2[2 × 1 + (n – 1) × 2]
= n[1 + (n – 1)]
= n^{2}
^{}
30. Two Aps have the same common difference. If the first terms of these Aps be 3 and 8 respectively. Find the difference between the sums of their first 50 terms.
Solution
Let a_{1} and a_{2} be the first terms of the two APs.
Here, a_{1} = 8 and a_{2} = 3
Suppose d be the common difference of the two Aps
Let S_{50} and S’_{50} = 50/2[2a_{1} + (50 – 1)d] – 50/2[2a_{2} + (50 – 1)d]
= 25(2 × 8 × 49d) – 25(2 × 3 + 49d)
= 25 × (16 – 6)
= 250
Hence, the required difference between the two sums is 250.
31. The sum first 10 terms of an AP is 150 and the sum of its next 10 terms is 550. Find the AP.
Solution
Let a be the first term and d be the common difference of the AP. Then,
S_{10} =  150 (Given)
⇒ 10/2 (2a + 9d) = 150 {S_{n} = n/2[2a + (n – 1)d}
⇒ 5(2a + 9d) =  150
⇒ 2a + 9d =  30 ....(1)
It is given that the sum of its next 1o terms is – 550.
Now,
S_{20} = Sum of first 20 terms = Sum of first 10 terms + Sum of the next 10 terms =  150 + (550)
= 700
∴ S_{20} = 700
⇒ 20/2(2a + 19d) =  700
⇒ 10(2a + 19d) =  700
⇒ 2a + 19d =  70 ...(2)
Subtracting (1) from (2), we get
(2a + 19d) – (2a + 9d) =  70 – (30)
⇒ 10d =  40
⇒ d =  4
Putting d =  4 in (1), we get
2a + 9 × (4) =  30
⇒ 2a =  30 + 36 = 6
⇒ a = 3
Hence, the required AP is 3, 1, 5, 9, ....
32. The 13^{th} terms of an AP is 4 times its 3^{rd} term. If its 5^{th} term is 16. Find the sum of its first 10 terms.
Solution
Let a be the first term and d be the common difference of the AP. Then,
a_{13} = 4 × a_{3} (Given)
⇒ a + 12d = 4(2 + 2d) [a_{n} = a + (n – 1)d]
⇒ a + 12d = 4a + 8d
⇒ 3a = 4d ...(1)
Also,
a_{5} = 16 (Given)
⇒ a + 4d = 16 ...(2)
Solving (1) and (2), we get
a + 3a = 16
⇒ 4a = 16
⇒ a = 4
Putting a = 4 in (1), we get
4d = 3 × 4 = 12
⇒ d = 3
Using the formula, S_{4} = n/2[2a + (n – 1)d], we get
S_{10} = 10/2[2 × 4 + (10 – 1) × 3]
= 5 × (8 + 27)
= 5 × 35
= 175
Hence, the required sum is 175.
Solution
Let a be the first term and d be the common difference of the AP. Then,
a_{16} = 5 × a_{3} (Given)
Let a_{1} and a_{2} be the first terms of the two APs.
Here, a_{1} = 8 and a_{2} = 3
Suppose d be the common difference of the two Aps
Let S_{50} and S’_{50} = 50/2[2a_{1} + (50 – 1)d] – 50/2[2a_{2} + (50 – 1)d]
= 25(2 × 8 × 49d) – 25(2 × 3 + 49d)
= 25 × (16 – 6)
= 250
Hence, the required difference between the two sums is 250.
31. The sum first 10 terms of an AP is 150 and the sum of its next 10 terms is 550. Find the AP.
Solution
Let a be the first term and d be the common difference of the AP. Then,
S_{10} =  150 (Given)
⇒ 10/2 (2a + 9d) =  150 {S_{n} = n/2[2a + (n – 1)d}]
⇒ 5(2a + 9d) =  150
⇒ 2a + 9d =  30 ...(1)
It is given that the sum of its next 1o terms is – 550.
Now,
S_{20} = Sum of first 20 terms = Sum of first 10 terms + Sum of the next 10 terms =  150 + (550)
= 700
∴ S_{20} = 700
⇒ 20/2(2a + 19d) =  700
⇒ 10(2a + 19d) =  700
⇒ 2a + 19d =  70 ...(2)
Subtracting (1) from (2), we get
(2a + 19d) – (2a + 9d) =  70 – (30)
⇒ 10d =  40
⇒ d =  4
Putting d =  4 in (1), we get
2a + 9 × (4) = 30
⇒ 2a =  30 + 36 = 6
⇒ a = 3
Hence, the required AP is 3, 1, 5, 9, ....
32. The 13^{th} terms of an AP is 4 times its 3^{rd} term. If its 5^{th} term is 16. Find the sum of its first 10 terms.
Solution
Let a be the first term and d be the common difference of the AP. Then,
a_{13} = 4 × a_{3} (Given)
⇒ a + 12d = 4(2 + 2d) [a_{n} = a + (n – 1)d]
⇒ a + 12d = 4a + 8d
⇒ 3a = 4d ...(1)
Also,
a_{5} = 16 (Given)
⇒ a + 4d = 16 ...(2)
Solving (1) and (2), we get
a + 3a = 16
⇒ 4a = 16
⇒ a = 4
Putting a = 4 in (1), we get
4d = 3 × 4 = 12
⇒ d = 3
Using the formula, S_{4} = n/2[2a + (n – 1)d], we get
S_{10} = 10/2[2 × 4 + (10 – 1) × 3]
= 5 × (8 + 27)
= 5 × 35
= 175
Hence, the required sum is 175.
33. The 16^{th} term of an AP is 5 times its 3^{rd} term. If its 10^{th} term is 41, find the sum of its first 15 terms.
Solution
Let a be the first term and d be the common difference of the AP. Then,
a_{16} = 5 × a_{3 } (Given)
⇒ a + 15d = 5(a + 2d) [a_{n} = a + (n – 1)d]
⇒ a + 15d = 5a + 10d
⇒ 4a = 5d ...(1)
Also,
a_{10 }= 41 (Given)
⇒ a + 9d – 41 ...(2)
Solving (1) and (2), we get
a + 9 × 4a/5 = 41
⇒ (5a + 36a)/5 = 41
⇒ 41a/5 = 41
⇒ a = 5
Putting a = 5 in (1), we get
5d = 4 × 5 = 20
⇒ d = 4
Using the formula, S_{n} = n/2[2a + (n – 1)d], we get
S_{n} = 15/2[2 × 5 + (15 – 1) × 4]
= 15/2 × (10 + 56)
= 15/2 × 66
= 495
Hence, the required sum is 495.
34. An AP 5, 12, 19, ..... has 50 term. Find its last term. Hence, find the sum of its last 15 terms.
Solution
The given AP is 5, 12, 19, .....
Here, a = 5, d = 12 – 5 = 7 and n = 50.
Since, there are 50 terms in the AP, so the last term of the AP is a50.
l = a_{50} = 5+ (50 – 1) × 7 [a_{n} = a + (n – 1)d]
= 5 + 343
= 348
Thus, the last term of the AP is 348.
Now,
Sum of the last 15 terms of the AP.
= S_{50} – S_{35}
= 50/2[2 × 5 + (50 – 1) × 7] – 35/2[2 × 5 + (35 – 1) × 7]
{S_{n} = n/2[2a + (n – 1)d]}
= 50/2 × (10 + 343) – 35/2 × (10 + 238)
= 50/2 × 353 – 35/2 × 248
= (17650 – 8680)/2
= 8970/2
= 4485
Hence, the require sum is 4485.
35. An AP 8, 10, 12, ...has 60 terms. Find its last term. Hence, find the sum of its terms.
Solution
The given AP is 8, 10, 12, ......
Here, a = 8, d = 10 – 8 = 2 and n = 60
Since, there are 60 terms in the AP, so the last term of the AP is a60.
l = a_{60} = 8 + (60 – 1)×2 [a_{n} = a + (n – 1)d]
= 8 + 118
= 126
Thus, the last term of the AP is 126.
Now,
Sum of the last 10 terms of AP
= S_{60} – S_{50}
= 60/2[2 × 8 + (60 – 1) × 2] – 50/2[2 × 8 + (50 – 1) × 2]
{S_{n} = n/2[2a + (n – 1)d]}
= 30 × (16 + 118) – 25 × (16 + 98)
= 30 × 134 – 25 × 114
= 4020 – 2850
= 1170
Hence, the required sum is 1170.
36. The sum of the 4^{th} and 8^{th} terms of an AP is 24 and the sum of its 6^{th} and 10^{th }terms is 44.
Find the sum of its first terms.
Solution
Let a be the first and d be the common difference of the AP.
∴ a_{4} + a_{8} = 24 (Given)
⇒ (a + 3d) + (a + 7d) = 24 [a_{n} = a + (n – 1)d]
⇒ 2a + 10d = 24
⇒ a + 5d = 12 ...(1)
Also,
∴ a_{6 }+ a_{10} = 44 (Given)
⇒ (a + 5d) + (a + 9d) = 44 [a_{n} = a + (n – 1)d]
⇒ 2a + 14d = 44
⇒ a + 7d = 22 ...(2)
Subtracting (1) from (2), we get
(a + 7d) – (a + 5d) = 22 – 12
⇒ 2d = 10
⇒ d = 5
Putting d = 5 in (1), we get
a + 5 × 5 = 12
⇒ a = 12 – 25 = 13
Using the formula, S_{n} = n/2[2a + (n – 1)d], we get
S_{10} = 10/2[2 × (13) + (10 – 1) × 5]
= 5 × (26 + 45)
= 5 × 19
= 95
Hence, the required sum is 95.
37. The sum of first m terms of an AP is (4m^{2} – m). If its nth term is 107, find the value of n. Also, find the 21^{st} term of this AP.
Solution
Let S_{m} denotes the sum of the first m terms of the AP. Then,
S_{m }= 4m^{2} – m
⇒ S_{m1} = 4(m – 1)^{2} – (m – 1)
= 4(m^{2} – 2m + 1) – (m – 1)
= 4m^{2} – 9m + 5
Suppose am denote the m^{th} term of the AP.
∴ a_{m} = S_{m }– S_{m1}
= (4m^{2} – m) – (4m^{2} – 9m + 5)
= 8m – 5 ...(1)
Now,
a_{n} = 107 (Given)
⇒ 8n – 5 = 107 [From (1)]
⇒ 8n = 107 + 5 = 112
⇒ n = 14
Thus, the value of n is 14.
Putting m = 21 in (1), we get
a_{21} = 8 × 21 – 5 = 168 – 5 = 163
Hence, the 21^{st} term of the AP is 163.
38. The sum of first q terms of an AP is (63q – 3q^{2}). If its pth term is 60, find the value of p. Also, find the 11^{th} term of its AP.
Solution
Let S_{q} denote the sum of the first q terms of the AP. Then,
S_{q} = 63q – 3q2
⇒ S_{q1} = 63(q – 1) – 3(q – 1)^{2} = 63q – 63 – 3(q^{2} – 2q + 1)
=  3q^{2} + 69q – 66
Suppose a_{q} denote the q^{th} term of the AP.
∴ a_{q} = S_{q} – S_{q1}
= (63q – 3q^{2}) – (3q^{2} + 69q – 66)
= 6q + 66 ...(1)
Now,
a_{p} =  60 (Given)
⇒  6p + 66 = 60 [From (1)]
⇒ 6p = 60 – 66 = 126
⇒ p = 21
Thus, the value of p is 21.
Putting q = 11 in (1), we get
a_{11} = 6×11 + 66 =  66 + 66 = 0
Hence, the 11^{th} term of the AP is 0.
39. Find the number of terms of the AP 12, 9, 6, ...., 21. If 1 is added to each term of this AP then the sum of all terms of the AP thus obtained.
Solution
The given AP is 12, 9, 6, ....., 21.
Here, a =  12d, d = 9 – (12)
= 9 + 12 = 3 and l = 2l
Suppose there are n terms in the AP.
∴ l = a_{n }= 21
⇒  12 + (n – 1) × 3 = 21 [a_{n} = a + (n – 1)d]
⇒ 3n – 15 = 21
⇒ 3n = 21 + 15 = 36
⇒ n= 12
Thus, there are 12 terms in the AP.
If 1 is added to each term of the AP, then the new AP so obtained is 11, 8, 5, ....., 22.
Here, first term, A =  11; last term, L = 22 and n = 12
∴ Sum of the terms of this AP
= 12/2 (11 + 22) [S_{n} = n/2(a + 1)]
= 6 × 11
= 66
Hence, the required sum is 56.
40. Sum of the first 14 terms of an AP is 1505 and its first term is 10. Find its 25^{th} term.
Solution
Here, a = 10 and n = 14
Now,
S_{14} = 1505 (Given)
⇒ 14/2 [2 × 10 + (14 – 1) × d] = 1505 {S_{n} = n/2[2a + (n – 1)d]}
⇒ 7(20 + 13d) = 1505
⇒ 20 + 13d = 215
⇒ 13d = 215 – 20 = 195
⇒ d = 15
∴ 25^{th} term of the AP, a_{25}
= (10 + (25 – 1) × 15 [a_{n} = a + (n – 1)d]
= 10 + 360
= 370
Hence, the required term is 370.
41. Find the sum of first 51 terms of an AP whose second and third terms are 14 and 18 respectively.
Solution
Let a be the first term and d be the common difference of the AP. Then,
d = a_{3} – a_{2} = 18 – 14 = 4
Now,
a_{2} = 14 (Given)
⇒ a + d = 14 [a_{n} = a + (n – 1)d]
⇒ a + 4 = 14
⇒ a = 14 – 4 = 10
Using the formula, S_{n} = n/2[2a + (n – 1)d], we get
S_{51} = 51/2[2 × 10 + (51 – 1) × 4]
= 51/2(20 + 200)
= 51/2 × 220
= 5610
Hence, the required sum is 5610.
42. In a school, students decided to plant 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 double of the class in which they are studying. If there are 1 to 12 classes in the school and each class how two section, find how many trees were planted by student. Which value is shown in the question?
Solution
Number of trees planted by the students of each section of class 1 = 2
There are two section of class 1.
∴ Number of trees planted by the students of class l = 2 × 2 = 4
Number of trees planted by the students of each section of class 2 = 4
There are two sections of class 2.
∴ Number of trees planted by the students of class 2 = 2 × 4 = 8
Similarly,
Number of trees planted by the students of class 3 = 2 × 6 = 12
So, the number of trees planted by the students of different classes are 4, 8, 12, .....
∴ Total number of trees planted by the students = 4 + 8 + 12 + ...... up to 12 terms
This series is an arithmetic series.
Here, a = 4, d = 8 – 4 = 4 and n = 12
Using the formula, S_{n} = n/2[2a + (n – 1)d], we get
S_{12} = 12/2[2 × 4 + (12 – 1) × 4]
= 6 × (8 + 44)
= 6 × 52
= 312
Hence, the total number of trees planted by the students is 312. The values shown in the question are social responsibility and awareness for conserving nature.
43. In a potato race, a bucket is placed at the starting point, which is 5 m from the potato, and the other potatoes are placed 3 m apart in a straight line. There are 10 potatoes in the line. 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 he continues in the same way until all the potatoes are in the bucket. what is the total distance the competitor has to run?
SolutionDistance covered by the competitor to pick and drop the first potato = 2 × 5m
= 10 m
Distance covered by the competitor to pick and drop the second potato = 2 × (5 + 3)m = 2 × 8 m = 16 m
Distance covered by the competitor to pick and drop the third potato = 2 × (5 + 3 + 3)m
= 2 × 11m
= 22m and so on.
∴ Total distance covered by the competitor = 10m + 16m + 22m + ....... upto 10 terms
This is an arithmetic series.
Here, a = 10, d = 16 – 10 = 6 and n = 10
Using the formula, S_{n} = n/2[2a + (n – 1)d], we get
S_{10} = 10/2[2 × 10 + (10 – 1) × 6]
= 5 × (20 + 54)
= 5 × 74
= 370
Hence, the total distance the competitor has to run is 370 m.
44. There ere 25 trees at equal distance of 5 m in a line with a water tank, the distance of the water tank from the nearest tree being 10 m. A gardener waters all the trees separately, starting from the water tank and returning back to the water tank after watering each tree to get water for the next. Find the total distance covered by the gardener in order to water all the trees.
SolutionDistance covered by the gardener to water the first tress and return to the water tank = 10 m + 10 m
= 20 m
Distance covered by the gardener to water the second tree and return to the water tank = 15m + 15m = 30 m
Distance covered by the gardener to water the third tree and return to the water tank = 20 m + 20m
= 40 m and so on.
∴ Total distance covered by the gardener to water all the trees = 20m + 30 m + 40 m + .... up to 25 terms.
This series is an arithmetic series.
Here, a = 20, d = 30 – 20 = 10 and n = 25
Using the formula, S_{n} = n/2[2a + (n – 1)d], we get
S_{25} = 25/2[2 × 20 + (25 – 1) × 10]
= 25/2(40 + 240)
= 25/2
= 280
= 3500
Hence, the total distance covered by the gardener to water all the trees 3500 m.
45. A sum of ₹700 is to be 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 prize.
Solution
Let the value of the first prize be a.
Since the value of each prize is 20 less than its preceding prize, so the values of the prizes are in AP with common difference = ₹20.
⇒ 40/2[2a + (40 – 1)d] = 36000
∴ d = ₹
⇒ 20(2a + 39d) = 36000
⇒ 2a + 39d = 1800 ......(2)
Number of cash prizes to be given to the students, n = 7
Total sum of the prizes, S_{7 }= ₹700
Using the formula, S_{n} = n/2[2a + (n – 1)d], we get
S_{7} = 7/2[2a + (7 – 1) × (20)] = 700
⇒ 7/2(2a – 120) = 700
⇒ 7a – 420 = 700
⇒ 7a = 700 + 420 = 1120
⇒ a = 160
Thus, the value of the first prize is ₹160.
Hence, the value of each prize is ₹160, ₹140, ₹120, ₹100, ₹80, ₹ 60 and ₹40.
46. A man saved ₹33000 in 10 months. In each month after the first, he saved ₹ 100 more than he did in the preceding month. How much did he save in the first month?
Solution
Let the money saved by the man in the first month be ₹a
It is given that in each month after the first, he saved ₹100 more than he did in the preceding month. So, the money saved by the man every month is in AP with common difference ₹ 100.
∴ d = ₹100
Number of months, n = 10
Sum of money saved in 10 months, S_{10} = ₹ 330,000
Using the formula, S_{n }= n/2[2a + (n – 1)d], we get
S_{10} = 10/2[2a + (10 – 1) × 100] = 33000
⇒ 5(2a + 900) = 33000
⇒ 2a + 900 = 6600
⇒ 2a = 6600 – 900 = 5700
⇒ a = 2850
Hence, the money saved by the man in the first month is ₹2,850.
47. A man arranges to pay off debt of ₹ 36000 by 40 monthly instalments which form an arithmetic series. When 30 of the installment are paid, he dies leaving onthird of the debt unpaid. Find the value of the first instalment.
Solution
Let the value of the first installment be ₹a.
Since the monthly installments form an arithmetic series, so let us suppose the man increases the value of each installment be ₹ d every month.
∴ Common difference of the arithmetic series = ₹d
Amount paid in 30 installments = ₹36,000 – 1/3 × ₹ 36,000
= ₹36,000  ₹12,000
= ₹ 24,000
Let S_{n} denote the total amount of money paid in the n installments. Then,
S_{30} = 24,000
⇒ 30/2 [2a + (30 – 1)d] = 24000 {S_{n} = n/2[2a + (n – 1)d]}
⇒ 15(2a + 29d) = 24000
⇒ 2a + 29d = 1600 ....(1)
Also,
S_{40} = ₹36,000
⇒ 40/2[2a + (40 – 1)d] = 36000
⇒ 20(2a + 39d) = 36000
⇒ 2a + 39d = 1800 ...(2)
Subtracting (1) from (2), we get
(2a + 39d) – (2a + 29d) = 1800 – 1600
⇒ 10d = 200
⇒ d = 20
Putting d = 20 in (1), we get
2a + 29 × 20 = 1600
⇒ 2a + 580 = 1600
⇒ 2a = 1600 – 580 = 1020
⇒ a = 150
Thus, the value of the first installment is ₹510.
48. A contact on construction job specifies a penalty for delay of completion beyond a certain date as follows: ₹200 just 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?
Solution
It is given that the penalty for each succeeding day is 50 more than for the preceding day, so the amount of penalties are in AP with common difference ₹50.
Number of days in the delay of the work = 30
The amount of penalties are ₹200, ₹ 250, ₹300, .... upto 30 terms.
∴ Total amount of money paid by the contractor as penalty,
S_{30} = ₹ 200 + ₹250 + ₹ 300 + .... upto 30 terms
Here. a = ₹ 200, d = ₹ 50 and n = 30
Using the formula, S_{n} = n/2[2a + (n – 1)d)], we get
S_{30} = 30/2[2 × 200 + (30 – 1) × 50]
= 15(400 – 1450)
= 15 × 1850
= 27750
Hence, the contractor has to pay ₹27,750 as penalty.