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 Page 1


 
1. Find the sum of each of the following Aps:
(iii) -37, -33, -
(iv) 
1 1 1
, , ,.....
15 12 10
to 11 terms. 
Sol:
(i)
Here, 2 a and 7 2 5 d
Using the formula, 2 1 ,
2
n
n
S a n d we have
19
19
2 2 19 1 5
2
19
4 90
2
19
94
2
893
S
(ii)
Here, 9 a and 7 9 2 d
Using the formula, 2 1 ,
2
n
n
S a n d we have
14
14
2 9 14 1 2
2
7 18 26
7 8
56
S
(iii) The given AP is -37, -33, -
Here, 37 a and 33 37 33 37 4 d
Using the formula, 2 1 ,
2
n
n
S a n d we have
12
12
2 37 12 1 4
2
S
6 74 44
6 30
Page 2


 
1. Find the sum of each of the following Aps:
(iii) -37, -33, -
(iv) 
1 1 1
, , ,.....
15 12 10
to 11 terms. 
Sol:
(i)
Here, 2 a and 7 2 5 d
Using the formula, 2 1 ,
2
n
n
S a n d we have
19
19
2 2 19 1 5
2
19
4 90
2
19
94
2
893
S
(ii)
Here, 9 a and 7 9 2 d
Using the formula, 2 1 ,
2
n
n
S a n d we have
14
14
2 9 14 1 2
2
7 18 26
7 8
56
S
(iii) The given AP is -37, -33, -
Here, 37 a and 33 37 33 37 4 d
Using the formula, 2 1 ,
2
n
n
S a n d we have
12
12
2 37 12 1 4
2
S
6 74 44
6 30
 
180
(iv) The given AP is
1 1 1
, , ,......
15 12 10
Here, 
1
15
a and 
1 1 5 4 1
12 15 60 60
d
Using the formula, 2 1 ,
2
n
n
S a n d we have
11
11 1 1
2 11 1
2 15 60
11 2 10
2 15 60
S
11 18
2 60
33
20
(v) The given AP is 0.6,1.7,2.8,..........
Here, 0.6 a and 1.7 0.6 1.1 d
Using formula, 2 1 ,
2
n
n
S a n d we have
100
100
2 0.6 100 1 1.1
2
S
50 1.2 108.9
50 110.1
5505
2. Find the sum of each of the following arithmetic series:
(i) 
1
7 10 14 ... 84
2
(ii) 34 32 30 ... 10
(iii) ( 5) ( 8) ( 11) ... ( 230)
Sol:
(i) The given arithmetic series is 
1
7 10 14 ..... 84.
2
Here, 
1 21 21 4 7
7, 10 7 7
2 2 2 2
a d and 84. l
Let the given series contains n terms. Then,
Page 3


 
1. Find the sum of each of the following Aps:
(iii) -37, -33, -
(iv) 
1 1 1
, , ,.....
15 12 10
to 11 terms. 
Sol:
(i)
Here, 2 a and 7 2 5 d
Using the formula, 2 1 ,
2
n
n
S a n d we have
19
19
2 2 19 1 5
2
19
4 90
2
19
94
2
893
S
(ii)
Here, 9 a and 7 9 2 d
Using the formula, 2 1 ,
2
n
n
S a n d we have
14
14
2 9 14 1 2
2
7 18 26
7 8
56
S
(iii) The given AP is -37, -33, -
Here, 37 a and 33 37 33 37 4 d
Using the formula, 2 1 ,
2
n
n
S a n d we have
12
12
2 37 12 1 4
2
S
6 74 44
6 30
 
180
(iv) The given AP is
1 1 1
, , ,......
15 12 10
Here, 
1
15
a and 
1 1 5 4 1
12 15 60 60
d
Using the formula, 2 1 ,
2
n
n
S a n d we have
11
11 1 1
2 11 1
2 15 60
11 2 10
2 15 60
S
11 18
2 60
33
20
(v) The given AP is 0.6,1.7,2.8,..........
Here, 0.6 a and 1.7 0.6 1.1 d
Using formula, 2 1 ,
2
n
n
S a n d we have
100
100
2 0.6 100 1 1.1
2
S
50 1.2 108.9
50 110.1
5505
2. Find the sum of each of the following arithmetic series:
(i) 
1
7 10 14 ... 84
2
(ii) 34 32 30 ... 10
(iii) ( 5) ( 8) ( 11) ... ( 230)
Sol:
(i) The given arithmetic series is 
1
7 10 14 ..... 84.
2
Here, 
1 21 21 4 7
7, 10 7 7
2 2 2 2
a d and 84. l
Let the given series contains n terms. Then,
 
84
7
7 1 84 1
2
7 7
84
2 2
7 7 161
84
2 2 2
161
23
7
n
n
a
n a a n d
n
n
n
Required sum 
23
7 84
2 2
n
n
S a l
23
91
2
2030
2
1
1046
2
(ii) The given arithmetic series is 34 32 30 ..... 10.
Here, 34, 32 34 2 a d and 10. l
Let the given series contain n terms. Then,
10
34 1 2 10 1
2 36 10
2 10 36 26
13
n
n
a
n a a n d
n
n
n
Required sum 
13
34 10
2 2
n
n
S a l
13
44
2
286
(iii) The given arithmetic series is 5 8 11 ....... 230 .
Here, 5, 8 5 8 5 3 a d and 230. l
Let the given series contain n terms. Then,
Page 4


 
1. Find the sum of each of the following Aps:
(iii) -37, -33, -
(iv) 
1 1 1
, , ,.....
15 12 10
to 11 terms. 
Sol:
(i)
Here, 2 a and 7 2 5 d
Using the formula, 2 1 ,
2
n
n
S a n d we have
19
19
2 2 19 1 5
2
19
4 90
2
19
94
2
893
S
(ii)
Here, 9 a and 7 9 2 d
Using the formula, 2 1 ,
2
n
n
S a n d we have
14
14
2 9 14 1 2
2
7 18 26
7 8
56
S
(iii) The given AP is -37, -33, -
Here, 37 a and 33 37 33 37 4 d
Using the formula, 2 1 ,
2
n
n
S a n d we have
12
12
2 37 12 1 4
2
S
6 74 44
6 30
 
180
(iv) The given AP is
1 1 1
, , ,......
15 12 10
Here, 
1
15
a and 
1 1 5 4 1
12 15 60 60
d
Using the formula, 2 1 ,
2
n
n
S a n d we have
11
11 1 1
2 11 1
2 15 60
11 2 10
2 15 60
S
11 18
2 60
33
20
(v) The given AP is 0.6,1.7,2.8,..........
Here, 0.6 a and 1.7 0.6 1.1 d
Using formula, 2 1 ,
2
n
n
S a n d we have
100
100
2 0.6 100 1 1.1
2
S
50 1.2 108.9
50 110.1
5505
2. Find the sum of each of the following arithmetic series:
(i) 
1
7 10 14 ... 84
2
(ii) 34 32 30 ... 10
(iii) ( 5) ( 8) ( 11) ... ( 230)
Sol:
(i) The given arithmetic series is 
1
7 10 14 ..... 84.
2
Here, 
1 21 21 4 7
7, 10 7 7
2 2 2 2
a d and 84. l
Let the given series contains n terms. Then,
 
84
7
7 1 84 1
2
7 7
84
2 2
7 7 161
84
2 2 2
161
23
7
n
n
a
n a a n d
n
n
n
Required sum 
23
7 84
2 2
n
n
S a l
23
91
2
2030
2
1
1046
2
(ii) The given arithmetic series is 34 32 30 ..... 10.
Here, 34, 32 34 2 a d and 10. l
Let the given series contain n terms. Then,
10
34 1 2 10 1
2 36 10
2 10 36 26
13
n
n
a
n a a n d
n
n
n
Required sum 
13
34 10
2 2
n
n
S a l
13
44
2
286
(iii) The given arithmetic series is 5 8 11 ....... 230 .
Here, 5, 8 5 8 5 3 a d and 230. l
Let the given series contain n terms. Then,
 
230
5 1 3 230 1
3 2 230
3 230 2 228
76
n
n
a
n a a n d
n
n
n
Required sum
76
5 230
2 2
n
n
S a l
76
235
2
8930
3. Find the sum of first n terms of an AP whose nth term is (5 - 6n). Hence, find the sum of its 
first 20 terms.
Sol:
Let 
n
a be the nth term of the AP.
5 6
n
a n
Putting 1, n we get
First term,
1
5 6 1 1 a a
Putting 2, n we get
2
5 6 2 7 a
Let d be the common difference of the AP.
2 1
7 ( 1) 7 1 6 d a a
Sum of first n tern of the AP, 
n
S
2
2 1 1 6 2 1
2 2
2 6 6
2
2 3
2 3
n
n n
n S a n d
n
n
n n
n n
Putting 20, n we get
2
20
2 20 3 20 40 1200 1160 S
4. The sum of the first n terms of an AP is 
2
3 6 . n n Find the nth term and the 15
th
 term of this 
AP.
Sol:
Page 5


 
1. Find the sum of each of the following Aps:
(iii) -37, -33, -
(iv) 
1 1 1
, , ,.....
15 12 10
to 11 terms. 
Sol:
(i)
Here, 2 a and 7 2 5 d
Using the formula, 2 1 ,
2
n
n
S a n d we have
19
19
2 2 19 1 5
2
19
4 90
2
19
94
2
893
S
(ii)
Here, 9 a and 7 9 2 d
Using the formula, 2 1 ,
2
n
n
S a n d we have
14
14
2 9 14 1 2
2
7 18 26
7 8
56
S
(iii) The given AP is -37, -33, -
Here, 37 a and 33 37 33 37 4 d
Using the formula, 2 1 ,
2
n
n
S a n d we have
12
12
2 37 12 1 4
2
S
6 74 44
6 30
 
180
(iv) The given AP is
1 1 1
, , ,......
15 12 10
Here, 
1
15
a and 
1 1 5 4 1
12 15 60 60
d
Using the formula, 2 1 ,
2
n
n
S a n d we have
11
11 1 1
2 11 1
2 15 60
11 2 10
2 15 60
S
11 18
2 60
33
20
(v) The given AP is 0.6,1.7,2.8,..........
Here, 0.6 a and 1.7 0.6 1.1 d
Using formula, 2 1 ,
2
n
n
S a n d we have
100
100
2 0.6 100 1 1.1
2
S
50 1.2 108.9
50 110.1
5505
2. Find the sum of each of the following arithmetic series:
(i) 
1
7 10 14 ... 84
2
(ii) 34 32 30 ... 10
(iii) ( 5) ( 8) ( 11) ... ( 230)
Sol:
(i) The given arithmetic series is 
1
7 10 14 ..... 84.
2
Here, 
1 21 21 4 7
7, 10 7 7
2 2 2 2
a d and 84. l
Let the given series contains n terms. Then,
 
84
7
7 1 84 1
2
7 7
84
2 2
7 7 161
84
2 2 2
161
23
7
n
n
a
n a a n d
n
n
n
Required sum 
23
7 84
2 2
n
n
S a l
23
91
2
2030
2
1
1046
2
(ii) The given arithmetic series is 34 32 30 ..... 10.
Here, 34, 32 34 2 a d and 10. l
Let the given series contain n terms. Then,
10
34 1 2 10 1
2 36 10
2 10 36 26
13
n
n
a
n a a n d
n
n
n
Required sum 
13
34 10
2 2
n
n
S a l
13
44
2
286
(iii) The given arithmetic series is 5 8 11 ....... 230 .
Here, 5, 8 5 8 5 3 a d and 230. l
Let the given series contain n terms. Then,
 
230
5 1 3 230 1
3 2 230
3 230 2 228
76
n
n
a
n a a n d
n
n
n
Required sum
76
5 230
2 2
n
n
S a l
76
235
2
8930
3. Find the sum of first n terms of an AP whose nth term is (5 - 6n). Hence, find the sum of its 
first 20 terms.
Sol:
Let 
n
a be the nth term of the AP.
5 6
n
a n
Putting 1, n we get
First term,
1
5 6 1 1 a a
Putting 2, n we get
2
5 6 2 7 a
Let d be the common difference of the AP.
2 1
7 ( 1) 7 1 6 d a a
Sum of first n tern of the AP, 
n
S
2
2 1 1 6 2 1
2 2
2 6 6
2
2 3
2 3
n
n n
n S a n d
n
n
n n
n n
Putting 20, n we get
2
20
2 20 3 20 40 1200 1160 S
4. The sum of the first n terms of an AP is 
2
3 6 . n n Find the nth term and the 15
th
 term of this 
AP.
Sol:
 
Let 
n
S denotes the sum of first n terms of the AP.
2
2
1
2
2
3 6
3 1 6 1
3 2 1 6 1
3 3
n
n
S n n
S n n
n n n
n
t h
n term of the AP, 
n
a
1
2 2
3 6 3 3
6 3
n n
S S
n n n
n
Putting 15, n we get
15
6 15 3 90 3 93 a
Hence, the 
t h
n term is 6 3 n and 15
th
term is 93.
5. The sum of the  first n terms of an AP is given by 
2
3
n
S n n . Find its 
(i) nth term, 
(ii) first term and  
(iii) common difference.
Sol:
Given: 
2
3 ......
n
S n n i
Replacing n by 1 n in (i), we get:
2
1
2
2
3 1 1
3 2 1 1
3 7 4
n
S n n
n n n
n n
(i) Now, 
1 n n n
T S S
2 2
3 3 7 4 6 4 n n n n n
t h
n term, 6 4 .......
n
T n i i
(ii) Putting 1 n in (ii), we get:
1
6 1 4 2 T
(iii) Putting 2 n in (ii), we get:
2
6 2 4 8 T
Common difference, 
2 1
8 2 6 d T T

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Document Description: RD Sharma Solutions: Arithmetic Progressions (Exercise 5D) for Class 10 2024 is part of Mathematics (Maths) Class 10 preparation. The notes and questions for RD Sharma Solutions: Arithmetic Progressions (Exercise 5D) have been prepared according to the Class 10 exam syllabus. Information about RD Sharma Solutions: Arithmetic Progressions (Exercise 5D) covers topics like and RD Sharma Solutions: Arithmetic Progressions (Exercise 5D) Example, for Class 10 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises and tests below for RD Sharma Solutions: Arithmetic Progressions (Exercise 5D).

Introduction of RD Sharma Solutions: Arithmetic Progressions (Exercise 5D) in English is available as part of our Mathematics (Maths) Class 10 for Class 10 & RD Sharma Solutions: Arithmetic Progressions (Exercise 5D) in Hindi for Mathematics (Maths) Class 10 course. Download more important topics related with notes, lectures and mock test series for Class 10 Exam by signing up for free. Class 10: Arithmetic Progressions (Exercise 5D) RD Sharma Solutions | Mathematics (Maths) Class 10

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