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Exercise 5A Solutions: 30 Questions (Short Answers)

Exercise 5B Solutions: 16 Questions (Short Answers)

Exercise 5D Solutions: 20 Questions (Long Answers)

**Exercise 5C Page No: 281**

**Question 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,….12 terms.**

**(iv) 1/15 ,1/12,1/10,…to 11 terms**

**(v) 0.6, 1.7, 2.8,…. to 100 terms.**

**Solution:**

Sum of n terms of AP formula:

â€¦â€¦(1)

Where

First term = a

Common difference = d

Number of terms = n

(i) 2, 7, 12, 17,….. to 19 terms.

a = 2

d = 7 – 2 = 5

Using (1)

S_{19} = 19/2(2(2) + (19 – 1)5)

= (19)(4 + 90)/2

= (19 Ã— 94)/2

= 893

Sum of 19 terms of this AP is 893.

(ii) 9, 7, 5, 3,…..to 14 terms.

a = 9

d = 7 – 9 = -2

Using (1)

S_{14} = 14/2 [2(9) + (14 – 1)(-2)]

= (7)(18 – 26)

= – 56

Sum of 14 terms of this AP is – 56.

(iii) -37, -33, -29,….12 terms.

a = -37

d = (-33) – (-37) = 4

Using (1)

S_{12} = 12/2 [2(-37) + (12 – 1)(4)]

= (6)(-74 + 44)

= 6 Ã— (-30)

= – 180

Sum of 12 terms of this AP is – 180.

(iv) 1/15 ,1/12,1/10,…to 11 terms

a = 1/15

d = (1/12) – (1/15) = 1/60

Using (1)

S_{11} = 11/2 [2(1/15) + (11 – 1)(1/60)]

= (11/2) Ã— [(2/15) + (1/6)]

=33/20

Sum of 11 terms of this AP is 33/20.

(v) 0.6, 1.7, 2.8,…. to 100 terms.

a = 0.6

d = 1.7 – 0.6 = 1.1

Using (1)

S_{100} = 100/2 [2(0.6) + (100 – 1)(1.1)]

= (50) Ã— [1.2 + (99 Ã— 1.1) ]

= 50 Ã— 110.1

= 5505

Sum of 100 terms of this AP is 5505.

**Question 2: Find the sum of each of the following arithmetic series:**

**(i) 7 + 10 1/2 + 14 + â€¦ + 84**

**(ii) 34 + 32 + 30 + â€¦ + 10**

**(iii) (-5) + (-8) + (-11) + â€¦ + (-230)**

**(iv) 5 + (-41) + 9 + (-39) + 13 + (-37) + 17 + â€¦ + (-5) + 81 + (-3)**

**Solution:**

(i) 7 + 10 1/2 + 14 + â€¦ + 84

First term = a = 7

Common difference = d = (21/2) – 7 = (7/2)

Last term = l = 84

Now, using formula:

84 = a + (n – 1)d

84 = 7 + (n – 1)(7/2)

77 = (n – 1)(7/2)

154 = 7n – 7

7n = 161

n = 23

Thus, there are 23 terms in AP.

Now,

Find Sum of these 23 terms:

S_{23} = 23/2 [2(7) + (23 – 1)(7/2)]

= (23/2) [14 + (22)(7/2) ]

= (23/2) [91]

= 2093/2

Sum of 23 terms of this AP is 2093/2.

(ii) 34 + 32 + 30 + â€¦ + 10

First term = a = 34

Common difference = d = 34 – 32 = – 2

Last term = l = 10

Now, using formula:

10 = a + (n – 1)d

10 = 34 + (n – 1)(-2)

10 – 34 = (n – 1)(-2)

n = 13

Thus, there are 13 terms in AP.

Now,

Find Sum of these 13 terms:

S_{13} = 13/2 [2(34) + (13 – 1)(-2)]

= (13/2) [68 + (12)(-2)]

= (13/2) x 44

= 286

Sum of 13 terms of this AP is 286.

(iii) (-5) + (-8) + (-11) + â€¦ + (-230)

First term = a = -5

Common difference = d = – 8 – (-5) = – 3

Last term = l = -230

Now, using formula:

-230 = a + (n – 1)d

-230 = -5 + (n – 1)(-3)

– 230 + 5 = (n – 1)(-3)

n = 76

Thus, there are 76 terms in AP.

Now,

Find Sum of these 76 terms:

S_{76} = 76/2 [2(-5) + (76 – 1)(-3)]

= 38 Ã— [(-10) + (75)(-3)]

= – 8930

Sum of 76 terms of this AP is -8930.

(iv) 5 + (-41) + 9 + (-39) + 13 + (-37) + 17 + â€¦ + (-5) + 81 + (-3)

The given series is combination of two AP’s.

Let A_{1} = 5 + 9 +13 + ……+ 77 + 81

and A_{2} = -41 – 39 – 37 – ……(-3)

For A_{1}:

First term = a = 5

Common difference = d = 2

Last term = l = 81

Now, using formula:

81 = 5 + (n – 1)4

n = 20

Thus, there are 20 terms in AP.

Now,

Find Sum of these 76 terms:

S_{20} = n/2 [a + l]

= 20/2 (5 + 81)

= 860

Sum of 20 terms of this AP is 860.

For A_{2}:

First term = a = -41

Common difference = d = 2

Last term = l = -3

Now, using formula:

-3 = -41 + (n – 1)2

n = 20

Thus, there are 20 terms in AP.

Now,

Find Sum of these 76 terms:

S_{20} = n/2 [a + l]

= 20/2 (-41- 3)

=-440

Sum of 20 terms of this AP is -440.

Therefore, Sum of total terms: 860 – 440 = 420.

**Question 3: Find the sum of first n terms of an AP whose nth term is (5 â€“ 6n). Hence, find the sum of its first 20 terms.**

**Solution**:

Given: a_{n} = 5 – 6n

Find some of the terms of AP:

Put n = 1, we get a_{1} = – 1 = first term

Put n = 2, we get a_{2} = – 7 = second term

Common difference = d = a_{2} – a_{1} = – 7 – (-1) = – 6

Sum of first n terms:

S_{n}= n/2 [2a + (n – 1)d]

= n/2[- 2 + (n – 1)(-6)]

= n(2 – 3n)

sum of first 20 terms:

S_{20} = 20/2[2(-1) + (20 – 1)(-6)]

= 10 [ – 2 – 114]

= – 1160

Sum of its first 20 terms of AP is -1160.

**Question 4:**

**The sum of the first n terms of an AP is (3nÂ² + 6n). Find the nth term and the 15th term of this AP.**

**Solution:**

Given: S_{n} = 3nÂ² + 6n

**Question 5: The sum of the first n terms of an AP is given by S _{n} = (3nÂ² â€“ n). Find its**

**(i) nth term,**

**(ii) first term and**

**(iii) common difference.**

**Solution:**

S_{n} = 3nÂ² â€“ n

S_{1} = 3(1)Â² â€“ 1 = 3 â€“ 1 = 2

S_{2} = 3(2)Â² â€“ 2 = 12 â€“ 2 = 10

a_{1} = 2

a_{2} = 10 â€“ 2 = 8

(i) a_{n} = a + (n â€“ 1) d

= 2 + (n â€“ 1) x 6

= 2 + 6n â€“ 6

= 6n â€“ 4

(ii) First term = 2

(iii) Common difference = 8 â€“ 2 = 6

**Question 6: (i) The sum of the first n terms of an AP is (5n ^{2}/2 + 3n/2). Find the nth term and the 20th term of this AP. **

**(ii) The sum of the first n terms of an AP is (3n ^{2}/2 + 5n/2). Find its nth term and the 25th term. **

**Solution**:

**(ii)**

**Question 7: If mth term of an AP is 1/n and nth term is 1/m then find the sum of its first mn terms.**

**Solution:**

Let a be first term and d be the common difference of an AP.

mth term = 1/n

so,

a_{m} = a + (m â€“ 1) d = 1/n ……….(1)

nth term = 1/m

a_{n} = a + (n â€“ 1) d = 1/m ……….(2)

Subtract (2) from (1)

Sum of first mn terms:

S_{mn} = mn/2 [2(1/mn) + (mn-1) (1/mn)]

= mn/2 [1/mn + 1]

= (1+mn)/2

**Question 8:**

**How many terms of the AP 21, 18, 15, â€¦ must be added to get die sum 0?**

**Solution:**

AP is 21, 18, 15,â€¦

a = 21,

d = 18 â€“ 21 = -3

Sum of terms = S_{n} = 0

(n/2) [2a + (n – 1)d] = 0

(n/2) [2(21) + (n – 1)(-3)] = 0

(n/2) [45 – 3n] = 0

[45 – 3n] = 0n = 15 (number of terms)

Thus, 15 terms of the given AP sums to zero.

**Question 9: How many terms of the AP 9, 17, 25, â€¦ must be taken so that their sum is 636?**

**Solution:**

AP is 9, 17, 25,â€¦

a = 9, d = 17 â€“ 9 = 8

Sum of terms = S_{n} = 636

(n/2) [2a + (n – 1)d] = 636

(n/2) [2(9) + (n – 1)(8)] = 636

(n/2)[10 + 8n] = 636

4n^{2} + 5n – 636 = 0 (which is a quadratic equation)

(n- 12)(4n + 53) = 0

Either (n- 12) = 0 or (4n + 53) = 0

n = 12 or n = – 53/4

Since n canâ€™t be negative and fraction, so

n = 12

Number of terms = 12 terms.

**Question 10: How many terms of the AP 63, 60, 57, 54, â€¦ must be taken so that their sum is 693? Explain the double answer.**

**Solution:**

AP is 63, 60, 57, 54,â€¦

Here, a = 63, d = 60 â€“ 63 = -3 and sum = S_{n} = 693

Which is a quadratic equation

Which shows that, 22th term of AP is zero.

Number of terms are 21 or 22. So there will be no effect on the sum.

**Question 11:**

**How many terms of the AP 20, 19 1/3 , 18 2/3 , â€¦ must be taken so that their sum is 300? Explain the double answer.**

**Solution**:

Here, a = 20, d = -2/3 and sum = s_{n} = 300 (for n number of terms)

n(n-25) – 36(n-25) = 0

(n – 25)(n – 36) = 0

Either (n – 25) = 0 or (n – 36) = 0

n = 25 or n = 36

For n = 25

= 300

Which is true.

Result is true for both the values of n

So both the numbers are correct.

Therefore, Sum of 11 terms is zero. (36-25 = 11)

**Question 12: Find the sum of all odd numbers between 0 and 50. **

**Solution**:

Odd numbers between 0 and 50 are 1, 3, 5, 7, 9, â€¦, 49

Here, a = 1, d = 3 â€“ 1 = 2, l = 49

Sum of odd numbers:

= n/2(a + l)

= 25/2 (l + 49)

= 25/2(50)

= 625

**Question 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, 217,……,399.

Here,AP is 203, 210, 217,……,399

First term = a = 203,

Common difference = d = 7 and

Last term = l = 399

We know that, l = a + (n-1)d

Sum of all 29 terms is 8729.

**Question 14: Find the sum of first forty positive integers divisible by 6. **

**Solution:**

First forty positive integers which are divisible by 6 are

6, 12, 18, 24,…… to 40 terms

Here, a = 6, d = 12 â€“ 6 = 6, and n = 40.

Sum of 40 terms:

**Question 15: Find the sum of the first 15 multiples of 8.**

**Solution:**

First 15 multiples of 8 are as given below:

8, 16, 24, 32,….. to 15 terms

Here, a = 8, d = 16 â€“ 8 = 8, and n = 15

Sum of 15 terms:

**Question 16: Find the sum of all multiples of 9 lying between 300 and 700.**

**Solution:**

Multiples of 9 lying between 300 and 700 = 306, 315, 324, 333, â€¦, 693

Here, a = 306, d = 9, and l = 693

We know that, l = a + (n-1)d

There are 44 terms.

Find the sum of 44 terms:

**Question 17: Find the sum of all three-digit natural numbers which are divisible by 13. **

**Solution:**

3-digit natural numbers: 100, 101, 102,……., 999.

3-digit natural numbers divisible by 13:

104, 117, 130,………, 988

Which is an AP.

Here, a = 104, d = 13, l = 988

l = a + (n â€“ 1) d

988 = 104 + (n-1)13

n = 69

There are 69 terms.

Find the sum of 69 terms:

S_{n} = n/2(a + l)

= 69/2(104 + 988)

= 37674

**Question 18: Find the sum of first 100 even natural numbers which are divisible by 5.**

**Solution:**

Even natural numbers: 2, 4, 6, 8, 10, â€¦

Even natural numbers divisible by 5: 10, 20, 30, 40, â€¦ to 100 terms

Here, a = 10, d = 20 â€“ 10 = 10, and n = 100

S_{n} = n/2(2a + (n-1)d)

= 100/2 [2 x 10 + (100-1)10]

= 50500

**Question 19: Find the sum of the following.**

**(4 â€“ 1/n) + (4-2/n) + (4â€“ 3/n) + â€¦.**

**Solution: **

Given sum can be written as (4 + 4 + 4 + 4+….) – (1/n, 2/n, 3/n, …..)

Now, We have two series:

First series: = 4 + 4 + 4 + …… up to n terms

= 4n

Second series: 1/n, 2/n, 3/n, …..

Here, first term = a = 1/n

Common difference = d = (2/n) – (1/n) = (1/n)

Sum of n terms formula:

S_{n} = n/2[2a + (n – 1)d]

Sum of n terms of second series:

S_{n} = n/2 [2(1/n) + (n – 1)(1/n)]

= n/2 [(2/n) + 1 – (1/n)]

= (n + 1)/2

Hence,

Sum of n terms of the given series = Sum of n terms of first series – Sum of n terms of second series

= 4n – (n + 1)/2

= (8n – n – 1)/2

= 1/2 (7n – 1)

**Question 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 the AP.

Which implies: a = 1 and d = 5

Therefore, the AP is 1, 6, 11, 16,â€¦â€¦

**Question 21: In an AP, the first term is 2, the last term is 29 and the sum of all the terms is 155. Find the common difference. **

**Solution:**

Let d be the common difference.

Given:

first term = a = 2

last term = l = 29

Sum of all the terms = S_{n} = 155

S_{n} = n/2[a + l]

155 = n/2[2 + 29]

n = 10

There are 10 terms in total.

Therefore, 29 is the 10th term of the AP.

Now, 29 = a + (10 – 1)d

29 = 2 + 9d

27 = 9d

d = 3

The common difference is 3.

**Question 22: In an AP, the first term is -4, the last term is 29 and the sum of all its terms is 150. Find its common difference. **

**Solution:**

Let d be the common difference.

Given:

first term = a = -4

last term = l = 29

Sum of all the terms = S_{n} = 150

S_{n} = n/2[a + l]

150 = n/2[-4 + 29]

n = 12

There are 12 terms in total.

Therefore, 29 is the 12th term of the AP.

Now, 29 = -4 + (12 – 1)d

29 = -4 + 11d

d = 3

The common difference is 3.

**Question 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:**

Let n be the total number of terms.

Given:

First term = a = 17

Last term = l = 350

Common difference = d = 9

l = a + (n-1)d

350 = 17 + (n-1)9

n = 38

Again,

S_{n} = n/2[a + l]

= 38/2[17 + 350]

= 6973

There are 38 terms in total and their sum is 6973.

**Question 24: The first and the 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:**

Let n be the total number of terms and d be the common difference.

Given:

first term = a = 5

last term = l = 45

Sum of all terms = S_{n} = 400

S_{n} = n/2[a + l]

400 = n/2[5 + 45]

n/2 = 400/50

n = 16

There are 16 terms in the AP.

Therefore, 45 is the 16th term of the AP.

45 = a + (16 – 1)d

45 = 5 + 15d

40 = 15d

15d = 40

d = 8/3

Common difference = d = 8/3

Common difference is 8/3 and the number of terms are 16.

**Question 25: In an AP, the first term is 22, nth term is -11 and sum of first n terms is 66. Find n and hence find the common difference.**

**Solution:**

Let n be the total number of terms and d be the common difference.

Given:

first term = a = 22

nth term = -11

Sum of all terms = S_{n} = 66

S_{n} = n/2[a + l]

66 = n/2[22 + (-11)]

n = 12

There are 12 terms in the AP.

Since nth term is -11, so

a_{n} = a + (n – 1)d

-11 = 22 + (12-1)d

d = -3

Therefore, Common difference is -3 and the number of terms are 12.

**R S Aggarwal Solutions for Class 10 Maths Chapter 5 Arithmetic Progression Exercise 5C**

Class 10 Maths Chapter 5 Arithmetic Progression Exercise 5C is based on the following topic:

- Sum of n terms of an AP

Sum of n terms of AP formula:

First term = a

Common difference = d

Number of terms = n

R S Aggarwal Solutions for Class 10 will help students to practice more questions and master various concepts.