NCERT Solutions for Class 10 Maths Ch 5 Arithmetic Progressions
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Page No: 99Exercise 5.1
1. In which of the following situations, does the list of numbers involved make as arithmetic progression and why?
(i) The taxi fare after each km when the fare is Rs 15 for the first km and Rs 8 for each additional km.
Answer
It can be observed that
Taxi fare for 1st km = 15
Taxi fare for first 2 km = 15 + 8 = 23
Taxi fare for first 3 km = 23 + 8 = 31
Taxi fare for first 4 km = 31 + 8 = 39
Clearly 15, 23, 31, 39 … forms an A.P. because every term is 8 more than the preceding term.
(ii) The amount of air present in a cylinder when a vacuum pump removes 1/4 of the air remaining in the cylinder at a time.
Answer
Let the initial volume of air in a cylinder be V litres. In each stroke, the vacuum pump removes 1/4 of air remaining in the cylinder at a time. In other words, after every stroke, only 1  1/4 = 3/4th part of air will remain.
Therefore, volumes will be V, 3V/4 , (3V/4)^{2} , (3V/4)^{3}...
Clearly, it can be observed that the adjacent terms of this series do not have the same difference between them. Therefore, this is not an A.P.
(iii) The cost of digging a well after every metre of digging, when it costs Rs 150 for the first metre and rises by Rs 50 for each subsequent metre.
Answer
Cost of digging for first metre = 150
Cost of digging for first 2 metres = 150 + 50 = 200
Cost of digging for first 3 metres = 200 + 50 = 250
Cost of digging for first 4 metres = 250 + 50 = 300
Clearly, 150, 200, 250, 300 … forms an A.P. because every term is 50 more than the preceding term.
(iv) The amount of money in the account every year, when Rs 10000 is deposited at compound interest at 8% per annum.
Answer
We know that if Rs P is deposited at r% compound interest per annum for n years, our money will be
(i) a = 10, d = 10
(ii) a = 2, d = 0
(iii) a = 4, d =  3
(iv) a = 1 d = 1/2
(v) a =  1.25, d =  0.25
Answer
(i) a = 10, d = 10
Let the series be a_{1}, a_{2}, a_{3}, a_{4}, a_{5} …
a_{1} = a = 10
a_{2} = a_{1} + d = 10 + 10 = 20
a_{3} = a_{2} + d = 20 + 10 = 30
a_{4} = a_{3} + d = 30 + 10 = 40
a_{5} = a_{4} + d = 40 + 10 = 50
Therefore, the series will be 10, 20, 30, 40, 50 …
First four terms of this A.P. will be 10, 20, 30, and 40.
(ii) a =  2, d = 0
Let the series be a_{1}, a_{2}, a_{3}, a_{4} …
a_{1} = a = 2
a_{2} = a_{1} + d =  2 + 0 =  2
a_{3} = a_{2} + d =  2 + 0 =  2
a_{4} = a_{3} + d =  2 + 0 =  2
Therefore, the series will be  2,  2,  2,  2 …
First four terms of this A.P. will be  2,  2,  2 and  2.
(iii) a = 4, d =  3
Let the series be a_{1}, a_{2}, a_{3}, a_{4} …
a_{1} = a = 4
a_{2} = a_{1} + d = 4  3 = 1
a_{3} = a_{2} + d = 1  3 =  2
a_{4} = a_{3} + d =  2  3 =  5
Therefore, the series will be 4, 1,  2  5 …
First four terms of this A.P. will be 4, 1,  2 and  5.
(iv) a =  1, d = 1/2
Let the series be a_{1}, a_{2}, a_{3}, a_{4} …a_{1} = a = 1
a_{2} = a_{1} + d = 1 + 1/2 = 1/2
a_{3} = a_{2} + d = 1/2 + 1/2 = 0
a_{4} = a_{3} + d = 0 + 1/2 = 1/2
Clearly, the series will be1, 1/2, 0, 1/2
First four terms of this A.P. will be 1, 1/2, 0 and 1/2.
(v) a =  1.25, d =  0.25
Let the series be a_{1}, a_{2}, a_{3}, a_{4} …
a_{1} = a =  1.25
a_{2} = a_{1} + d =  1.25  0.25 =  1.50
a_{3} = a_{2} + d =  1.50  0.25 =  1.75
a_{4} = a_{3} + d =  1.75  0.25 =  2.00
Clearly, the series will be 1.25,  1.50,  1.75,  2.00 ……..
First four terms of this A.P. will be  1.25,  1.50,  1.75 and  2.00.
3. For the following A.P.s, write the first term and the common difference.
(i) 3, 1,  1,  3 …
(ii) 5,  1, 3, 7 …
(iii) 1/3, 5/3, 9/3, 13/3 ....
(iv) 0.6, 1.7, 2.8, 3.9 …
Answer
(i) 3, 1,  1,  3 …
Here, first term, a = 3
Common difference, d = Second term  First term
= 1  3 =  2
(ii)  5,  1, 3, 7 …
Here, first term, a =  5
Common difference, d = Second term  First term
= (  1)  (  5) =  1 + 5 = 4
(iii) 1/3, 5/3, 9/3, 13/3 ....
Here, first term, a = 1/3
Common difference, d = Second term  First term
= 5/3  1/3 = 4/3
(iv) 0.6, 1.7, 2.8, 3.9 …
Here, first term, a = 0.6
Common difference, d = Second term  First term
= 1.7  0.6
= 1.1
4. Which of the following are APs? If they form an A.P. find the common difference d and write three more terms.
(i) 2, 4, 8, 16 …
(ii) 2, 5/2, 3, 7/2 ....
(iii) 1.2, 3.2, 5.2, 7.2 …
(iv) 10,  6,  2, 2 …
(v) 3, 3 + √2, 3 + 2√2, 3 + 3√2
(vi) 0.2, 0.22, 0.222, 0.2222 ….
(vii) 0,  4,  8,  12 …
(viii) 1/2, 1/2, 1/2, 1/2 ....
(ix) 1, 3, 9, 27 …
(x) a, 2a, 3a, 4a …
(xi) a, a^{2}, a^{3}, a^{4} …
(xii) √2, √8, √18, √32 ...
(xiii) √3, √6, √9, √12 ...
(xiv) 1^{2}, 3^{2}, 5^{2}, 7^{2} …
(xv) 1^{2}, 5^{2}, 7^{2}, 7^{3} …
Answer
(i) 2, 4, 8, 16 …
Here,
a_{2}  a_{1} = 4  2 = 2
a_{3}  a_{2} = 8  4 = 4
a_{4}  a_{3} = 16  8 = 8
⇒ a_{n+1}  a_{n} is not the same every time.
Therefore, the given numbers are forming an A.P.
(ii) 2, 5/2, 3, 7/2 ....
Here,
a_{2}  a_{1} = 5/2  2 = 1/2(ii) 2, 5/2, 3, 7/2 ....
Here,
a_{3}  a_{2} = 3  5/2 = 1/2
a_{4}  a_{3} = 7/2  3 = 1/2
⇒ a_{n+1}  a_{n} is same every time.
Therefore, d = 1/2 and the given numbers are in A.P.
Three more terms are
a_{5} = 7/2 + 1/2 = 4
a_{6} = 4 + 1/2 = 9/2
a_{7} = 9/2 + 1/2 = 5
(iii) 1.2,  3.2, 5.2, 7.2 …
Here,
a_{2}  a_{1} = ( 3.2)  ( 1.2) = 2
a_{3}  a_{2} = ( 5.2)  ( 3.2) = 2
a_{4}  a_{3} = ( 7.2)  ( 5.2) = 2
⇒ a_{n+1}  a_{n} is same every time.
Therefore, d = 2 and the given numbers are in A.P.
Three more terms are
a_{5} =  7.2  2 =  9.2
a_{6} =  9.2  2 =  11.2
a_{7} =  11.2  2 =  13.2
(iv) 10,  6,  2, 2 …
Here,
a_{2}  a_{1} = (6)  (10) = 4
a_{3}  a_{2} = (2)  (6) = 4
a_{4}  a_{3} = (2)  (2) = 4
⇒ a_{n+1}  a_{n} is same every time.
Therefore, d = 4 and the given numbers are in A.P.
Three more terms are
a_{5} = 2 + 4 = 6
a_{6} = 6 + 4 = 10
a_{7} = 10 + 4 = 14
(v) 3, 3 + √2, 3 + 2√2, 3 + 3√2
Here,
a_{2}  a_{1} = 3 + √2  3 = √2
a_{3}  a_{2} = (3 + 2√2)  (3 + √2) = √2
a_{4}  a_{3} = (3 + 3√2)  (3 + 2√2) = √2
⇒ a_{n+1}  a_{n} is same every time.
Therefore, d = √2 and the given numbers are in A.P.
Three more terms are
a_{5} = (3 + √2) + √2 = 3 + 4√2
a_{6} = (3 + 4√2) + √2 = 3 + 5√2
a_{7} = (3 + 5√2) + √2 = 3 + 6√2
(vi) 0.2, 0.22, 0.222, 0.2222 ….
Here,
a_{2}  a_{1} = 0.22  0.2 = 0.02
a_{3}  a_{2} = 0.222  0.22 = 0.002
a_{4}  a_{3} = 0.2222  0.222 = 0.0002
⇒ a_{n+1}  a_{n} is not the same every time.
Therefore, the given numbers are forming an A.P.
(vii) 0, 4, 8, 12 …
Here,
a_{2}  a_{1} = (4)  0 = 4
a_{3}  a_{2} = (8)  (4) = 4
a_{4}  a_{3} = (12)  (8) = 4
⇒ a_{n+1}  a_{n} is same every time.
Therefore, d = 4 and the given numbers are in A.P.
Three more terms are
a_{5} = 12  4 = 16
a_{6} = 16  4 = 20
a_{7} = 20  4 = 24
(viii) 1/2, 1/2, 1/2, 1/2 ....
Here,
a_{2}  a_{1} = (1/2)  (1/2) = 0
a_{3}  a_{2} = (1/2)  (1/2) = 0
a_{4}  a_{3} = (1/2)  (1/2) = 0
⇒ a_{n+1}  a_{n} is same every time.
Therefore, d = 0 and the given numbers are in A.P.
Three more terms are
a_{5} = (1/2)  0 = 1/2
a_{6} = (1/2)  0 = 1/2
a_{7} = (1/2)  0 = 1/2
(ix) 1, 3, 9, 27 …
Here,
a_{2}  a_{1} = 3  1 = 2
a_{3}  a_{2} = 9  3 = 6
a_{4}  a_{3} = 27  9 = 18
⇒ a_{n+1}  a_{n} is not the same every time.
(vii) 0, 4, 8, 12 …
Here,
a_{2}  a_{1} = (4)  0 = 4
a_{3}  a_{2} = (8)  (4) = 4
a_{4}  a_{3} = (12)  (8) = 4
⇒ a_{n+1}  a_{n} is same every time.
Therefore, d = 4 and the given numbers are in A.P.
Three more terms are
a_{5} = 12  4 = 16
a_{6} = 16  4 = 20
a_{7} = 20  4 = 24
(viii) 1/2, 1/2, 1/2, 1/2 ....
Here,
a_{2}  a_{1} = (1/2)  (1/2) = 0
a_{3}  a_{2} = (1/2)  (1/2) = 0
a_{4}  a_{3} = (1/2)  (1/2) = 0
⇒ a_{n+1}  a_{n} is same every time.
Therefore, d = 0 and the given numbers are in A.P.
Three more terms are
a_{5} = (1/2)  0 = 1/2
a_{6} = (1/2)  0 = 1/2
a_{7} = (1/2)  0 = 1/2
(ix) 1, 3, 9, 27 …
Here,
a_{2}  a_{1} = 3  1 = 2
a_{3}  a_{2} = 9  3 = 6
a_{4}  a_{3} = 27  9 = 18
⇒ a_{n+1}  a_{n} is not the same every time.
Therefore, the given numbers are forming an A.P.
(x) a, 2a, 3a, 4a …
Here,
a_{2}  a_{1} = 2a  a = a
a_{3}  a_{2} = 3a  2a = a
a_{4}  a_{3} = 4a  3a = a
⇒ a_{n+1}  a_{n} is same every time.
Therefore, d = a and the given numbers are in A.P.
Three more terms are
a_{5} = 4a + a = 5a
a_{6} = 5a + a = 6a
a_{7} = 6a + a = 7a
(xi) a, a^{2}, a^{3}, a^{4} …
Here,
a_{2}  a_{1} = a^{2 } a = (a  1)
a_{3}  a_{2} = a^{3 }^{ }a^{2 }= a^{2 }(a  1)
a_{4}  a_{3} = a^{4}  a^{3 }= a^{3}(a  1)
⇒ a_{n+1}  a_{n} is not the same every time.
(xii) √2, √8, √18, √32 ...
Here,
a_{2}  a_{1} = √8  √2 = 2√2  √2 = √2
a_{3}  a_{2} = √18  √8 = 3√2  2√2 = √2
a_{4}  a_{3} = 4√2  3√2 = √2
⇒ a_{n+1}  a_{n} is same every time.
Therefore, d = √2 and the given numbers are in A.P.
Three more terms are
a_{5} = √32 + √2 = 4√2 + √2 = 5√2 = √50
a_{6} = 5√2 +√2 = 6√2 = √72
a_{7} = 6√2 + √2 = 7√2 = √98
(x) a, 2a, 3a, 4a …
Here,
a_{2}  a_{1} = 2a  a = a
a_{3}  a_{2} = 3a  2a = a
a_{4}  a_{3} = 4a  3a = a
⇒ a_{n+1}  a_{n} is same every time.
Therefore, d = a and the given numbers are in A.P.
Three more terms are
a_{5} = 4a + a = 5a
a_{6} = 5a + a = 6a
a_{7} = 6a + a = 7a
(xi) a, a^{2}, a^{3}, a^{4} …
Here,
a_{2}  a_{1} = a^{2 } a = (a  1)
a_{3}  a_{2} = a^{3 }^{ }a^{2 }= a^{2 }(a  1)
a_{4}  a_{3} = a^{4}  a^{3 }= a^{3}(a  1)
⇒ a_{n+1}  a_{n} is not the same every time.
Therefore, the given numbers are forming an A.P.
(xii) √2, √8, √18, √32 ...
Here,
a_{2}  a_{1} = √8  √2 = 2√2  √2 = √2
a_{3}  a_{2} = √18  √8 = 3√2  2√2 = √2
a_{4}  a_{3} = 4√2  3√2 = √2
⇒ a_{n+1}  a_{n} is same every time.
Therefore, d = √2 and the given numbers are in A.P.
Three more terms are
a_{5} = √32 + √2 = 4√2 + √2 = 5√2 = √50
a_{6} = 5√2 +√2 = 6√2 = √72
a_{7} = 6√2 + √2 = 7√2 = √98
(xiii) √3, √6, √9, √12 ...
Here,
a_{2}  a_{1} = √6  √3 = √3 × 2 √3 = √3(√2  1)
a_{3}  a_{2} = √9  √6 = 3  √6 = √3(√3  √2)
a_{4}  a_{3} = √12  √9 = 2√3  √3 × 3 = √3(2  √3)
⇒ a_{n+1}  a_{n} is not the same every time.
Therefore, the given numbers are forming an A.P.
(xiv) 1^{2}, 3^{2}, 5^{2}, 7^{2} …
Or, 1, 9, 25, 49 …..(xiv) 1^{2}, 3^{2}, 5^{2}, 7^{2} …
Here,
a_{2} − a_{1} = 9 − 1 = 8
a_{3} − a_{2 }= 25 − 9 = 16
a_{4} − a_{3} = 49 − 25 = 24
⇒ a_{n+1}  a_{n} is not the same every time.
Therefore, the given numbers are forming an A.P.
(xv) 1^{2}, 5^{2}, 7^{2}, 73 …
Or 1, 25, 49, 73 …
Here,
a_{2} − a_{1} = 25 − 1 = 24
a_{3} − a_{2 }= 49 − 25 = 24
a_{4} − a_{3} = 73 − 49 = 24
i.e., a_{k}_{+1 }− a_{k} is same every time.
⇒ a_{n+1}  a_{n} is same every time.
Therefore, d = 24 and the given numbers are in A.P.
Three more terms are
a_{5} = 73+ 24 = 97
a_{6} = 97 + 24 = 121
a_{7 }= 121 + 24 = 145
Page No: 105
Exercise 5.2
1. Fill in the blanks in the following table, given that a is the first term, d the common difference and a_{n} the n^{th} term of the A.P.
a

d

n

a_{n}


(i)

7

3

8

…...

(ii)

− 18

…..

10

0

(iii)

…..

− 3

18

− 5

(iv)

− 18.9

2.5

…..

3.6

(v)

3.5

0

105

…..

Answer
(i) a = 7, d = 3, n = 8, a_{n} = ?
We know that,
For an A.P. a_{n} = a + (n − 1) d
= 7 + (8 − 1) 3
= 7 + (7) 3
= 7 + 21 = 28
Hence, a_{n} = 28
(ii) Given that
a = −18, n = 10, a_{n} = 0, d = ?
We know that,
a_{n} = a + (n − 1) d
0 = − 18 + (10 − 1) d
18 = 9d
d = 18/9 = 2
Hence, common difference, d = 2
(iii) Given that
d = −3, n = 18, a_{n} = −5
We know that,
a_{n} = a + (n − 1) d
−5 = a + (18 − 1) (−3)
−5 = a + (17) (−3)
−5 = a − 51
a = 51 − 5 = 46
Hence, a = 46
(iv) a = −18.9, d = 2.5, a_{n} = 3.6, n = ?
We know that,
a_{n} = a + (n − 1) d
3.6 = − 18.9 + (n − 1) 2.5
3.6 + 18.9 = (n − 1) 2.5
22.5 = (n − 1) 2.5
(n  1) = 22.5/2.5
n  1 = 9
n = 10
Hence, n = 10
(v) a = 3.5, d = 0, n = 105, a_{n} = ?
We know that,
a_{n} = a + (n − 1) d
a_{n} = 3.5 + (105 − 1) 0
a_{n} = 3.5 + 104 × 0
a_{n} = 3.5
Hence, a_{n} = 3.5
Page No: 106
Choose the correct choice in the following and justify
(i) 30^{th} term of the A.P: 10, 7, 4, …, is
(A) 97 (B) 77 (C) −77 (D.) −87
(ii) 11^{th }term of the A.P. 3, 1/2, ,2 .... is
(A) 28 (B) 22 (C)  38 (D)
Answer
(i) Given that
A.P. 10, 7, 4, …
First term, a = 10
Common difference, d = a_{2} − a_{1 }= 7 − 10 = −3
We know that, a_{n} = a + (n − 1) d
a_{30} = 10 + (30 − 1) (−3)
a_{30} = 10 + (29) (−3)
a_{30} = 10 − 87 = −77
Hence, the correct answer is option C.
(ii) Given that A.P. is 3, 1/2, ,2 ...
First term a =  3
Common difference, d = a_{2} − a_{1} = (1/2)  (3)
= (1/2) + 3 = 5/2
We know that, a_{n} = a + (n − 1) d
a_{11} = 3 + (11 1)(5/2)
a_{11} = 3 + (10)(5/2)
a_{11} = 3 + 25
a_{11} = 22
Hence, the answer is option B.
3. In the following APs find the missing term in the boxes.
Answer
(i) For this A.P.,
a = 2a_{3} = 26
We know that, a_{n} = a + (n − 1) d
a_{3} = 2 + (3  1) d
26 = 2 + 2d
24 = 2d
d = 12
a_{2} = 2 + (2  1) 12
= 14
Therefore, 14 is the missing term.
(ii) For this A.P.,
a_{2} = 13 and
a_{4} = 3
We know that, a_{n} = a + (n − 1) d
a_{2} = a + (2  1) d
13 = a + d ... (i)
a_{4} = a + (4  1) d
3 = a + 3d ... (ii)
On subtracting (i) from (ii), we get
 10 = 2d
d =  5
From equation (i), we get
13 = a + (5)
a = 18
a_{3} = 18 + (3  1) (5)
= 18 + 2 (5) = 18  10 = 8
Therefore, the missing terms are 18 and 8 respectively.
(iii) For this A.P.,
a = 5 and
a_{4} = 19/2
We know that, a_{n} = a + (n − 1) d
a_{4} = a + (4  1) d
19/2 = 5 + 3d
19/2  5 = 3d3d = 9/2
d = 3/2
a_{2} = a + (2  1) d
a_{2} = 5 + 3/2
a_{2} = 13/2
a_{3} = a + (3  1) d
a_{3} = 5 + 2×3/2
a_{3} = 8
Therefore, the missing terms are 13/2 and 8 respectively.(iv) For this A.P.,
a = −4 and
a_{6} = 6
We know that,
a_{n} = a + (n − 1) d
a_{6} = a + (6 − 1) d
6 = − 4 + 5d
10 = 5d
d = 2
a_{2} = a + d = − 4 + 2 = −2
a_{3} = a + 2d = − 4 + 2 (2) = 0
a_{4} = a + 3d = − 4 + 3 (2) = 2
a_{5} = a + 4d = − 4 + 4 (2) = 4
Therefore, the missing terms are −2, 0, 2, and 4 respectively.
(v)
For this A.P.,
a_{2} = 38
a_{6} = −22
We know that
a_{n} = a + (n − 1) d
a_{2} = a + (2 − 1) d
38 = a + d ... (i)
a_{6} = a + (6 − 1) d
−22 = a + 5d ... (ii)
On subtracting equation (i) from (ii), we get
− 22 − 38 = 4d
−60 = 4d
d = −15
a = a_{2} − d = 38 − (−15) = 53
a_{3} = a + 2d = 53 + 2 (−15) = 23
a_{4} = a + 3d = 53 + 3 (−15) = 8
a_{5} = a + 4d = 53 + 4 (−15) = −7
Therefore, the missing terms are 53, 23, 8, and −7 respectively.
4. Which term of the A.P. 3, 8, 13, 18, … is 78?
Answer
3, 8, 13, 18, …
For this A.P.,
a = 3
d = a_{2} − a_{1} = 8 − 3 = 5
Let n^{th} term of this A.P. be 78.
a_{n} = a + (n − 1) d
78 = 3 + (n − 1) 5
75 = (n − 1) 5
(n − 1) = 15
n = 16
Hence, 16^{th} term of this A.P. is 78.
5. Find the number of terms in each of the following A.P.
(i) 7, 13, 19, …, 205
(ii) 18,, 13,...., 47
Answer
(i) For this A.P.,
a = 7
d = a_{2} − a_{1} = 13 − 7 = 6
Let there are n terms in this A.P.
a_{n} = 205
We know that
a_{n} = a + (n − 1) d
Therefore, 205 = 7 + (n − 1) 6
198 = (n − 1) 6
33 = (n − 1)
n = 34
Therefore, this given series has 34 terms in it.
(ii) For this A.P.,
a = 18
Let there are n terms in this A.P.
a_{n} = 205
a_{n} = a + (n − 1) d
47 = 18 + (n  1) (5/2)
47  18 = (n  1) (5/2)
65 = (n  1)(5/2)
(n  1) = 130/5
(n  1) = 26
n = 27
Therefore, this given A.P. has 27 terms in it.
65 = (n  1)(5/2)
(n  1) = 130/5
(n  1) = 26
n = 27
Therefore, this given A.P. has 27 terms in it.
6. Check whether 150 is a term of the A.P. 11, 8, 5, 2, …
Answer
For this A.P.,a = 11
d = a_{2} − a_{1} = 8 − 11 = −3
Let −150 be the n^{th} term of this A.P.
We know that,
a_{n} = a + (n − 1) d
150 = 11 + (n  1)(3)
150 = 11  3n + 3
164 = 3n
n = 164/3
Clearly, n is not an integer.
Therefore,  150 is not a term of this A.P.
7. Find the 31^{st} term of an A.P. whose 11^{th} term is 38 and the 16^{th} term is 73.
Answer
Given that,
a_{11} = 38
a_{16} = 73
We know that,
a_{n} = a + (n − 1) d
a_{11} = a + (11 − 1) d
38 = a + 10d ... (i)
Similarly,
a_{16} = a + (16 − 1) d
73 = a + 15d ... (ii)
On subtracting (i) from (ii), we get
35 = 5d
d = 7
From equation (i),
38 = a + 10 × (7)
38 − 70 = a
a = −32
a_{31} = a + (31 − 1) d
= − 32 + 30 (7)
= − 32 + 210
= 178
Hence, 31^{st} term is 178.
8. An A.P. consists of 50 terms of which 3^{rd} term is 12 and the last term is 106. Find the 29^{th} term.
Answer
Given that,
a_{3} = 12
a_{50} = 106
We know that,
a_{n} = a + (n − 1) d
a_{3} = a + (3 − 1) d
12 = a + 2d ... (i)
Similarly, a_{50 }= a + (50 − 1) d
106 = a + 49d ... (ii)
On subtracting (i) from (ii), we get
94 = 47d
d = 2
From equation (i), we get
12 = a + 2 (2)
a = 12 − 4 = 8
a_{29} = a + (29 − 1) d
a_{29} = 8 + (28)2
a_{29} = 8 + 56 = 64
Therefore, 29^{th} term is 64.
9. If the 3^{rd} and the 9^{th} terms of an A.P. are 4 and − 8 respectively. Which term of this A.P. is zero.
Answer
Given that,
a_{3} = 4Answer
Given that,
a_{11} = 38
a_{16} = 73
We know that,
a_{n} = a + (n − 1) d
a_{11} = a + (11 − 1) d
38 = a + 10d ... (i)
Similarly,
a_{16} = a + (16 − 1) d
73 = a + 15d ... (ii)
On subtracting (i) from (ii), we get
35 = 5d
d = 7
From equation (i),
38 = a + 10 × (7)
38 − 70 = a
a = −32
a_{31} = a + (31 − 1) d
= − 32 + 30 (7)
= − 32 + 210
= 178
Hence, 31^{st} term is 178.
8. An A.P. consists of 50 terms of which 3^{rd} term is 12 and the last term is 106. Find the 29^{th} term.
Answer
Given that,
a_{3} = 12
a_{50} = 106
We know that,
a_{n} = a + (n − 1) d
a_{3} = a + (3 − 1) d
12 = a + 2d ... (i)
Similarly, a_{50 }= a + (50 − 1) d
106 = a + 49d ... (ii)
On subtracting (i) from (ii), we get
94 = 47d
d = 2
From equation (i), we get
12 = a + 2 (2)
a = 12 − 4 = 8
a_{29} = a + (29 − 1) d
a_{29} = 8 + (28)2
a_{29} = 8 + 56 = 64
Therefore, 29^{th} term is 64.
9. If the 3^{rd} and the 9^{th} terms of an A.P. are 4 and − 8 respectively. Which term of this A.P. is zero.
Answer
Given that,
a_{9} = −8
We know that,
a_{n} = a + (n − 1) d
a_{3} = a + (3 − 1) d
4 = a + 2d ... (i)
a_{9} = a + (9 − 1) d
−8 = a + 8d ... (ii)
On subtracting equation (i) from (ii), we get,
−12 = 6d
d = −2
From equation (i), we get,
4 = a + 2 (−2)
4 = a − 4
a = 8
Let n^{th} term of this A.P. be zero.
a_{n }= a + (n − 1) d
0 = 8 + (n − 1) (−2)
0 = 8 − 2n + 2
2n = 10
n = 5
Hence, 5^{th} term of this A.P. is 0.
10. If 17^{th} term of an A.P. exceeds its 10^{th} term by 7. Find the common difference.
Answer
We know that,
For an A.P., a_{n} = a + (n − 1) d
a_{17} = a + (17 − 1) d
a_{17} = a + 16d
Similarly, a_{10} = a + 9d
It is given that
a_{17} − a_{10} = 7
(a + 16d) − (a + 9d) = 7
7d = 7
d = 1
Therefore, the common difference is 1.
11. Which term of the A.P. 3, 15, 27, 39, … will be 132 more than its 54^{th} term?
Answer
Given A.P. is 3, 15, 27, 39, …
a = 3
d = a_{2} − a_{1} = 15 − 3 = 12
a_{54} = a + (54 − 1) d
= 3 + (53) (12)
= 3 + 636 = 639
132 + 639 = 771
We have to find the term of this A.P. which is 771.
Let n^{th} term be 771.
a_{n} = a + (n − 1) d
771 = 3 + (n − 1) 12
768 = (n − 1) 12
(n − 1) = 64
n = 65
Therefore, 65^{th} term was 132 more than 54^{th} term.
Or
Let n^{th} term be 132 more than 54^{th} term.
n = 54 + 132/2
= 54 + 11 = 65^{th} term
12. Two APs have the same common difference. The difference between their 100^{th} term is 100, what is the difference between their 1000^{th} terms?
Answer
Let the first term of these A.P.s be a_{1} and a_{2} respectively and the common difference of these A.P.s be d.
For first A.P.,
a_{100} = a_{1} + (100 − 1) d
= a_{1} + 99d
a_{1000} = a_{1} + (1000 − 1) d
a_{1000} = a_{1} + 999d
For second A.P.,
a_{100} = a_{2} + (100 − 1) d
= a_{2} + 99d
a_{1000} = a_{2} + (1000 − 1) d
= a_{2} + 999d
Given that, difference between
100^{th} term of these A.P.s = 100
Therefore, (a_{1} + 99d) − (a_{2} + 99d) = 100
a_{1} − a_{2} = 100 ... (i)
Difference between 1000^{th} terms of these A.P.s
(a_{1} + 999d) − (a_{2} + 999d) = a_{1} − a_{2}
From equation (i),
This difference, a_{1} − a_{2 }= 100
Hence, the difference between 1000^{th} terms of these A.P. will be 100.
13. How many three digit numbers are divisible by 7?
Answer
First threedigit number that is divisible by 7 = 105
Next number = 105 + 7 = 112
Therefore, 105, 112, 119, …
All are three digit numbers which are divisible by 7 and thus, all these are terms of an A.P. having first term as 105 and common difference as 7.
The maximum possible threedigit number is 999. When we divide it by 7, the remainder will be 5. Clearly, 999 − 5 = 994 is the maximum possible threedigit number that is divisible by 7.
The series is as follows.
105, 112, 119, …, 994
Let 994 be the nth term of this A.P.
a = 105
d = 7
a_{n} = 994
n = ?
a_{n} = a + (n − 1) d
994 = 105 + (n − 1) 7
889 = (n − 1) 7
(n − 1) = 127
n = 128
Therefore, 128 threedigit numbers are divisible by 7.
Or
Three digit numbers which are divisible by 7 are 105, 112, 119, .... 994 .
These numbers form an AP with a = 105 and d = 7.
Let number of threedigit numbers divisible by 7 be n, a_{n} = 994
⇒ a + (n  1) d = 994
⇒ 105 + (n  1) × 7 = 994
⇒7(n  1) = 889
⇒ n  1 = 127
⇒ n = 128
14. How many multiples of 4 lie between 10 and 250?
Answer
First multiple of 4 that is greater than 10 is 12. Next will be 16.
Therefore, 12, 16, 20, 24, …
All these are divisible by 4 and thus, all these are terms of an A.P. with first term as 12 and common difference as 4.
When we divide 250 by 4, the remainder will be 2. Therefore, 250 − 2 = 248 is divisible by 4.
The series is as follows.
12, 16, 20, 24, …, 248
Let 248 be the n^{th} term of this A.P.
a = 12
d = 4
a_{n} = 248
a_{n} = a + (n  1) d
248 = 12 + (n  1) × 4
236/4 = n  1
59 = n  1
n = 60
Therefore, there are 60 multiples of 4 between 10 and 250.
Or
Multiples of 4 lies between 10 and 250 are 12, 16, 20, ...., 248.
These numbers form an AP with a = 12 and d = 4.
Let number of threedigit numbers divisible by 4 be n, a_{n} = 248
⇒ a + (n  1) d = 248
⇒ 12 + (n  1) × 4 = 248
⇒4(n  1) = 248
⇒ n  1 = 59
⇒ n = 60
15. For what value of n, are the n^{th} terms of two APs 63, 65, 67, and 3, 10, 17, … equal?
Answer
63, 65, 67, …
a = 63
d = a_{2} − a_{1} = 65 − 63 = 2
n^{th} term of this A.P. = a_{n} = a + (n − 1) d
a_{n}= 63 + (n − 1) 2 = 63 + 2n − 2
a_{n} = 61 + 2n ... (i)
3, 10, 17, …
a = 3
d = a_{2} − a_{1} = 10 − 3 = 7
n^{th} term of this A.P. = 3 + (n − 1) 7
a_{n}_{ }= 3 + 7n − 7
a_{n} = 7n − 4 ... (ii)
It is given that, n^{th} term of these A.P.s are equal to each other.
Equating both these equations, we obtain
61 + 2n = 7n − 4
61 + 4 = 5n
5n = 65
n = 13
Therefore, 13^{th} terms of both these A.P.s are equal to each other.
16. Determine the A.P. whose third term is 16 and the 7^{th} term exceeds the 5^{th} term by 12.
Answer
a_{3} = 16
a + 2d = 16 ... (i)
a_{7} − a_{5} = 12
[a+ (7 − 1) d] − [a + (5 − 1) d]= 12
(a + 6d) − (a + 4d) = 12
2d = 12
d = 6
From equation (i), we get,
a + 2 (6) = 16
a + 12 = 16
a = 4
Therefore, A.P. will be
4, 10, 16, 22, …
Page No: 107
17. Find the 20^{th} term from the last term of the A.P. 3, 8, 13, …, 253.
Answer
Given A.P. is
3, 8, 13, …, 253
Common difference for this A.P. is 5.
Therefore, this A.P. can be written in reverse order as
253, 248, 243, …, 13, 8, 5
For this A.P.,
a = 253
d = 248 − 253 = −5
n = 20
a_{20} = a + (20 − 1) d
a_{20} = 253 + (19) (−5)
a_{20} = 253 − 95
a = 158
Therefore, 20^{th} term from the last term is 158.
18. The sum of 4^{th} and 8^{th} terms of an A.P. is 24 and the sum of the 6^{th} and 10^{th} terms is 44. Find the first three terms of the A.P.
Answer
We know that,
a_{n} = a + (n − 1) d
a_{4} = a + (4 − 1) d
a_{4} = a + 3d
Similarly,
a_{8} = a + 7d
a_{6} = a + 5d
a_{10} = a + 9d
Given that, a_{4} + a_{8} = 24
a + 3d + a + 7d = 24
2a + 10d = 24
a + 5d = 12 ... (i)
a_{6} + a_{10} = 44
a + 5d + a + 9d = 44
2a + 14d = 44
a + 7d = 22 ... (ii)
On subtracting equation (i) from (ii), we get,
2d = 22 − 12
2d = 10
d = 5
From equation (i), we get
a + 5d = 12
a + 5 (5) = 12
a + 25 = 12
a = −13
a_{2} = a + d = − 13 + 5 = −8
a_{3} = a_{2} + d = − 8 + 5 = −3
Therefore, the first three terms of this A.P. are −13, −8, and −3.
19. Subba Rao started work in 1995 at an annual salary of Rs 5000 and received an increment of Rs 200 each year. In which year did his income reach Rs 7000?
Answer
It can be observed that the incomes that Subba Rao obtained in various years are in A.P. as every year, his salary is increased by Rs 200.
Therefore, the salaries of each year after 1995 are
5000, 5200, 5400, …
Here, a = 5000
d = 200
Let after n^{th} year, his salary be Rs 7000.
Therefore, a_{n} = a + (n − 1) d
7000 = 5000 + (n − 1) 200
200(n − 1) = 2000
(n − 1) = 10
n = 11
Therefore, in 11th year, his salary will be Rs 7000.
20. Ramkali saved Rs 5 in the first week of a year and then increased her weekly saving by Rs 1.75. If in the n^{th} week, her week, her weekly savings become Rs 20.75, find n.
Answer
Given that,
a = 5
d = 1.75
a_{n }= 20.75
n = ?
a_{n} = a + (n − 1) d
20.75 = 5 + (n  1) × 1.75
15.75 = (n  1) × 1.75
(n  1) = 15.75/1.75 = 1575/175
= 63/7 = 9
n  1 = 9
n = 10
Hence, n is 10.
Page No: 112
Exercise 5.3
1. Find 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) 1/15, 1/12, 1/10, ...... , to 11 terms
Answer
(i) 2, 7, 12 ,…, to 10 terms
For this A.P.,
a = 2
d = a_{2} − a_{1} = 7 − 2 = 5
n = 10
We know that,
S_{n} = n/2 [2a + (n  1) d]
S_{10} = 10/2 [2(2) + (10  1) × 5]
= 5[4 + (9) × (5)]
= 5 × 49 = 245
(ii) −37, −33, −29 ,…, to 12 terms
For this A.P.,
a = −37
d = a_{2} − a_{1} = (−33) − (−37)
= − 33 + 37 = 4
n = 12
We know that,
S_{n} = n/2 [2a + (n  1) d]
S_{12} = 12/2 [2(37) + (12  1) × 4]
= 6[74 + 11 × 4]
= 6[74 + 44]
= 6(30) = 180
(iii) 0.6, 1.7, 2.8 ,…, to 100 terms
For this A.P.,
a = 0.6
d = a_{2} − a_{1} = 1.7 − 0.6 = 1.1
n = 100
We know that,
S_{n} = n/2 [2a + (n  1) d]
S_{12} = 50/2 [1.2 + (99) × 1.1]
= 50[1.2 + 108.9]
= 50[110.1]
= 5505
Exercise 5.3
1. Find 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) 1/15, 1/12, 1/10, ...... , to 11 terms
Answer
(i) 2, 7, 12 ,…, to 10 terms
For this A.P.,
a = 2
d = a_{2} − a_{1} = 7 − 2 = 5
n = 10
We know that,
S_{n} = n/2 [2a + (n  1) d]
S_{10} = 10/2 [2(2) + (10  1) × 5]
= 5[4 + (9) × (5)]
= 5 × 49 = 245
(ii) −37, −33, −29 ,…, to 12 terms
For this A.P.,
a = −37
d = a_{2} − a_{1} = (−33) − (−37)
= − 33 + 37 = 4
n = 12
We know that,
S_{n} = n/2 [2a + (n  1) d]
S_{12} = 12/2 [2(37) + (12  1) × 4]
= 6[74 + 11 × 4]
= 6[74 + 44]
= 6(30) = 180
(iii) 0.6, 1.7, 2.8 ,…, to 100 terms
For this A.P.,
a = 0.6
d = a_{2} − a_{1} = 1.7 − 0.6 = 1.1
n = 100
We know that,
S_{n} = n/2 [2a + (n  1) d]
S_{12} = 50/2 [1.2 + (99) × 1.1]
= 50[1.2 + 108.9]
= 50[110.1]
= 5505
(iv) 1/15, 1/12, 1/10, ...... , to 11 terms
For this A.P.,
2. Find the sums given below
(i) 7 + + 14 + .................. +84
(ii)+ 14 + ………… + 84
(ii) 34 + 32 + 30 + ……….. + 10
(iii) − 5 + (− 8) + (− 11) + ………… + (− 230)
Answer
(i) For this A.P.,
a = 7
l = 84
d = a_{2} − a_{1} =  7 = 21/2  7 = 7/2
Let 84 be the n^{th }term of this A.P.
l = a (n  1)d
84 = 7 + (n  1) × 7/2
77 = (n  1) × 7/2
22 = n − 1
n = 23
We know that,
S_{n} = n/2 (a + l)
S_{n} = 23/2 (7 + 84)
= (23×91/2) = 2093/2
=
(ii) 34 + 32 + 30 + ……….. + 10
For this A.P.,
a = 34
d = a_{2} − a_{1} = 32 − 34 = −2
l = 10
Let 10 be the n^{th} term of this A.P.
l = a + (n − 1) d
10 = 34 + (n − 1) (−2)
−24 = (n − 1) (−2)
12 = n − 1
n = 13
S_{n} = n/2 (a + l)
= 13/2 (34 + 10)
= (13×44/2) = 13 × 22
= 286
(iii) (−5) + (−8) + (−11) + ………… + (−230) For this A.P.,
a = −5
l = −230
d = a_{2} − a_{1} = (−8) − (−5)
= − 8 + 5 = −3
Let −230 be the n^{th} term of this A.P.
l = a + (n − 1)d
−230 = − 5 + (n − 1) (−3)
−225 = (n − 1) (−3)
(n − 1) = 75
n = 76
And,
S_{n} = n/2 (a + l)
= 76/2 [(5) + (230)]
= 38(235)
= 8930
3. In an AP
(i) Given a = 5, d = 3, a_{n} = 50, find n and S_{n}.
(ii) Given a = 7, a_{13} = 35, find d and S_{13}.
(iii) Given a_{12} = 37, d = 3, find a and S_{12}.
(iv) Given a_{3} = 15, S_{10} = 125, find d and a_{10}.
(v) Given d = 5, S_{9} = 75, find a and a_{9}.
(vi) Given a = 2, d = 8, S_{n} = 90, find n and a_{n}.
(vii) Given a = 8, a_{n} = 62, S_{n} = 210, find n and d.
(viii) Given a_{n} = 4, d = 2, S_{n} = − 14, find n and a.
(ix) Given a = 3, n = 8, S = 192, find d.
(x) Given l = 28, S = 144 and there are total 9 terms. Find a.
Answer
(i) Given that, a = 5, d = 3, a_{n} = 50
As a_{n} = a + (n − 1)d,
⇒ 50 = 5 + (n  1) × 3
⇒ 3(n  1) = 45
⇒ n  1 = 15
⇒ n = 16
Now, S_{n} = n/2 (a + a_{n})
S_{n} = 16/2 (5 + 50) = 440
(ii) Given that, a = 7, a_{13} = 35
As a_{n} = a + (n − 1)d, ⇒ 35 = 7 + (13  1)d
⇒ 12d = 28
⇒ d = 28/12 = 2.33
Now, S_{n} = n/2 (a + a_{n})
S_{13} = 13/2 (7 + 35) = 273
(iii)Given that, a_{12} = 37, d = 3 As a_{n} = a + (n − 1)d,
⇒ a_{12} = a + (12 − 1)3
⇒ 37 = a + 33
⇒ a = 4
S_{n} = n/2 (a + a_{n})
S_{n} = 12/2 (4 + 37)
= 246
(iv) Given that, a_{3} = 15, S_{10} = 125
As a_{n} = a + (n − 1)d,
a_{3} = a + (3 − 1)d
15 = a + 2d ... (i)
S_{n} = n/2 [2a + (n  1)d]
S_{10} = 10/2 [2a + (10  1)d]
125 = 5(2a + 9d)
25 = 2a + 9d ... (ii)
On multiplying equation (i) by (ii), we get
30 = 2a + 4d ... (iii)
On subtracting equation (iii) from (ii), we get
−5 = 5d
d = −1
From equation (i),
15 = a + 2(−1)
15 = a − 2
a = 17
a_{10} = a + (10 − 1)d
a_{10} = 17 + (9) (−1)
a_{10} = 17 − 9 = 8
(v) Given that, d = 5, S_{9} = 75
As S_{n} = n/2 [2a + (n  1)d]
S_{9} = 9/2 [2a + (9  1)5]
25 = 3(a + 20)
25 = 3a + 60
3a = 25 − 60
a = 35/3
a_{n} = a + (n − 1)d
a_{9} = a + (9 − 1) (5)
= 35/3 + 8(5)
= 35/3 + 40
= (35+120/3) = 85/3
(vi) Given that, a = 2, d = 8, S_{n} = 90
As S_{n} = n/2 [2a + (n  1)d]
90 = n/2 [2a + (n  1)d]
⇒ 180 = n(4 + 8n  8) = n(8n  4) = 8n^{2}  4n
⇒ 8n^{2}  4n  180 = 0
⇒ 2n^{2}  n  45 = 0
⇒ 2n^{2}  10n + 9n  45 = 0
⇒ 2n(n 5) + 9(n  5) = 0
⇒ (2n  9)(2n + 9) = 0
So, n = 5 (as it is positive integer)
∴ a_{5 }= 8 + 5 × 4 = 34
(vii) Given that, a = 8, a_{n} = 62, S_{n} = 210
As S_{n} = n/2 (a + a_{n})
210 = n/2 (8 + 62)
⇒ 35n = 210
⇒ n = 210/35 = 6
Now, 62 = 8 + 5d
⇒ 5d = 62  8 = 54
⇒ d = 54/5 = 10.8
(viii) Given that, a_{n} = 4, d = 2, S_{n} = −14
a_{n} = a + (n − 1)d
4 = a + (n − 1)2
4 = a + 2n − 2
a + 2n = 6
a = 6 − 2n ... (i)
S_{n} = n/2 (a + a_{n})
14 = n/2 (a + 4)
−28 = n (a + 4)
−28 = n (6 − 2n + 4) {From equation (i)}
−28 = n (− 2n + 10)
−28 = − 2n^{2} + 10n
2n^{2} − 10n − 28 = 0
n^{2} − 5n −14 = 0
n^{2} − 7n + 2n − 14 = 0
n (n − 7) + 2(n − 7) = 0
(n − 7) (n + 2) = 0
Either n − 7 = 0 or n + 2 = 0
n = 7 or n = −2
However, n can neither be negative nor fractional.
Therefore, n = 7
From equation (i), we get
a = 6 − 2n
a = 6 − 2(7)
= 6 − 14
= −8
(ix) Given that, a = 3, n = 8, S = 192
As S_{n} = n/2 [2a + (n  1)d]
192 = 8/2 [2 × 3 + (8  1)d]
192 = 4 [6 + 7d]
48 = 6 + 7d
42 = 7d
d = 6
(x) Given that, l = 28, S = 144 and there are total of 9 terms.
S_{n} = n/2 (a + l)
144 = 9/2 (a + 28)
(16) × (2) = a + 28
32 = a + 28
a = 4
Page No: 113
4. How many terms of the AP. 9, 17, 25 … must be taken to give a sum of 636?
Answer
Let there be n terms of this A.P.
For this A.P., a = 9
d = a_{2} − a_{1} = 17 − 9 = 8
As S_{n} = n/2 [2a + (n  1)d]
636 = n/2 [2 × a + (8  1) × 8]
636 = n/2 [18 + (n 1) × 8]
636 = n [9 + 4n − 4]
636 = n (4n + 5)
4n^{2} + 5n − 636 = 0
4n^{2} + 53n − 48n − 636 = 0
n (4n + 53) − 12 (4n + 53) = 0
(4n + 53) (n − 12) = 0
Either 4n + 53 = 0 or n − 12 = 0
n = (53/4) or n = 12
n cannot be (53/4). As the number of terms can neither be negative nor fractional, therefore, n = 12 only.
5. The 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.
Answer
Given that,
a = 5
l = 45
S_{n} = 400
S_{n} = n/2 (a + l)
400 = n/2 (5 + 45)
400 = n/2 (50)
n = 16
l = a + (n − 1) d
45 = 5 + (16 − 1) d
40 = 15d
d = 40/15 = 8/3
6. The first and the last term of an AP are 17 and 350 respectively. If the common difference is 9, how many terms are there and what is their sum?
Answer
Given that,
a = 17
l = 350
d = 9
Let there be n terms in the A.P.
l = a + (n − 1) d
350 = 17 + (n − 1)9
333 = (n − 1)9
(n − 1) = 37
n = 38
S_{n} = n/2 (a + l)
S_{38} = 13/2 (17 + 350)
= 19 × 367
= 6973
Thus, this A.P. contains 38 terms and the sum of the terms of this A.P. is 6973.
7. Find the sum of first 22 terms of an AP in which d = 7 and 22^{nd} term is 149.
Answer
d = 7
a_{22} = 149
S_{22} = ?
a_{n} = a + (n − 1)d
a_{22} = a + (22 − 1)d
149 = a + 21 × 7
149 = a + 147
a = 2
S_{n} = n/2 (a + a_{n})
= 22/2 (2 + 149)
= 11 × 151
= 1661
8. Find the sum of first 51 terms of an AP whose second and third terms are 14 and 18 respectively.
Answer
Given that,
a_{2} = 14
a_{3} = 18
d = a_{3} − a_{2} = 18 − 14 = 4
a_{2} = a + d
14 = a + 4
a = 10
S_{n} = n/2 [2a + (n  1)d]
S_{51} = 51/2 [2 × 10 + (51  1) × 4]
= 51/2 [2 + (20) × 4]
= 51×220/2
= 51 × 110
= 5610
9. If 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.
Answer
Given that,
S_{7} = 49l = 350
d = 9
Let there be n terms in the A.P.
l = a + (n − 1) d
350 = 17 + (n − 1)9
333 = (n − 1)9
(n − 1) = 37
n = 38
S_{n} = n/2 (a + l)
S_{38} = 13/2 (17 + 350)
= 19 × 367
= 6973
Thus, this A.P. contains 38 terms and the sum of the terms of this A.P. is 6973.
7. Find the sum of first 22 terms of an AP in which d = 7 and 22^{nd} term is 149.
Answer
d = 7
a_{22} = 149
S_{22} = ?
a_{n} = a + (n − 1)d
a_{22} = a + (22 − 1)d
149 = a + 21 × 7
149 = a + 147
a = 2
S_{n} = n/2 (a + a_{n})
= 22/2 (2 + 149)
= 11 × 151
= 1661
8. Find the sum of first 51 terms of an AP whose second and third terms are 14 and 18 respectively.
Answer
Given that,
a_{2} = 14
a_{3} = 18
d = a_{3} − a_{2} = 18 − 14 = 4
a_{2} = a + d
14 = a + 4
a = 10
S_{n} = n/2 [2a + (n  1)d]
S_{51} = 51/2 [2 × 10 + (51  1) × 4]
= 51/2 [2 + (20) × 4]
= 51×220/2
= 51 × 110
= 5610
9. If 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.
Answer
Given that,
S_{17} = 289
S_{7} = 7/2 [2a + (n  1)d]
S_{7} = 7/2 [2a + (7  1)d]
49 = 7/2 [2a + 16d]
7 = (a + 3d)
a + 3d = 7 ... (i)
Similarly,
S_{17} = 17/2 [2a + (17  1)d]
289 = 17/2 (2a + 16d)
17 = (a + 8d)
a + 8d = 17 ... (ii)
Subtracting equation (i) from equation (ii),
5d = 10
d = 2
From equation (i),
a + 3(2) = 7
a + 6 = 7
a = 1
S_{n} = n/2 [2a + (n  1)d]
= n/2 [2(1) + (n  1) × 2]
= n/2 (2 + 2n  2)
= n/2 (2n)
= n^{2}
^{} 10. Show that a_{1}, a_{2 }… , a_{n} , … form an AP where a_{n} is defined as below
(i) a_{n} = 3 + 4n
(ii) a_{n} = 9 − 5n
Also find the sum of the first 15 terms in each case.
Answer
(i) a_{n} = 3 + 4n
a_{1} = 3 + 4(1) = 7
a_{2} = 3 + 4(2) = 3 + 8 = 11
a_{3} = 3 + 4(3) = 3 + 12 = 15
a_{4} = 3 + 4(4) = 3 + 16 = 19
It can be observed that
a_{2} − a_{1} = 11 − 7 = 4
a_{3} − a_{2} = 15 − 11 = 4
a_{4} − a_{3} = 19 − 15 = 4
i.e., a_{k}_{ + 1} − a_{k} is same every time. Therefore, this is an AP with common difference as 4 and first term as 7.
S_{n} = n/2 [2a + (n  1)d]
S_{15 }= 15/2 [2(7) + (15  1) × 4]
= 15/2 [(14) + 56]
= 15/2 (70)
= 15 × 35
= 525
(ii) a_{n} = 9 − 5n
a_{1} = 9 − 5 × 1 = 9 − 5 = 4
a_{2} = 9 − 5 × 2 = 9 − 10 = −1
a_{3} = 9 − 5 × 3 = 9 − 15 = −6
a_{4} = 9 − 5 × 4 = 9 − 20 = −11
It can be observed that
a_{2} − a_{1} = − 1 − 4 = −5
a_{3} − a_{2} = − 6 − (−1) = −5
a_{4} − a_{3} = − 11 − (−6) = −5
i.e., a_{k}_{ + 1} − a_{k} is same every time. Therefore, this is an A.P. with common difference as −5 and first term as 4.
S_{n} = n/2 [2a + (n  1)d]
S_{15 }= 15/2 [2(4) + (15  1) (5)]
= 15/2 [8 + 14(5)]
= 15/2 (8  70)
= 15/2 (62)
= 15(31)
= 465
11. If the sum of the first n terms of an AP is 4n − n^{2}, what is the first term (that is S_{1})? What is the sum of first two terms? What is the second term? Similarly find the 3^{rd}, the10^{th} and the n^{th} terms.
Answer
Given that,
S_{n} = 4n − n^{2}
First term, a = S_{1} = 4(1) − (1)^{2} = 4 − 1 = 3
Sum of first two terms = S_{2}
= 4(2) − (2)^{2} = 8 − 4 = 4
Second term, a_{2} = S_{2} − S_{1} = 4 − 3 = 1
d = a_{2} − a = 1 − 3 = −2
a_{n} = a + (n − 1)d
= 3 + (n − 1) (−2)
= 3 − 2n + 2
= 5 − 2n
Therefore, a_{3} = 5 − 2(3) = 5 − 6 = −1
a_{10} = 5 − 2(10) = 5 − 20 = −15
Hence, the sum of first two terms is 4. The second term is 1. 3^{rd}, 10^{th}, and n^{th} terms are −1, −15, and 5 − 2n respectively.
12. Find the sum of first 40 positive integers divisible by 6.
Answer
The positive integers that are divisible by 6 are
6, 12, 18, 24 …
It can be observed that these are making an A.P. whose first term is 6 and common difference is 6.
a = 6
d = 6
S_{40} = ?
S_{n} = n/2 [2a + (n  1)d]
S_{40 }= 40/2 [2(6) + (40  1) 6]
= 20[12 + (39) (6)]
= 20(12 + 234)
= 20 × 246
= 4920
13. Find the sum of first 15 multiples of 8.
Answer
The multiples of 8 are
8, 16, 24, 32…
These are in an A.P., having first term as 8 and common difference as 8.
Therefore, a = 8
d = 8
S_{15} = ?
S_{n} = n/2 [2a + (n  1)d]
S_{15} = 15/2 [2(8) + (15  1)8]
= 15/2[6 + (14) (8)]
= 15/2[16 + 112]
= 15(128)/2
= 15 × 64
= 960
14. Find the sum of the odd numbers between 0 and 50.
Answer
The odd numbers between 0 and 50 are
1, 3, 5, 7, 9 … 49
Therefore, it can be observed that these odd numbers are in an A.P.
a = 1
d = 2
l = 49
l = a + (n − 1) d
49 = 1 + (n − 1)2
48 = 2(n − 1)
n − 1 = 24
n = 25
S_{n} = n/2 (a + l)
S_{25} = 25/2 (1 + 49)
= 25(50)/2
=(25)(25)
= 625
15. A contract on construction job specifies a penalty for delay of completion beyond a certain dateas 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.
Answer
It can be observed that these penalties are in an A.P. having first term as 200 and common difference as 50.
a = 200
d = 50
Penalty that has to be paid if he has delayed the work by 30 days = S_{30}
= 30/2 [2(200) + (30  1) 50]
= 15 [400 + 1450]
= 15 (1850)
= 27750
Therefore, the contractor has to pay Rs 27750 as penalty.
16. A sum of Rs 700 is to be used to give seven cash prizes to students of a school for their overall academic performance. If each prize is Rs 20 less than its preceding prize, find the value of each of the prizes.
Answer
Let the cost of 1^{st} prize be P.
Cost of 2^{nd} prize = P − 20
And cost of 3^{rd} prize = P − 40
It can be observed that the cost of these prizes are in an A.P. having common difference as −20 and first term as P.
a = P
d = −20
Given that, S_{7} = 700
7/2 [2a + (7  1)d] = 700
a + 3(−20) = 100
a − 60 = 100
a = 160
Therefore, the value of each of the prizes was Rs 160, Rs 140, Rs 120, Rs 100, Rs 80, Rs 60, and Rs 40.
17. In 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?
Answer
It can be observed that the number of trees planted by the students is in an AP.
1, 2, 3, 4, 5………………..12
First term, a = 1
Common difference, d = 2 − 1 = 1
S_{n} = n/2 [2a + (n  1)d]
S_{12} = 12/2 [2(1) + (12  1)(1)]
= 6 (2 + 11)
= 6 (13)
= 78
Therefore, number of trees planted by 1 section of the classes = 78
Number of trees planted by 3 sections of the classes = 3 × 78 = 234
Therefore, 234 trees will be planted by the students.
18. A spiral is made up of successive semicircles, with centres alternately at A and B, starting with centre at A of radii 0.5, 1.0 cm, 1.5 cm, 2.0 cm, ……… as shown in figure. What is the total length of such a spiral made up of thirteen consecutive semicircles? (Take Ï€ = 22/7)
perimeter of semicircle = Ï€r
_{P1} = Ï€(0.5) = Ï€/2 cm
_{P2} = Ï€(1) = Ï€ cm
_{P3} = Ï€(1.5) = 3Ï€/2 cm
_{P1}, P_{2}, P_{3} are the lengths of the semicircles
Ï€/2, Ï€, 3Ï€/2, 2Ï€, ....
P1 = Ï€/2 cm
_{P2} = Ï€ cm
d = P2 P1 = Ï€  Ï€/2 = Ï€/2
First term = P1 = a = Ï€/2 cm
S_{n} = n/2 [2a + (n  1)d]
Therefor, Sum of the length of 13 consecutive circles
S_{13} = 13/2 [2(Ï€/2) + (13  1)Ï€/2]
= 13/2 [Ï€ + 6Ï€]
=13/2 (7Ï€) = 13/2 × 7 × 22/7
= 143 cm
Page No: 114
19. 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. In how many rows are the 200 logs placed and how many logs are in the top row?
It can be observed that the numbers of logs in rows are in an A.P.
20, 19, 18…
For this A.P.,
a = 20
d = a_{2} − a_{1} = 19 − 20 = −1
Let a total of 200 logs be placed in n rows.
S_{n} = 200
S_{n} = n/2 [2a + (n  1)d]
S_{12} = 12/2 [2(20) + (n  1)(1)]
400 = n (40 − n + 1)
400 = n (41 − n)
400 = 41n − n^{2}
n^{2} − 41n + 400 = 0
n^{2} − 16n − 25n + 400 = 0
n (n − 16) −25 (n − 16) = 0
(n − 16) (n − 25) = 0
Either (n − 16) = 0 or n − 25 = 0
n = 16 or n = 25
a_{n} = a + (n − 1)d
a_{16} = 20 + (16 − 1) (−1)
a_{16} = 20 − 15
a_{16} = 5
Similarly,
a_{25} = 20 + (25 − 1) (−1)
a_{25} = 20 − 24
= −4
Clearly, the number of logs in 16^{th} row is 5. However, the number of logs in 25^{th} row is negative, which is not possible.
Therefore, 200 logs can be placed in 16 rows and the number of logs in the 16^{th} row is 5.
20. In a potato race, a bucket is placed at the starting point, which is 5 m from the first potato and other potatoes are placed 3 m apart in a straight line. There are ten potatoes in the line.
[Hint: to pick up the first potato and the second potato, the total distance (in metres) run by a competitor is 2 × 5 + 2 ×(5 + 3)]
Answer
The distances of potatoes from the bucket are 5, 8, 11, 14…
Distance run by the competitor for collecting these potatoes are two times of the distance at which the potatoes have been kept because first she has to first pick the potato and again return back to the same place in order to start picking the second potato.. Therefore, distances to be run are
10, 16, 22, 28, 34,……….
a = 10
d = 16 − 10 = 6
S_{10} =?
S_{10} = 10/2 [2(20) + (n  1)(1)]
= 5[20 + 54]
= 5 (74)
= 370
Therefore, the competitor will run a total distance of 370 m.
Exercise 5.4 (Optional)
1. Which term of the AP : 121, 117, 113, . . ., is its first negative term?
[Hint : Find n for a_{n} < 0]
Answer
We have the A.P. having a = 121 and d = 117  121 =  4
∴ an = a + (n  1) d
= 121 + (n  1) × ( 4)
= 121  4n + 4
= 125  4n
For the first negative term, we have
an < 0
⇒ (125  4n) < 0
⇒ 125 < 4n
⇒ 125/4 <n
⇒ n > 31 1⁄4
Thus, the first negative term is 32nd term.
2. The 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.
Answer
3. A 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 21⁄2 m apart, what is the length of the wood required for the rungs?
[Hint : Number of rungs = 250/25 + 1]
Answer
4. 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 numbered x is equal to the sum of the numbers of the houses following it. Find this value of x.
[Hint : S_{x1} = S_{49} – S_{x}]
Answer
5. A 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 1⁄4 m and a tread of 1⁄2 m. (see Fig. 5.8). Calculate the total volume of concrete required to build the terrace.
[Hint : Volume of concrete required to build the first step = 1/4 × 1/2 × 50m^{3}]
Answer
Go Back To NCERT Solutions for Class 10 Maths
NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions
Chapter 5 Arithmetic Progressions NCERT Solutions is provided here which will be essential in knowing the topics that could come in the examinations. A sequence of numbers in which the successive terms increase or decrease by a constant number is called an Arithmetic Progression (AP).
There are total 4 sections in this chapter that will widen your perspective. It will develop you thinking skills and application of formulas to yield results.
• Introduction: This section will introduce you to some real life examples where arithmetic progressions can be practically used.
• Arithmetic Progressions: You will find definition of Arithmetic progressions and terms like general form of an AP, finite and infinite Arithmetic Progressions.
• nth Term of an AP: We will learn to calculate nth term of an AP. The nth term a n of the AP with first term a and common difference d is given by a_{n} = a + (n – 1) d.
• Sum of First n Terms of an AP: We will know how to find the sum of the first n terms of an AP is which is given by S = n/2 [2a + (n1)d]. If l is the last term of the finite AP, say the nth term, then the sum of all terms of the AP which is given by S = n/2(a+l).
There are total 4 exercises in the chapter in which last one is optional which can be useful in revising the chapter. We have also provided exercisewise NCERT Solutions of chapter 5 Arithmetic Progressions.
NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions
Chapter 5 Arithmetic Progressions NCERT Solutions is provided here which will be essential in knowing the topics that could come in the examinations. A sequence of numbers in which the successive terms increase or decrease by a constant number is called an Arithmetic Progression (AP).
There are total 4 sections in this chapter that will widen your perspective. It will develop you thinking skills and application of formulas to yield results.
• Introduction: This section will introduce you to some real life examples where arithmetic progressions can be practically used.
• Arithmetic Progressions: You will find definition of Arithmetic progressions and terms like general form of an AP, finite and infinite Arithmetic Progressions.
• nth Term of an AP: We will learn to calculate nth term of an AP. The nth term a n of the AP with first term a and common difference d is given by a_{n} = a + (n – 1) d.
• Sum of First n Terms of an AP: We will know how to find the sum of the first n terms of an AP is which is given by S = n/2 [2a + (n1)d]. If l is the last term of the finite AP, say the nth term, then the sum of all terms of the AP which is given by S = n/2(a+l).
There are total 4 exercises in the chapter in which last one is optional which can be useful in revising the chapter. We have also provided exercisewise NCERT Solutions of chapter 5 Arithmetic Progressions.
NCERT Solutions for Class 10 Maths Chapters:
FAQ on Chapter 5 Arithmetic Progressions
How many exercises in Chapter 5 Arithmetic Progressions
There are only 4 exercises in the Chapter 5 Class 10. You will find the answers of every questions in detailed so you can complete your work on time and crosscheck your answers. These NCERT Solutions for Chapter 5 Arithmetic Progresssions Class 10 Maths will help you a lot in board exams.
What do you mean by Sequence?
When some numbers are arranged in a definite order, according to a definite rule, they are said to form a sequence.
The number occurring at the 1st place is called the 1st term, denoted by T_{1}.
The number occurring at the nth place is called the nth term, denoted by T_{n}.
The number occurring at the 1st place is called the 1st term, denoted by T_{1}.
The number occurring at the nth place is called the nth term, denoted by T_{n}.
If 9th term of an A.P. is zero, prove that its 29th term is double of its 19th term.
Let ‘a’ be the first term and ‘d’ be the common difference.
Now, Using T_{n} = a + (n  1) d, we have
T_{9} = a + 8d ⇒ a + 8d = 0 ...(1) [∵ T9 = 0 Given]
T_{19} = a + 18d = (a + 8d) + 10d = (0) + 10d = 10d ...(2)
[∵ a + 8d = 0]
T_{29} = a + 28d
= (a + 8d) + 20d
= 0 + 20d = 20d [∵ a + 8d = 0]
= 2 × (10d) = 2 (T_{19}) [∵ T_{19} = 10d]
⇒ T_{29} = 2 (T_{19})
Thus, the 29th term of the A.P. is double of its 19th term.
Now, Using T_{n} = a + (n  1) d, we have
T_{9} = a + 8d ⇒ a + 8d = 0 ...(1) [∵ T9 = 0 Given]
T_{19} = a + 18d = (a + 8d) + 10d = (0) + 10d = 10d ...(2)
[∵ a + 8d = 0]
T_{29} = a + 28d
= (a + 8d) + 20d
= 0 + 20d = 20d [∵ a + 8d = 0]
= 2 × (10d) = 2 (T_{19}) [∵ T_{19} = 10d]
⇒ T_{29} = 2 (T_{19})
Thus, the 29th term of the A.P. is double of its 19th term.
What is Common Difference?
The constant difference between two consecutive terms of an AP is called “Common Difference”.