Arithmetic Progressions

An Arithmetic Progression (AP) is a sequence of numbers in which the difference between any two consecutive terms remains constant

Key Concepts:

• Common Difference (d): The constant difference between consecutive terms.
• First Term (a): The initial term of the sequence.
• nth term (an): The general formula for finding any term of an AP is: an = a + (n-1)d
• Sum of n terms (Sn): The formula for finding the sum of n terms is: Sn = n/2 [2a + (n-1)d] or Sn = n/2 [a + l], where l is the last term.
• Arithmetic Mean (AM): The middle term of an AP when the number of terms is odd.

Applications

Arithmetic Progressions have various real-life applications, such as:

• Calculating compound interest
• Predicting population growth
• Analyzing physical phenomena involving constant rates of change

In essence, this chapter provides a foundational understanding of sequences and series, with a specific focus on arithmetic progressions. It equips students with the tools to analyze patterns, find specific terms, and calculate the sum of a given number of terms in an arithmetic sequence.

Exercise 5.1

1. In which of the following situations, does the list of numbers involved make an arithmetic progression and why?

(i) The taxi fare after each km when the fare is ₹ 15 for the first km and ₹ 8 for each additional km.

(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.

(iii) The cost of digging a well after every meter of digging, when it costs ₹ 150 for the first meter and rises by ₹ 50 for each subsequent meter.

(iv) The amount of money in the account every year, when ₹ 10000 is deposited at compound interest at 8% per annum.

Ans :

2. Write first four terms of the AP, when the first term a and the common difference d are given as follows:

(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

Ans : 

(i) a = 10, d = 10

• First term (a₁) = 10
• Second term (a₂) = 10 + (2 – 1) * 10 = 20
• Third term (a₃) = 10 + (3 – 1) * 10 = 30
• Fourth term (a₄) = 10 + (4 – 1) * 10 = 40

Therefore, the first four terms are 10, 20, 30, 40.

(ii) a = -2, d = 0

• First term (a₁) = -2
• Second term (a₂) = -2 + (2 – 1) * 0 = -2
• Third term (a₃) = -2 + (3 – 1) * 0 = -2
• Fourth term (a₄) = -2 + (4 – 1) * 0 = -2

Therefore, the first four terms are -2, -2, -2, -2.

(iii) a = 4, d = -3

• First term (a₁) = 4
• Second term (a₂) = 4 + (2 – 1) * (-3) = 1
• Third term (a₃) = 4 + (3 – 1) * (-3) = -2
• Fourth term (a₄) = 4 + (4 – 1) * (-3) = -5

Therefore, the first four terms are 4, 1, -2, -5.

(iv) a = -1, d = 1/2

• First term (a₁) = -1
• Second term (a₂) = -1 + (2 – 1) * (1/2) = -0.5
• Third term (a₃) = -1 + (3 – 1) * (1/2) = 0
• Fourth term (a₄) = -1 + (4 – 1) * (1/2) = 0.5

Therefore, the first four terms are -1, -0.5, 0, 0.5.

(v) a = -1.25, d = -0.25

• First term (a₁) = -1.25
• Second term (a₂) = -1.25 + (2 – 1) * (-0.25) = -1.5
• Third term (a₃) = -1.25 + (3 – 1) * (-0.25) = -1.75
• Fourth term (a₄) = -1.25 + (4 – 1) * (-0.25) = -2

3. For the following APs, 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, …….

Ans : 

i) 3, 1, -1, -3, …

• First term (a) = 3
• Common difference (d) = 1 – 3 = -2

ii) -5, -1, 3, 7, …

• First term (a) = -5
• Common difference (d) = -1 – (-5) = 4

iii) 1/3, 5/3, 9/3, 13/3, …

• First term (a) = 1/3
• Common difference (d) 
• = 5/3 – 1/3 = 4/3

iv) 0.6, 1.7, 2.8, 3.9, …

• First term (a) = 0.6
• Common difference (d) = 1.7 – 0.6 = 1.1

4. Which of the following are APs ? If they form an AP, 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, a2, a3, a4, …….

xii) √2, √8, √18, √32, …

xiii) √3, √6, √9, √12, ….

(xiv) 12, 32, 52, 72, ……

(xv) 12, 52, 72, 73, ……

Ans : 

Exercise 5.2

1. Fill in the blanks in the following table, given that a is the first term, d the common difference and the nth term of the AP:

Ans : 

Case (i):

• The first term (a) is 7.
• The common difference (d) is 3.
• The number of terms (n) is 8.
• To find the 8th term (a₈), we use the formula: aₙ = a + (n – 1)d.
• Substituting the values, we get: a₈ = 7 + (8 – 1) * 3 = 7 + 21 = 28.

Case (ii):

• The first term (a) is -18.
• The number of terms (n) is 10.
• The nth term (a₁₀) is 0.
• To find the common difference (d), we use the formula: aₙ = a + (n – 1)d.
• Substituting the values, we get: 0 = -18 + (10 – 1)d.
• Solving for d, we get:
•  d = 2.

Case (iii):

• The common difference (d) is -3.
• The number of terms (n) is 18.
• The nth term (a₁₈) is -5.
• To find the first term (a), we use the formula: aₙ = a + (n – 1)d.
• Substituting the values, we get: -5 = a + (18 – 1)(-3).
• Solving for a, we get: a = 54.

Case (iv):

• The first term (a) is -18.9.
• The common difference (d) is 2.5.
• The nth term (an) is 3.6.
• To find the number of terms (n), we use the formula: aₙ = a + (n – 1)d.
• Substituting the values, we get: 3.6 = -18.9 + (n – 1)2.5.
• Solving for n, we get: n = 7.

Case (v):

• The first term (a) is 3.5.
• The common difference (d) is 0.
• The number of terms (n) is 105.
• Since the common difference is 0, all terms in the AP will be the same as the first term.
• Therefore, the 105th term (a₁₀₅) is also 3.5.

	a	d	n	an
(i)	7	3	8	28
(ii)	-18	2	10	0
(iii)	54	-3	18	-5
(iv)	-18.9	2.5	7	3.6
(v)	3.5	0	105	3.5

2. Choose the correct choice in the following and justify:

(i) 30th term of the AP: 10, 7, 4, …, is

(a) 97

(b) 77

(c) -77

(d) -87

Ans : 

First, let’s identify the common difference (d):

• d = T₂ – T₁ 
• = 7 – 10 = -3

Next, we’ll use the formula for the nth term of an AP:

• Tₙ = a + (n – 1)d

Where:

• Tₙ is the nth term
• a is the first term
• d is the common difference
• n is the number of terms

Substituting the given values:

• T₃₀ = 10 + (30 – 1) * (-3)
• T₃₀ = 10 + 29 * (-3)
• T₃₀ = 10 – 87
• T₃₀ = -77

So, the correct answer is (c) -77.

(ii) 11th term of the AP: -3, −1/2 , 2, …, is

(a) 28

(b) 22

(c) -38

(d) -48

Ans : 

First, let’s find the common difference (d):

• d = T₂ – T₁ = (-1/2) – (-3) = 5/2

Next, we’ll use the formula for the nth term of an AP:

• Tₙ = a + (n – 1)d

Where:

• Tₙ is the nth term
• a is the first term
• d is the common difference
• n is the number of terms

Substituting the given values:

• T₁₁ = -3 + (11 – 1) * (5/2)
• T₁₁ = -3 + 10 * (5/2)
• T₁₁ = -3 + 25
• T₁₁ = 22

So, the correct answer is (b) 22.

3. In the following APs, find the missing terms in the boxes:

Ans : 

4. Which term of the AP: 3, 8, 13, 18, …, is 78?

Ans : 

First, let’s find the common difference (d):

• d = T₂ – T₁ = 8 – 3 = 5

Next, we’ll use the formula for the nth term of an AP:

• Tₙ = a + (n – 1)d

Where:

• Tₙ is the nth term
• a is the first term
• d is the common difference
• n is the number of terms

We know Tₙ = 78, a = 3, and d = 5. Substituting these values into the formula, 

we get:

• 78 = 3 + (n – 1) * 5

Solving for n:

• 75 = (n – 1) * 5
• 15 = n – 1
• n = 16

78 is the 16th term of the AP.

5. Find the number of terms in each of the following APs:

(i) 7, 13, 19, …, 205

ii) 18, 15/2, 13, …, -47

Ans : 

6. Check, whether -150 is a term of the AP: 11, 8, 5, 2, ….

Ans : 

First, let’s find the common difference (d):

• d = T₂ – T₁ = 8 – 11 = -3

Next, let’s assume -150 is the nth term of the AP:

• Tₙ = a + (n – 1)d

Where:

• Tₙ is the nth term (-150 in this case)
• a is the first term (11)
• d is the common difference (-3)
• n is the number of terms (to be found)

Substituting the values, we get:

• -150 = 11 + (n – 1)(-3)

Simplifying the equation:

• -161 = -3(n – 1)
• 53.67 ≈ n

Since the number of terms (n) must be a positive integer, and we got a decimal value, -150 cannot be a term of the given AP.

7. Find the 31st term of an AP whose 11th term is 38 and the 16th term is 73.

Ans : 

We are given an AP where:

• T₁₁ (11th term) = 38
• T₁₆ (16th term) = 73

We need to find T₃₁ (31st term).

Solution

• Tₙ = a + (n – 1)d

Where:

• Tₙ is the nth term
• a is the first term
• d is the common difference
• n is the number of terms

1. T₁₁ = a + 10d = 38
2. T₁₆ = a + 15d = 73

Subtracting equation 1 from equation 2, we get:

• 5d = 35
• d = 7

Substituting d = 7 in equation 1:

• a + 10(7) = 38
• a + 70 = 38
• a = -32

Now, we can find T₃₁:

• T₃₁ = a + (31 – 1)d
• T₃₁ = -32 + 30 * 7
• T₃₁ = -32 + 210
• T₃₁ = 178

8. An AP consists of 50 terms of which 3rd term is 12 and the last term is 106. Find the 29th term.

Ans : 

We are given an AP with 50 terms:

• The 3rd term (T₃) is 12.
• The last term (T₅₀) is 106.

We need to find the 29th term (T₂₉).

Solution

• Tₙ = a + (n – 1)d

Where:

• Tₙ is the nth term
• a is the first term
• d is the common difference
• n is the number of terms

1. T₃ = a + 2d = 12
2. T₅₀ = a + 49d = 106

Subtracting equation 1 from equation 2, we get:

• 47d = 94
• d = 2

Substituting d = 2 in equation 1:

• a + 2(2) = 12
• a + 4 = 12
• a = 8

Now, we can find T₂₉:

• T₂₉ = a + (29 – 1)d
• T₂₉ = 8 + 28 * 2
• T₂₉ = 8 + 56
• T₂₉ = 64

Therefore, the 29th term of the AP is 64.

9. If the 3rd and the 9th term of an AP are 4 and -8 respectively, which term of this AP is zero?

Ans : 

• Tₙ = a + (n – 1)d

Where:

• Tₙ is the nth term
• a is the first term
• d is the common difference
• n is the number of terms

1. T₃ = a + 2d = 4
2. T₉ = a + 8d = -8

Subtracting equation 1 from equation 2, we get:

• 6d = -12
• d = -2

Substituting d = -2 in equation 1:

• a + 2(-2) = 4
• a – 4 = 4
• a = 8

Now, we need to find n when Tₙ = 0:

• 0 = 8 + (n – 1)(-2)
• -8 = -2n + 2
• 2n = 10
• n = 5

Therefore, the 5th term of the AP is zero.

10.  The 17th term of an AP exceeds its 10th term by 7. Find the common difference.

Ans : 

• Tₙ = a + (n – 1)d

Where:

• Tₙ is the nth term
• a is the first term
• d is the common difference
• n is the number of terms
• T₁₇ = T₁₀ + 7

Substituting the formula for Tₙ:

• a + 16d = a + 9d + 7

Simplifying the equation:

• 7d = 7
• d = 1

Therefore, the common difference (d) of the AP is 1.

11. Which term of the AP: 3, 15, 27, 39, … will be 132 more than its 54th term?

Ans :

We are given an AP: 3, 15, 27, 39, … We need to find the term which is 132 more than the 54th term.

Solution

First, let’s find the common difference (d):

• d = T₂ – T₁ = 15 – 3 = 12

Next, let’s find the 54th term (T₅₄):

• Tₙ = a + (n – 1)d
• T₅₄ = 3 + (54 – 1) * 12
• T₅₄ = 3 + 53 * 12
• T₅₄ = 639

Now, we need to find the term which is 132 more than T₅₄:

• Required term = T₅₄ + 132 = 639 + 132 = 771

Let the required term be Tₙ:

• Tₙ = 771

Using the formula for the nth term again:

• 771 = 3 + (n – 1) * 12
• 768 = (n – 1) * 12
• 64 = n – 1
• n = 65

Therefore, the 65th term of the AP is 132 more than its 54th term.

12. Two APs have the same common difference. The difference between their 100th terms is 100, what is the difference between their 1000th terms?

Ans : 

Let’s denote:

• The first term of the first AP as ‘a’
• The first term of the second AP as ‘b’
• The common difference of both APs as ‘d’

For the first AP:

• 100th term (T₁₀₀) = a + (100-1)d = a + 99d

For the second AP:

• 100th term (T₁₀₀’) = b + (100-1)d = b + 99d

Given that the difference between their 100th terms is 100:

• (a + 99d) – (b + 99d) = 100
• a – b = 100

Now, let’s find the difference between their 1000th terms:

• For the first AP, T₁₀₀₀ = a + 999d
• For the second AP, T₁₀₀₀’ = b + 999d

Difference between the 1000th terms:

• (a + 999d) – (b + 999d) = a – b

We already know that a – b = 100.

Therefore, the difference between their 1000th terms is also 100.

The difference between the 1000th terms of the two APs remains constant and is equal to 100.

13. How many three-digit numbers are divisible by 7?

Ans : 

14. How many multiples of 4 lie between 10 and 250?

Ans : 

15. For what value of n, the nth term of two APs: 63, 65, 61,… and 3, 10, 17,… are equal?

Ans : 

Let’s find the common difference for each AP:

• For the first AP, d₁ = 65 – 63 = 2
• For the second AP, d₂ = 10 – 3 = 7

Let the nth term of both APs be equal to x:

• For the first AP, x = 63 + (n – 1) * 2
• For the second AP, x = 3 + (n – 1) * 7

Since both expressions equal x, we can equate them:

• 63 + 2(n – 1) = 3 + 7(n – 1)

Simplifying the equation:

• 63 + 2n – 2 = 3 + 7n – 7
• 61 + 2n = 7n – 4
• 65 = 5n
• n = 13

16. Determine the AP whose 3rd term is 16 and 7th term exceeds the 5th term by 12.

Ans : 

• Tₙ = a + (n – 1)d

Where:

• Tₙ is the nth term
• a is the first term
• d is the common difference
• n is the number of terms

1. T₃ = a + 2d = 16
2. T₇ – T₅ = 12

Expanding T₇ and T₅ using the formula:

• (a + 6d) – (a + 4d) = 12
• 2d = 12
• d = 6

Substituting d = 6 in equation 1:

• a + 2(6) = 16
• a + 12 = 16
• a = 4

Therefore, the first term (a) is 4 and the common difference (d) is 6.

The AP is: 4, 10, 16, 22, …

17. Find the 20th term from the last term of the AP: 3, 8, 13, …, 253.

Ans : 

 Find the common difference (d):

• d = 8 – 3 = 5

• The last term is 253.
• Using the formula for the nth term of an AP:
  • L = a + (n – 1)d
  • 253 = 3 + (n – 1) * 5
  • 250 = (n – 1) * 5
  • n – 1 = 50
  • n = 51
• This is equivalent to finding the (51 – 20 + 1)th term from the beginning, which is the 32nd term.

 Calculate the 32nd term:

• T₃₂ = a + (32 – 1)d
• T₃₂ = 3 + 31 * 5
• T₃₂ = 3 + 155
• T₃₂ = 158

Therefore, the 20th term from the last term of the AP is 158.

18. The sum of the 4th and 8th terms of an AP is 24 and the sum of the 6th and 10th terms is 44. Find the first three terms of the AP

Ans : 

• Tₙ = a + (n – 1)d

Where:

• Tₙ is the nth term
• a is the first term
• d is the common difference
• n is the number of terms

1. T₄ + T₈ = 24
   • (a + 3d) + (a + 7d) = 24
   • 2a + 10d = 24
   • a + 5d = 12 –(Equation 1)
2. T₆ + T₁₀ = 44
   • (a + 5d) + (a + 9d) = 44
   • 2a + 14d = 44
   • a + 7d = 22 –(Equation 2)

Subtracting equation 1 from equation 2, we get:

• 2d = 10
• d = 5

Substituting d = 5 in equation 1:

• a + 5(5) = 12
• a + 25 = 12
• a = -13

Therefore, the first term (a) is -13 and the common difference (d) is 5.

The first three AP are -13, -8, and -3.

19. Subba Rao started work in 1995 at an annual salary of ₹ 5000 and received an increment of ₹ 200 each year. In which year did his income reach ₹ 7000 ?

Ans : 

20. Ramkali saved ₹ 5 in the first week of a year and then increased her weekly saving by ₹ 1.75. If in the nth week, her weekly saving become ₹ 20.75, find n.

Ans : 

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

Ans : 

2. Find the sums given below:

(i) 7 + 101/2 + 14 + … + 84

(ii) 34 + 32 + 30 + … + 10

(iii) -5 + (-8) + (-11) + ….. + (-230)

Ans : 

(i) 7 + 10 1/2 + 14 + … + 84

• Convert mixed numbers to improper fractions: 7 + 21/2 + 14 + … + 84
• First term (a) = 7
• Common difference (d) 
• = 21/2 – 7 = 7/2
• Last term (l) = 84

To find n, use the formula for the nth term:

• l = a + (n – 1)d
• 84 = 7 + (n – 1)(7/2)
• 77 = (n – 1)(7/2)
• n – 1 = 22
• n = 23

Now, find the sum using the formula:

• Sₙ = n/2 [a + l]
• S₂₃ = 23/2 [7 + 84]
• S₂₃ = 23/2 * 91
• S₂₃ = 1046.5

(ii) 34 + 32 + 30 + … + 10

• First term (a) = 34
• Common difference (d) = 32 – 34 = -2
• Last term (l) = 10

To find n, use the formula for the nth term:

• l = a + (n – 1)d
• 10 = 34 + (n – 1)(-2)
• -24 = -2(n – 1)
• n – 1 = 12
• n = 13

Now, find the sum using the formula:

• Sₙ = n/2 [a + l]
• S₁₃ = 13/2 [34 + 10]
• S₁₃ = 13/2 * 44
• S₁₃ = 286

(iii) -5 + (-8) + (-11) + … + (-230)

• First term (a) = -5
• Common difference (d) 
• = -8 – (-5) = -3
• Last term (l) = -230

To find n, use the formula for the nth term:

• l = a + (n – 1)d
• -230 = -5 + (n – 1)(-3)
• -225 = -3(n – 1)
• n – 1 = 75
• n = 76

Now, find the sum using the formula:

• Sₙ = n/2 [a + l]
• S₇₆ = 76/2 [-5 + (-230)]
• S₇₆ = 38 * (-235)
• S₇₆ = -8930

3. In an AP:

(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, S = 192, find d.

(x) given l = 28, S = 144, and there are total 9 terms. Find a.

Ans : 

4. How many terms of AP: 9, 17, 25, … must be taken to give a sum of 636?

Ans : 

First, let’s find the common difference (d):

• d = T₂ – T₁ = 17 – 9 = 8

Next, we’ll use the formula for the sum of n terms of an AP:

• Sₙ = n/2 [2a + (n – 1)d]

Where:

• Sₙ is the sum of n terms
• a is the first term (9)
• d is the common difference (8)
• n is the number of terms

We know Sₙ = 636. Substituting the values:

• 636 = n/2 [2 * 9 + (n – 1) * 8]
• 1272 = n [18 + 8n – 8]
• 1272 = n [10 + 8n]
• 1272 = 10n + 8n²
• 8n² + 10n – 1272 = 0
• 4n² + 5n – 636 = 0

This is a quadratic equation. 

(4n + 49)(n – 12) = 0

So, n = -49/4 or n = 12. Since the number of terms cannot be negative, n = 12.

Therefore, 12 terms of the AP must be taken to give a sum of 636.

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.

Ans : 

• Sn = n/2 [a + l]

Substituting the given values:

• 400 = n/2 [5 + 45]
• 400 = n/2 * 50
• n = 16

Now, we have the number of terms, n = 16.

• l = a + (n – 1)d

Substituting the values:

• 45 = 5 + (16 – 1)d
• 40 = 15d
• d = 8/3

Therefore, the number of terms (n) is 16 and the common difference (d) is 8/3.

6. 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?

Ans : 

Finding the number of terms (n):

l = a + (n – 1)d

Substituting the given values:

• 350 = 17 + (n – 1) * 9
• 333 = (n – 1) * 9
• n – 1 = 37
• n = 38

Therefore, there are 38 terms in the AP.

Finding the sum of the AP (Sn): 

We can use the formula for the sum of n terms of an AP:

• Sn = n/2 * (a + l)

Substituting the values:

• Sn = 38/2 * (17 + 350)
• Sn = 19 * 367
• Sn = 6973

Therefore, the sum of the AP is 6973.

7. Find the sum of first 22 terms of an AP in which d = 7 and 22nd term is 149.

Ans : 

We know that aₙ = a + (n-1)d

So, a22 = a + (22-1)7

149 = a + 147

a = 2

We use the formula: Sn = n/2 [2a + (n-1)d]

S22 = 22/2 [2*2 + (22-1)7]

S22 = 11 [4 + 147]

S22 = 11 * 151

S22 = 1661

Therefore, the sum of the first 22 terms of the AP is 1661.

8. Find the sum of first 51 terms of an AP whose second and third terms are 14 and 18 respectively.

Ans : 

Step 1: Find the common difference (d):

• d = a₃ – a₂
•  = 18 – 14 = 4

Step 2: Find the first term (a):

• We know a₂ = a + d
• So, 14 = a + 4
• a = 10

Step 3:

• We use the formula: Sn = n/2 [2a + (n-1)d]
• S₅₁ = 51/2 [2*10 + (51-1)4]
• S₅₁ = 51/2 [20 + 200]
• S₅₁ = 51/2 * 220
• S₅₁ = 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.

Ans :

10. Show that a1, a2, ……. an,…… form an AP where an is defined as below:

(i) an = 3 + 4n

(ii) an = 9 – 5n

Also find the sum of the first 15 terms in each case.

Ans : 

(i) an = 3 + 4n

To show that a sequence is an AP, we need to prove that the difference between consecutive terms is constant.

• a₁ = 3 + 4(1) = 7
• a₂ = 3 + 4(2) = 11
• a₃ = 3 + 4(3) = 15

The common difference, d = a₂ – a₁ = 11 – 7 = 4. Since the common difference is constant, the sequence is an AP.

To find the sum of the first 15 terms (S₁₅):

• We know a = 7, d = 4, and n = 15.
• Using the formula Sₙ = n/2 [2a + (n-1)d]:
  • S₁₅ = 15/2 [2(7) + (15-1)4]
  • S₁₅ = 15/2 [14 + 56]
  • S₁₅ = 15/2 * 70
  • S₁₅ = 525

Therefore, the sequence is an AP with a common difference of 4, and the sum of the first 15 terms is 525.

(ii) an = 9 – 5n

• a₁ = 9 – 5(1) = 4
• a₂ = 9 – 5(2) = -1
• a₃ = 9 – 5(3) = -6

The common difference, d = a₂ – a₁ = -1 – 4 = -5. Since the common difference is constant, the sequence is an AP.

To find the sum of the first 15 terms (S₁₅):

• We know a = 4, d = -5, and n = 15.
• Using the formula Sₙ = n/2 [2a + (n-1)d]:
  • S₁₅ = 15/2 [2(4) + (15-1)(-5)]
  • S₁₅ = 15/2 [8 – 70]
  • S₁₅ = 15/2 * (-62)
  • S₁₅ = -465

Therefore, the sequence is an AP with a common difference of -5, and the sum of the first 15 terms is -465.

11. If the sum of the first n terms of an AP is 4n – n2, what is the first term (that is S1)? What is the sum of first two terms? What is the second term? Similarly, find the 3rd, the 10th and the nth terms.

Ans : 

1. First term (S₁):

• S₁ = 4(1) – 1² = 3

2. Sum of first two terms (S₂):

• S₂ = 4(2) – 2² = 8 – 4 = 4

3. Second Term (a₂):

• a₂ = S₂ – S₁ 
• = 4 – 3 = 1

4. Third Term (a₃):

• S₃ = 4(3) – 3² = 12 – 9 = 3
• a₃ = S₃ – S₂
•  = 3 – 4 = -1

5. Tenth Term (a₁₀):

• S₁₀ = 4(10) – 10² = 40 – 100 = -60
• S₉ = 4(9) – 9² = 36 – 81 = -45
• a₁₀ = S₁₀ – S₉ = -60 – (-45) = -15

6. nth Term (aₙ):

• Sn = 4n – n²
• Sₙ₋₁ = 4(n-1) – (n-1)² = 4n – 4 – n² + 2n – 1 = -n² + 6n – 5
• aₙ = Sn – Sₙ₋₁ = (4n – n²) – (-n² + 6n – 5) = 5 – 2n

12. Find the sum of the first 40 positive integers divisible by 6.

Ans : 

The first few terms of the series are: 6, 12, 18, …

This is an arithmetic progression with:

• First term (a) = 6
• Common difference (d) = 6
• Number of terms (n) = 40
• Sn = n/2 [2a + (n-1)d]

Substituting the values:

• S₄₀ = 40/2 [2*6 + (40-1)6] = 20 [12 + 234] = 20 * 246 = 4920

13. Find the sum of the first 15 multiples of 8.

Ans : 

The multiples of 8 form an arithmetic progression (AP) with:

• First term (a) = 8
• Common difference (d) = 8
• Number of terms (n) = 15

We can use the formula for the sum of an AP:

• Sn = n/2 [2a + (n-1)d]

Substituting the values:

• S₁₅ = 15/2 [2*8 + (15-1)8] = 15/2 [16 + 112] = 15/2 * 128 = 15 * 64 = 960

14. Find the sum of the odd numbers between 0 and 50.

Ans : 

The odd numbers between 0 and 50 form an arithmetic progression (AP) with:

• First term (a) = 1
• Common difference (d) = 2
• Last term (l) = 49

To find the number of terms (n), we can use the formula for the nth term of an AP:

• l = a + (n – 1)d
• 49 = 1 + (n – 1)2
• 48 = 2(n – 1)
• n – 1 = 24
• n = 25

Now, we can use the formula for the sum of n terms of an AP:

• Sn = n/2 * (a + l)
• S₂₅ = 25/2 * (1 + 49)
• S₂₅ = 25/2 * 50
• S₂₅ = 25 * 25
• S₂₅ = 625

15. A contract on construction job specifies a penalty for delay of completion beyond a certain date as follows:

₹ 200 for the first day, ₹ 250 for the second day, ₹ 300 for the third day, etc. the penalty for each succeeding day being ₹ 50 more than for the preceding day. How much money the contractor has to pay as penalty, if he has delayed the work by 30 days?

Ans : 

• Sn = n/2 [2a + (n-1)d]

Substituting the values:

• S₃₀ = 30/2 [2*200 + (30-1)50]
• S₃₀ = 15 [400 + 1450]
• S₃₀ = 15 * 1850
• S₃₀ = 27750

Therefore, the contractor has to pay a penalty of ₹27,750 for delaying the work by 30 days.

16. A sum of ₹ 700 is to be used to give seven cash prizes to students of a school for their overall academic performance. If each prize is ₹ 20 less than its preceding prize, find the value of each of the prizes.

Ans : 

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, eg. 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?

Ans : 

This problem forms an arithmetic progression (AP) where:

• First term (a) = 3 (since there are 3 sections per class, and class I plants 1 tree each)
• Common difference (d) = 3 (as the number of trees increases by 3 for each subsequent class)
• Number of terms (n) = 12 (as there are 12 classes)

We can use the formula for the sum of an AP:

• Sn = n/2 [2a + (n-1)d]

Substituting the values:

• S₁₂ = 12/2 [2*3 + (12-1)3] = 6 [6 + 33] = 6 * 39 = 234

Therefore, a total of 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 cm, 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 π = 22/7)

[Hint: Length of successive semicircles is l1, l2, l3, l4, … with centres at A, B, respectively.]

Ans : 

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 (see Figure). In how many rows are the 200 logs placed and how many logs are in the top row?

Ans : 

Observations from the Image:

1. Number of logs per row: The number of logs decreases by 1 in each subsequent row.
2. Total number of logs: We are given that there are 200 logs in total.

Analysis and Solution:

The number of logs in each row forms an arithmetic sequence: 20, 19, 18, … Let’s denote:

• n: the number of rows
• a: the first term (20)
• d: the common difference (-1)
• Sn: the sum of the logs in n rows (Sn = 200)

We can use the formula for the sum of an arithmetic series:

Sn = (n/2)(2a + (n-1)d)

Substituting the values:

200 = (n/2)(2*20 + (n-1)(-1))

400 = n(40 – n + 1)

400 = n(41 – n)

n^2 – 41n + 400 = 0

Solving this quadratic equation, we get n = 16 or n = 25.

However, n = 25 leads to a negative value for the number of logs in the top row, which is not possible. Therefore, the number of rows is 16.

To find the number of logs in the top row, we can use the formula for the nth term of an arithmetic sequence:

an = a + (n-1)d

a16 = 20 + (16-1)(-1)

a16 = 20 – 15

a16 = 5

Conclusion:

The logs are placed in 16 rows, and the top row has 5 logs.

20. 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

A competitor starts from the bucket, picks up the nearest potato, runs back with it, drops it in the bucket, runs back to pick up the next potato, runs to the bucket to drop it in, and she continues in the same way until all the potatoes are in the bucket. What is the total distance the competitor has to run?

Ans :