5 ARITHMETIC PROGRESSION CHAPTER

CHAPTER
5
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ARITHMETIC PROGRESSION
Points to Remember :
1. A sequence is an arrangement of numbers or objects in a definite order.
2. A sequence a1, a2, a3, ......., an, ...... is called an Arithmetic Progression (A.P) if there exists a constant d
such that a2 – a1 = d, a3 – a2 = d, a4 – a3 = d, ....., an – an–1 = d and so on. The constant d is called the
common difference.
3. If ‘a’ is the first term and ‘d’ the common difference of an A.P., then the A.P. is a, a + d, a + 2d, a + 3d....
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4. The nth term of an A.P. with first term ‘a’ and common difference ‘d’ is given by an  a  (n  1)d .
5. The sum to n terms of an A.P. with first term ‘a’ and common difference ‘d’ is given by
n
Sn  [2a  (n  1)d ]
2
n
Also, S n  [a  l ] , where l = last term.
2
6. Sum of first n natural numbers = 1 + 2 + 3 + ..... n 
B
n(n  1)
.
2
ILLUSTRATIVE EXAMPLES
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Example 1. A sequence an is defined by an = 5 – 2n. Prove that it is an AP.
Solution.
Given an  5  2n

an 1  5  2(n  1)  5  2n  2  2n  7
Consider, an  an 1  (5  2n)  (2n  7)
= 5 – 2n + 2n – 7 = – 2.
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We observe that an  an 1 is independent of n and hence a constant.
 The given sequence is an A.P. with common difference –2.
Example 2. Which term of the A.P. 5, 15, 25, .... will be 130 more than its 31st term?
Solution.
[CBSE 2006(C)]
We have, a  5, d  10
 a31  a  (31  1)d  a  30d  5  30 10  305.
Let nth term of the given A.P. be 130 more than its 31st term.
Then, an = 130 + a31
 a + (n – 1) d = 130 + 305
 5 + 10 (n – 1) = 435  10 (n – 1) = 430  n – 1 = 43  n = 44.
Hence, 44th term of the given A.P. is 130 more than its 31st term.
1
2
Example 3. Which term of the A.P. 19, 18 , 17 ,.... is the first negative term?
5
5
Solution.
1
2
1 4
Here, a  19, d  18  19  17  18 
.
5
5
5 5
MATHEMATICS–X
ARITHMETIC PROGRESSIONS
71
Let the nth term of the given A.P. be the first negative term.
i.e. a + (n – 1) d < 0

 4
19  (n  1)      0
 5

95 – 4n + 4 < 0  99 < 4n  n 
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99
3
 n  24
4
4
3
is 25.
4
 The first negative term of given A.P. is 25th term.
Example 4. How many numbers of two digits are exactly divisible by 8?
Solution. We observe that 16 is the first two digit number divisible by 8 and 96 is the last two digit number
divisible by 8. Thus, we have to determine the number of terms in the sequence 16, 24, 32, ...., 96.
Clearly, it is an A.P. with first term = 16 and common difference = 8, i.e. a = 16, d = 8.
Let there be n terms in this A.P. Then, an = 96
 a + (n – 1) d = 96
 16 + (n – 1) (8) = 96
 8 (n – 1) = 80  n – 1 = 10  n = 11
Hence, there are 11 numbers of two digits which are divisible by 8.
Example 5. Find the sum of the first 25 terms of an A.P. whose nth term is given by an = 7 – 3n. [CBSE 2004]
 The smallest natural number satisfying n  24
Solution.
We have, an  7  3n
Putting n = 1, we get a1  7  3(1)  4
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Putting n = 25, we get a25  7  3(25)  68
so, for the given A.P., we have, first then = a = 4 and 25th term = – 68.
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 Required sum = S 25 

25
[a  a25 ]
2
25
(4  68)
2
n


 Sn  2 (a  l ) 


 25  (32)  800 Ans.
Example 6. How many terms of the series 54, 51, 48, ..... be taken so that their sum is 513?
[CBSE 2005]
Solution. Clearly, the given sequence is an A.P. with first term a ( = 54) and common difference d (= – 3). Let
the sum of n terms be 513. Then,
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S n  513





n
[2a  (n  1)d ]  513
2
n
[108  (n  1)(3)]  513
2
n2 – 37n + 342 = 0
(n – 18) (n – 19) = 0
n = 18 or n = 19
Here, the common difference is negative. So, 19th term is given by a19  54  (19  1)  (3)  0.
Thus, the sum of 18 terms as well as that of 19 terms is 513.
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ARITHMETIC PROGRESSIONS
MATHEMATICS–X
3n 2 5n
+ . Find its 25th term.
2
2
Let Sn denote the sum of n terms of an A.P. whose nth term is an.
Example 7. In an A.P., the sum of first n terms is
Solution.
We have, S n 
[CBSE 2006(C)]
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3n 2 5n

2
2
3
5
(n  1)2  (n  1)2
2
2

S n 1 

an  Sn  Sn 1
[ replacing n by (n – 1)]
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 3n 2 5n   3
5


    (n  1) 2  (n  1)2 
2  2
2

 2
3
5
 [n 2  (n  1)2 ]  [n  (n  1)]
2
2
3
5
 (2n  1) 
2
2
3
5
 a25  (2  25  1)   76 Ans.
2
2
Example 8. If the sum of the first 10 terms of an A.P. is 110 and the sum of first 20 terms is 420, find the sum
of first 30 terms of this A.P.
Solution. Let the first term be a and the common difference be d. Now,
Given S10  110
10
[2a  (10  1)d ]  110
2
 2a + 9d = 22
Also, S20 = 420

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20
[2a  (20  1)d ]  420

2
 2a + 19d = 42
Subtracting (1) from (2), we get
10d = 20  d = 2
Using d = 2 in (1), we get
2a + 9 (2) = 22  a = 2
n
Now, Sn  [2a  (n  1)d ]
2
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...(1)
...(2)
30
[2(2)  (30  1)(2)]
2
= 15 × 62 = 930 Ans.
Example 9. A contract on construction job specifies a penalty for delay of completion beyond a certain date
as follows : Rs. 200 for the first day, Rs. 250 for the second day, Rs. 300 for the third day, etc. the
penalty for each succeeding day being Rs. 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?
[NCERT]
Solution. Clearly, the amount of penalty for different days forms an A.P. with first term a (= 200) and common
difference d (=50). According to question, we have to find the sum of 30 terms of this A.P.

S30 
MATHEMATICS–X
ARITHMETIC PROGRESSIONS
73
n
Now, Sn  [2a  (n  1)d ]
2
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here, a  200, d  50, n  30
30
[2(200)  (30  1)(50)]
2
=15 (400 + 1450)
= 15 × 1850 = 27750
Thus, a delay of 30 days will cost the contractor of Rs. 27,750.
Example 10. A spiral is made up of successive semicircles, with centres alternatively at A and B, starting with
centre at A, of radii 0.5 cm, 1.0 cm, 2.0 cm, as shown in figure. What is the total length of such a

Required sum  S30 
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22 

spiral made up of thirteen consecutive semicircles?  Take   
7 

l3
l1
A B
l2
Solution.
B
r1 = 0.5 cm, r2 = 1.0 cm, r3 = 1.5 cm, ......
here, a = 0.5, d = 1.0 – 0.5 = 0.5
length of spiral made up of 13 consecutive semicircles
 r1  r2  ......  r13
 (r1  r2  ......  r13 )
=  (0.5 + 1.0 + 1.5 + .... upto 13th term)
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13

   [2(0.5)  (13  1)(0.5)]
2


[NCERT]
13  22 13
    7     7  143 cm Ans.
2
 7 2
Example 11. 200 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 diagram). In how many rows the 200 logs are placed and how
many logs are in the top row?
[NCERT]
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Solution.
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Here, an = 20, d = 1, Sn = 200
Now, an = 20
 a + (n – 1) d = 20
 a + (n – 1) (1) = 20
 a + n = 21
...(1)
ARITHMETIC PROGRESSIONS
MATHEMATICS–X
n
Also, S n  [a  an ]
2
n
200  [a  20]
2
 na + 20 n = 400
from (1) and (2), we get
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
n (21 – n) + 20 n = 400
...(2)
( a  21  n)
2
 n – 41n + 400 = 0
 (n – 16) (n – 25) = 0
 n = 16 or n = 25
`
(i) Taking n = 25, we get
a + (n – 1) d = 20
 a + (25 – 1) (1) = 20  a = – 4, which is not possible as number of logs cannot be negative.
 n = 25 is rejected.
(ii) Taking n = 16, we get
a + (n – 1) d = 20
 a + (16 – 1) (1) = 20
 a=5
 Number of logs in the first row = 5, and the number of rows = 16 Ans.
Example 12. In 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
diagram). Find the total distance travelled for placing the ten potatoes from a line into the
bucket.
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Solution.
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Clearly, the competitor is to cover double distance in order to get the potato from the place and to
be placed in the bucket.
Distance between bucket and Ist potato = 5 m
Distance between bucket and 2nd potato = (5 + 3) m = 8 m
Distance between bucket and 3rd potato = (8 + 3) m = 11 m
.......................................................
.......................................................
here, a = 5, d = 8 – 5 = 3
 Total distance travelled for placing 10 potatoes in the bucket
= 2(5 + 8 + 11 + ........ upto 10 terms)
10

 2  (2  5  (10  1)3) 
2

= 2 [5 (10 + 27)] = 370 m Ans.
MATHEMATICS–X
ARITHMETIC PROGRESSIONS
75
Example 13. A ladder has rungs 25 cm apart (see figure). The rungs decrease uniformly in length from 45 cm
at the bottom and 25 cm at the top. If the top and the bottom rungs are 2
length of the wood required for the rungs?
Solution.
1
Distance between rung at the top and the bottom  2 m=250 cm.
2
Distance between two consecutive rungs = 25 cm.
1
m apart, what is the
2
[NCERT]
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25 cm
250
 1  10  1  11.
25
Length of rungs is increasing uniformly and forming an A.P.
 a = 25, an = 45
 Length of wood required to make rungs
Number of rungs required 
n
 Sn  [a  l ]
2
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45 cm
m
2 1 cm
2
25 cm
11
 [25  45]  385 cm  3.85 m Ans.
2
Example 14. The 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 number x is equal to the sum of the
numbers of the houses following it. Find this value of x.
[NCERT]
Solution. Let the number of houses before house no. x = x – 1.
B
49 houses
1
2
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3
...
x–1
(x–1) houses
x
x+1
x+2
. . . 49
(49 – x) houses
Since, house numbers are consecutive, so sum of house numbers preceding house x
x ( x  1)
2
Now, the number of houses following x = 49 – x
So, sum of house numbers following x
= (x + 1) + (x + 2) + ....... + 49
 1  2  3  ....  ( x  1) 
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
49  x
[2( x  1)  (49  x  1)(1)]
2

49  x
(50  x)
2
n(n  1) 

 Sn 
2 
x( x  1) (49  x)

(50  x )
2
2
x2 – x = 49 (50 + x) – x (50 + x)
2x2 = 49 × 50
x2 = 49 × 25 = 72 × 52 = (7 × 5)2 = (35)2
x = 35 Ans.
According to the given question,




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ARITHMETIC PROGRESSIONS
MATHEMATICS–X
Example 15. A small terrace at a football ground comprises of 15 steps each of which is 50 m long and built
1
1
m and a tread of m (see figure). Calculate the total
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2
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Solution.
[NCERT]
Volume of concrete needed to make first step
1 1
25
 lbh    50 m3 = m3
4 2
4
Volume of concrete needed to make second step
1 1
50
 2    50 m3 = m3
4 2
4
Volume of concrete needed to make third step
1 1
75
 3    50 m3 = m3
4 2
4
1m
2
1m
4

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25
50 25 25
,d 


4
4
4
4
Total volume of concrete needed to make such 15 steps
here, a 
B

15  25
25 
2   (15  1)  

2
4
4

15  50 350  15 400

 
 750 m3 Ans.


24
4  2
4
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PRACTICE EXERCISE
Problem based on nth term of an A.P.
m
volume of concrete required to build the terrace.
50
of solid concrete. Each step has a rise of
Determine which of the following are A.Ps. If they form an AP, find the common difference d and write three
more terms (Qs. 1-6) :
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1. 0, –4, –8, –12, ............
3. 6, 6, 6, 6, 6, ............
5. 10, 10 + 51, 10 + 52, 10 + 53, ..............
2. 1, 2, 4, 8, 16, ...............
4. 5, 5  3, 5  2 3, 5  3 3,...........
6.
2, 8, 18, 32,..........
7. Find the sequence whose of nth term is given by :
(a) 2n – 7
8.
9.
10.
11.
(b) –3 + 5n2
(c)
4
 6n
3
(d)
n2
n 1
Also, determine which of these sequences are A.P’s.
Find the A.P. whose nth term is given by :
(a) 9 – 5n
(b) –n + 6
(c) 2n + 7
th
th
Find the 8 term of an A.P. whose 15 term is 47 and the common difference is 4.
The 10th term of an A.P. is 52 and 16th term is 82. Find the 32nd term and the general term.
The 7th term of an A.P. is – 4 and its 13th term is –16. Find the A.P.
[CBSE 2004]
MATHEMATICS–X
ARITHMETIC PROGRESSIONS
77
12. Which term of the A.P. 3, 10, 17, .... will be 84 more than its 13th term?
[CBSE 2004]
13. Find the A.P. whose third term is 16 and the seventh term exceeds its fifth term by 12.
14. For what value of n is the nth term of the following A.P.’s the same?
1, 7, 13, 19, ..... and 69, 68, 67...............
[CBSE 2006C]
15. If the 10th term of an A.P. is 52 and 17th term is 20 more than the 13th term, find the A.P.
16. If seven times the seventh term of an A.P. is equal to nine times the ninth term, show that 16th term is
zero.
17. (a) Which term of the A.P. 2, 6, 10, 14, ....... is 78?
(b) Which term of the A.P. 5, 9, 13, 17, ....... is 125?
(a) Which term of the A.P. –4, –1, 2, 5, ....... is 119?
(a) Which term of the A.P.
2, 18, 50 ....... is 21 2 ?
18. How many terms are there in each of the following finite A.P.’s?
(a) –3, –4, –5, –6, ....., – 107
(b) 7, 13, 19, ........, 211
(c) 8, 13, 18, ......., 208
19.
20.
21.
22.
23.
24.
25.
34.
35.
78
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(d) 20, 19 , 19,......, 6
2
(a) Is 216 a term of A.P. 3, 8, 13, 18, ......?
(b) Is 271 a term of A.P. 1, 4, 7, 10, .....?
(c) Is –52 a term of A.P. 10, 7, 4 ......?
(d) Is –190 a term of A.P. –3, –8, –13, ......?
Find the 8th term from the end of the A.P. 7, 10, 13, ..... , 184.
[CBSE 2005]
Find the 20th term from the end of the A.P. 3, 8, 13, ....., 253.
Find the value of k so that k – 3, 4k – 11 and 3k – 7 are three consecutive terms of an A.P.
Find the value of x so that 3x + 2, 7x – 1 and 6x + 6 are three consecutive terms of an A.P.
The 4th term of an A.P. is three times the first and the 7th term exceeds twice the third term by 1. Find the
first term and the common difference.
The 9th term of an A.P. is equal to 7 times the 2nd term and 12th term exceeds 5 times the 3rd term by 2. Find
the first term and the common difference.
Find the middle term of an A.P. with 17 terms whose 5th term is 23 and the common difference is –2.
For what value of n, the nth terms of the two A.P.’s are same? 2, –3, –8, –13, ...... and –26, –27, –28, –29, .......
For what value of n, the nth terms of the two A.P.’s are same? 3, 10, 17, ...... and 63, 65, 67, .....
Find the number of integers between 200 and 500 which are divisible by 7.
How many numbers of three digits are exactly divisible by 11?
The angles of a triangle are in A.P. If the greatest angle equals the sum of the other two, find the angles.
Three numbers are in A.P. If the sum of these numbers is 27 and the product is 648, find the numbers.
If m times the mth term of an A.P. is equal to n times its nth term, show that the (m + n)th term of the A.P. is
zero.
A sum of Rs. 1000 is invested at 8% simple interest per annum. Calculate the interest at the end of 1, 2, 3,
......years. Is the sequence of interest an A.P.? Find the interest at the end of 30 years.
Two A.P.’s have the same common difference. The difference between their 100th terms is 111222333.
What is the difference between their millionth terms?
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26.
27.
28.
29.
30.
31.
32.
33.
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ARITHMETIC PROGRESSIONS
MATHEMATICS–X
Problems based on sum to n terms of an A.P. (Sn) :
36. Find the sum of first n natural numbers.
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Find the sum of the following series (Qs. 37-40) :
37. 1 + 3 + 5 + ....... to 50 terms.
38. 6 + 6.5 + 7 + ....... to 20 terms.
39. –11 – 5 + 1 + ....... to 10 terms.
40.
 1  2  3
 1     1    1    .... upto n terms.
 n  n  n
41.
42.
43.
44.
45.
46.
Find the sum of first 30 even natural numbers.
Find the sum of first 25 odd natural numbers.
Show that the sum of the first n odd natural numbers equals n2.
Find the sum of n terms of an A.P. whose nth term is given by an = 5 – 6n.
In an A.P., an = 116, a = 2 and d = 6. Find Sn.
The nth term of an A.P. is given by tn = 4n – 5. Find the sum of the first 25 terms of the A.P. [CBSE 2004]
Find the sum of all the natural numbers (Qs. 47-50) :
47.
48.
49.
50.
51.
52.
53.
[CBSE 2007 (C)]
B
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between 250 and 1000 which are exactly divisible by 3.
between 50 and 500 which are divisible by 7.
between 100 and 1000 which are multiples of 5.
between 50 and 500 which are divisible by 3 and 5.
How many terms of the sequence –35, –28, –21, ...... should be taken so that their sum is zero?
How many terms of the A.P. 63, 60, 57,..... must be taken so that their sum is 693?
[CBSE 2005]
How many terms of an A.P. 1, 4, 7, ..... are needed to give the sum 1335?
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54. How many terms of an A.P. 6, 
11
, 5..... are needed to give the sum –25? Explain the double answer..
2
Find the following sum (Qs. 55-60) :
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55. 2 + 4 + 6 + ... + 150
57. 25 + 28 + 31 + ... + 100
59.
3 4 7
 1 
   ...   
2 3 6
 6 
56. (–5) + (–8) + (–11) + ... + (–230)
58. 1 + 3 + 5 + ... + 199
1
2
60. 2  5  8  12  ...  62
3
3
3n 2 13
 n. Find its 25th term.
[CBSE 2006 C]
2
2
62. If the sum of the first 6 terms of an A.P. is 36 and the sum of the first 21 terms is 441, find the first term and
the common difference and hence find the sum of n terms of this A.P.
63. Find the common difference of an A.P. whose first term is 1 and the sum of the first four terms is one-third
the sum of the next four terms.
64. The sum of the first 9 terms of an A.P. is 171 and that of the first 24 terms is 996, find the first term and the
common difference.
61. In an A.P., the sum of the first n terms is
MATHEMATICS–X
ARITHMETIC PROGRESSIONS
79
65. If the sum of the first 20 terms of an A.P. is 400 and the sum of the first 40 terms is 1600, find the sum of
its first 10 terms.
66. Find the sum of first 20 terms of an A.P., in which 3rd term is 7 and 7th term is two more than thrice of its 3rd
term.
67. The sums of n-terms of two A.P.’s are in the ratio 7n + 1 : 4n + 27. Find the ratio of their 11th terms.
68. (a) For an A.P., find Sn if given an = 3 – 5n.
(b) For an A.P., find an if given Sn = 2n2 + 5n.
69. A person borrows Rs. 4500 and promises to pay back (without any interest) in 30 instalments each of
value Rs. 10 more than last (preceding one). Find the first and the last instalments.
70. A man saved Rs. 32 during the first year, Rs. 36 in the second year and in this way be increases his
savings by Rs. 4 every year. Find in what time his savings will be Rs. 200.
71. Find the three numbers in A.P. such that their sum is 24 and the sum of their squares is 194.
72. 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 sum
of first 20 terms of the A.P.
73. The sum of the third and the seventh terms of an A.P. is 6 and their product is 8. Find the sum of first ten
terms of the A.P.
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A
1
1
and nth term be , then show that its (mn)th term is 1.
n
m
74. If the mth term of an A.P. be
B
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A
75. The 10th term of an A.P. is 29 and and sum of the first 20 terms is 610. Find the sum of the first 30 terms.
76. The sum of the first 15 terms of an A.P. is 105 and the sum of next 15 terms is 780. Find the first three terms
of the A.P.
1
1
1
and nth term is , show that the sum of mn terms is (mn  1).
n
m
2
78. Two cars start together in the same direction from the same place. The first goes with a uniform speed of
1
10 km/hr. The second goes with a speed of 8 km/hr in the first hour and increases the speed by km/hr
2
in each succeeding hour. After how many hours will the second car overtake the first if both cars go nonstop?
79. The ages of the students in a class are in A.P. whose common difference is 4 months. If the youngest
student is 8 years old and the sum of the ages of all the students is 168 years, find the number of students
in the class.
80. In an A.P.
(i) given a = 5, d = 3, an = 50, find n and Sn.
(ii) given a = 7, a13 = 35, find d and S13.
(iii) given a12 = 37, d = 3, find a and S12.
(iv) given a3 = 15, S10 = 125, find d and a10.
(v) given d = 5, S9 = 75, find a and a9.
(vi) given a = 2, d = 8, Sn = 90, find n and an.
(vii) given a = 8, an = 62, Sn = 210, find n and d.
(viii) given an = 4, d = 2, Sn = – 14, find n and a.
(ix) given a = 3, n = 8, Sn = 192, find d.
(x) given l = 28, Sn = 144, and there are total 9 terms. Find a.
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77. If mth term of A.P. is
M
A
80
ARITHMETIC PROGRESSIONS
MATHEMATICS–X
HINTS TO SELECTED QUESTIONS
25. Let first term be a and common difference be d.
J
A
Then, a9  7a2  a  8d  7(a  d )
and,
a12  5a3  2  (a  11d )  5(a  2d )  2.
31. Let the angles be a – d, a, a + d.
Then, (a – d) + a + (a + d) = 180°  3a = 180°  a = 60°
 angles are 60° – d, 60° and 60° + d.
Also, 60° + d = (60° – d) + 60°  d = 30°
 angles are 30°, 60° and 90°.
an  5  6n
44.
for
n = 1, a1 = a = 5 – 6(1) = – 1
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A
n
n
n
Now, S n  [a  an ]  [1  5  6n]  (6n  4)  n(3n  2) = 2n – 3n2
2
2
2
61. here, S n 
3n 2  13n
.
2
Now, an  Sn  Sn 1  a25  S25  S24

B
3(25) 2  13(25) 3(24) 2  13(24) 1
 [3(252  24 2 )  13(25  24)]

2
2
2
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1
 [3  49  13]  80
2
63. here, a = 1. Let common difference be d.
Let A.P. be a1, a2, a3, a4, a5, a6, .....
M
A
Then, a1 + a2 + a3 + a4  1 (a5  a6  a7  a8 )
3

3 (a1 + a2 + a3 + a4) = a5 + a6 + a7 + a8

3 (a1 + a2 + a3 + a4) = (a1 + a2 + ... + a7 + a8) – (a1 + a2 + ... + a4)

3 S4 = S8 – S4  4S4 = S8
4
8
 4. [2(1)  (4  1)d ]  [2(1)  (8  1)d ]  d = 2
2
2
67. Let a1, a2 be first terms and d1, d2 the common differences of the two given A.P.’s. Then,
 n 1 
n
a1  
[2a1  (n  1)d1 ]
 d1
Sn
2
a

(
n

1)
d
 2 
1
 2
 1

Sn n [2 a  (n  1)d ] 2a2  (n  1)d 2
 n 1 
a2  
2
2
 d2
2
 2 
MATHEMATICS–X
ARITHMETIC PROGRESSIONS
81
Sn
7n  1
Given, S   4n  27
n
 n 1 
a1  
 d1
 2   7n  1

4n  27
 n 1 
a2  
 d2
2


a11 a1  10d1
Clearly,

a11
a2  10d 2
Comparing (1) and (2), we get
...(2)
n 1
 10  n  21
2
a11
7(21)  1

 4:3.
a11
4(21)  27

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A
...(1)
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77. Let a be the first term and d be the common difference of the given A.P. Then,
am 
1
1
 a  ( m  1) d 
n
n
an 
1
1
 a  ( n  1) d 
m
m
...(2)
Solving (1) and (2), we get a  d 
Now, S mn 
B
...(1)
1
mn
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I
mn
[2a  (mn  1)d ]
2
1
Put a  d 
in Smn and solve.
mn
78. Let the second car overtakes the first car after t hours. Then, the two cars travel the same distance in t
hours.
M
A
1
Now, Distance travelled by the second car = sum of t terms of an A.P. with a = 8, d  .
2

t (t  31)
4
according to question,
t (t  31)
 40 t  t 2  9t  0  t = 9 [ t  0]
4
1
79. here, a = 8, d  , Sn  168
4
n
n
1 
Then, S n  [2a  ( n  1)d ]  168   2(8)  ( n  1)   
2
2
 4 
Simplify and get quadratic equation in n.
82
ARITHMETIC PROGRESSIONS
MATHEMATICS–X
MULTIPLE CHOICE QUESTIONS
Mark the correct alternative in each of the following :
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A
1. Which of the following sequence whose nth term is given by an, is not an A.P. :
(a) an = 3 – 2n
(c) an = 2n2 + 1
(b) an = 3n – 7
nd
th
(d) an = – 7n + 5
th
2. The 32 term of an AP, whose 10 term is 52 and 16 term is 82, is:
(a) 162
(b) 152
(c) 172
(d) none of these
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A
1
1
, and nth term is , then its (mn)th term is :
n
m
mn
(b) 1
(c)
mn
3. In mth term of an AP is
(a) 0
1
4. The total number of terms in an A.P. 18, 15 ,13,......, 47 are :
2
(a) 20
(b) 25
(c) 27
B
5. The total number of multiples of 4, between 10 and 250 are :
(a) 45
(b) 50
(c) 60
th
(d)
mn
mn
(d) 30
(d) none of these
6. If n term of an A.P. is (2n + 1), then the sum of first n terms of the A.P. is :
(a) n(n  1)
(b) n( n  2)
(c) n( n  3)
(d) none of these
7. The sum of all three digit natural numbers, which are divisible by 7, is :
(a) 70336
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(b) 70784
(c) 64064
(d) none of these
8. The total numbers of terms of A.P. 9, 17, 25, ..... that must be taken so that their sum is 636, is :
(a) 10
(b) 11
(c) 12
(d) 13
9. If the sum of n terms of an A.P. is 2n2 + 5n then its nth term is :
(a) 4n – 3
(b) 3n – 4
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A
(c) 4n + 3
(d) 3n + 4
10. If the first term of an A.P. is 2 and common difference is 4, then the sum of its 40 terms is :
(a) 3200
(b) 1600
(c) 2000
(d) 3000
VERY SHORT ANSWER TYPE QUESTIONS (1 MARK QUESTIONS)
1. Find a, such that –15, a, 35 are in A.P.
1
2. Find the 10th term of an A.P. –3,  , 2,.....
2
3. Write the 15th term of an A.P. whose first term is 7 and the common difference is –3.
4. Write the missing terms of the A.P.
–9,
, –19, –24,
5. For what value of k, the numbers k + 2, 4k + 4 and 9k + 4 are three consecutive terms of an A.P.
MATHEMATICS–X
ARITHMETIC PROGRESSIONS
83
6. Write the first three terms of an A.P. whose nth term is –3n + 5.
7. Write the A.P. whose nth term is 4n – 7.
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A
8. Which term of the A.P. 2, 6, 10, .... is 210?
9. Is 67 a term of the A.P. 7, 10, 13, ......?
10. Find the number of terms of A.P. 7, 13, 19, ....., 301.
11. The 17th term of an A.P. exceeds its 10th term by 7. Find the common difference.
12. How many 2-digit numbers are divisible by 3?
1
2
13. Which term of the A.P. 19, 18 , 17 ,.... is the first negative term?
5
5
14. Find the sum of first 50 natural numbers.
15. Find the following sum : 1 + 3 + 5 + ..... 20 terms.
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A
16. If the sum of first n terms of an A.P. is 2n2 + 5n, write the sum of its first 5 terms.
17. Find the sum of first 15 terms of an AP, whose nth term is 9–5n.
18. Given that the first term of an A.P. is 2 and its common difference is 4, find the sum of its first 40 terms.
B
19. Find the sum of the odd numbers between 0 and 50.
20. Find the sum to n terms of the A.P. whose rth term is 5r + 1.
PRACTICE TEST
M.M : 30
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General Instructions :
Time : 1 hour
Q. 1-4 carry 2 marks, Q. 5-8 carry 3 marks and Q. 9-10 carry 5 marks each.
1. Find the term of A.P. 9, 12, 15, 18, .... which is 39 more than its 36th term.
2. Find the number of integers between 50 and 500 which are divisible by 7.
3. Find the sum of first 15 multiples of 8.
M
A
4. Find the sum of first 25 terms of an A.P. whose nth term is given by an = 2 – 3n.
5. If the sum of 7 terms of an A.P. is 49 and that of 17 terms is 289, find the sum of n terms.
6. If the sum of m terms of an A.P. is the same as the sum of its n terms, show that the sum of its (m + n) terms
is zero.
7. Pranshi saved Rs. 5 in the first week of the year and then increased her weekly savings by Rs. 1.75 each
week. In what week will her weekly savings be Rs. 20.75?
8. The sum of the third and the seventh terms of an A.P. is 6 and their product is 8. Find the sum of first
sixteen terms of the A.P.
9. (a) Find the 12th term from the end of the A.P. 3, 8, 13, ..... , 253.
(b) How many three digit numbers are divisible by 7?
10. Salman Khan buys a shop for Rs. 1, 20,000. He pays half of the amount in cash and agrees to pay the
balance in 12 annual instalments of Rs. 5000 each. If the rate of interest is 12% and he pays with the
instalment the interest due on the unpaid amount, find the total cost of the shop.
84
ARITHMETIC PROGRESSIONS
MATHEMATICS–X
ANSWERS OF PRACTICE EXERCISE
1. Yes, d = – 4; –16, –20, –24
2. No
4. Yes, d  3;5  4 3, 5  5 3, 5  6 3
6. Yes, d  2; 50, 72, 98
8. (a) 4, – 1, –6, –11, .....
9. 19
5. No
7. (a) and (c) form an A.P.
(b) 5, 4, 3, 2, .......
10. a32 = 162, an = 5n + 2
(b) 31st term
18. (a) 105
(b) 35
(c) 41
20. 163
21. 158
(c) 42nd term
19. (a) No
22. k = 3
23. x = 2
26. 15
27. n = 8
31. 30°, 60°, 90° 32. 6, 9, 12
n(n  1)
2
37. 2500
(b) Yes (c) No
(d) No
24. a = 3, d = 2
28. n = 13
B
38. 215
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A
15. 7, 12, 17, 22.......
(d) 11th term
(d) 29
34. Rs. 80, Rs. 160, Rs. 240, and so on. Yes; Rs. 2400
36.
12. 25th
14. No value of n
17. (a) 20th term
30. 81
(c) 9, 11, 13, 15, ......
11. 8, 6, 4, .....
13. 4, 10, 16, ......
25. a = 1, d = 6
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A
3. Yes, d = 0; 6, 6, 6
29. 43
35. 111222333
n 1
2
39. 160
40.
45. 1180
46. 1175
41. 930
42. 625
44. 2n –3n2
47. 156375
48. 17696
49. 98450
50. 8325
51. 11
52. 21 or 22
53. 30
54. 5 or 20
55. 5700
56. –8930
57. 1625
58. 10000
59.
22
3
60. 608
64. a = 7, d = 3
M
A
61. 80
62. a = 1, d = 2; Sn = n2
63. 2
66. 740
67. 4 : 3
68. (a)
70. 5 years.
71. 7, 8 and 9
72. 690
76. –14, –11, –8
78. 9 hours
79. 16
80. (i) n = 16, Sn = 440
(iv) d  1, a10  8
(vii) n  6, d 
MATHEMATICS–X
54
5
(ii) d  7 , S13  273
3
(v) a  
35
85
, a9 
3
3
(viii) a = –8, n = 7
n
(1 –5n) (b) 4n + 3
2
73. 32.5 or 27.5
65. 100
69. Rs. 5, Rs. 295
75. 1365
(iii) a  4, S12  246
(vi) n = 5, an = 34
(ix) d = 6
ARITHMETIC PROGRESSIONS
(x) a = 4
85
ANSWERS OF MULTIPLE CHOICE QUESTIONS
1. (c)
2. (a)
3. (b)
4. (c)
5. (c)
6. (b)
7. (a)
8. (c)
9. (c)
10. (a)
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ANSWERS OF VERY SHORT ANSWER TYPE QUESTIONS
1. a = 10
2.
6. 2, –1, –4
39
2
3. –35
4. –14, –29
7. –3, 1, 5, .....
8. 53rd
9. yes
11. 1
12. 30
13. 25th
14. 1275
16. 75
17. –465
18. 3200
19. 625
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A
5. k = 1
10. 50
15. 400
20.
B
n
(5n  7)
2
ANSWERS OF PRACTICE TEST
1. 49th
2. 64
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5. n2
7. 10th week
10. Rs. 1,66,800
M
A
86
3. 960
8. 20 or 76
ARITHMETIC PROGRESSIONS
4. –925
9. (a) 198 (b) 128
MATHEMATICS–X