10 Math Worksheet CH-5 Arithmetic Progressions | PDF | Triangle ... - Free Printable

Problem: Solve the arithmetic progression (AP) problems from the worksheet and explain the solutions.

#### Arithmetic Progression Formulae:
1. General term of an AP:
\[
a_n = a + (n-1)d
\]
where:
- \( a \) is the first term,
- \( d \) is the common difference,
- \( n \) is the term number,
- \( a_n \) is the \( n \)-th term.

2. Sum of the first \( n \) terms of an AP:
\[
S_n = \frac{n}{2} \left[ 2a + (n-1)d \right]
\]

3. Common difference:
\[
d = a_{n+1} - a_n
\]

---

Solutions to the Problems:

#### 1. In an AP, if \( d = -4 \), \( n = 7 \), \( a_n = 4 \), find \( a \).
Using the general term formula:
\[
a_n = a + (n-1)d
\]
Substitute \( a_n = 4 \), \( d = -4 \), and \( n = 7 \):
\[
4 = a + (7-1)(-4)
\]
\[
4 = a + 6(-4)
\]
\[
4 = a - 24
\]
\[
a = 4 + 24 = 28
\]
Answer: (d) 28

---

#### 2. In an AP, if \( a = 3.5 \), \( d = 0 \), \( n = 101 \), then \( a_n \) will be:
Using the general term formula:
\[
a_n = a + (n-1)d
\]
Substitute \( a = 3.5 \), \( d = 0 \), and \( n = 101 \):
\[
a_n = 3.5 + (101-1)(0)
\]
\[
a_n = 3.5 + 0 = 3.5
\]
Answer: (b) 3.5

---

#### 3. The list of terms \(-10, -6, -2, 2, \ldots\) is:
To check if it is an AP, calculate the common difference \( d \):
\[
d = a_2 - a_1 = -6 - (-10) = -6 + 10 = 4
\]
\[
d = a_3 - a_2 = -2 - (-6) = -2 + 6 = 4
\]
Since the common difference is constant (\( d = 4 \)), the sequence is an AP.
Answer: (c) an A.P. with \( d = 4 \)

---

#### 4. The 11th term of the AP: \(-5, -\frac{5}{2}, \frac{5}{2}, \ldots\) is:
First, identify the first term \( a \) and the common difference \( d \):
\[
a = -5
\]
\[
d = a_2 - a_1 = -\frac{5}{2} - (-5) = -\frac{5}{2} + 5 = -\frac{5}{2} + \frac{10}{2} = \frac{5}{2}
\]
Using the general term formula for the 11th term (\( n = 11 \)):
\[
a_{11} = a + (11-1)d
\]
\[
a_{11} = -5 + 10 \left( \frac{5}{2} \right)
\]
\[
a_{11} = -5 + 25 = 20
\]
Answer: (b) 20

---

#### 5. The first four terms of an AP whose first term is \(-2\) and the common difference is \(-2\) are:
Using the general term formula:
\[
a_1 = a = -2
\]
\[
a_2 = a + d = -2 + (-2) = -4
\]
\[
a_3 = a + 2d = -2 + 2(-2) = -2 - 4 = -6
\]
\[
a_4 = a + 3d = -2 + 3(-2) = -2 - 6 = -8
\]
The first four terms are: \(-2, -4, -6, -8\).
Answer: (d) \(-2, -4, -6, -8\)

---

#### 6. The 21st term of the AP whose first two terms are \(-3\) and \(4\):
First, find the common difference \( d \):
\[
d = a_2 - a_1 = 4 - (-3) = 4 + 3 = 7
\]
Using the general term formula for the 21st term (\( n = 21 \)):
\[
a_{21} = a + (21-1)d
\]
\[
a_{21} = -3 + 20(7)
\]
\[
a_{21} = -3 + 140 = 137
\]
Answer: (b) 137

---

#### 7. If the 2nd term of an AP is 13 and the 5th term is 25, what is its 7th term?
Let the first term be \( a \) and the common difference be \( d \). Using the general term formula:
\[
a_2 = a + d = 13 \quad \text{(1)}
\]
\[
a_5 = a + 4d = 25 \quad \text{(2)}
\]
Subtract equation (1) from equation (2):
\[
(a + 4d) - (a + d) = 25 - 13
\]
\[
3d = 12
\]
\[
d = 4
\]
Substitute \( d = 4 \) into equation (1):
\[
a + 4 = 13
\]
\[
a = 9
\]
Now, find the 7th term:
\[
a_7 = a + 6d
\]
\[
a_7 = 9 + 6(4) = 9 + 24 = 33
\]
Answer: (b) 33

---

#### 8. Which term of the AP: \( 2, 14, 26, 38, \ldots \) is 210?
First, identify the first term \( a \) and the common difference \( d \):
\[
a = 2
\]
\[
d = a_2 - a_1 = 14 - 2 = 12
\]
Using the general term formula:
\[
a_n = a + (n-1)d
\]
Set \( a_n = 210 \):
\[
210 = 2 + (n-1)(12)
\]
\[
210 = 2 + 12(n-1)
\]
\[
210 = 2 + 12n - 12
\]
\[
210 = 12n - 10
\]
\[
220 = 12n
\]
\[
n = \frac{220}{12} = \frac{55}{3} \approx 18.33
\]
Since \( n \) must be an integer, there is no term in the AP that equals 210.
Answer: None of the options (re-evaluate the problem or options).

---

#### 9. If the common difference of an AP is 5, then what is \( a_{18} - a_{14} \)?
Using the general term formula:
\[
a_{18} = a + 17d
\]
\[
a_{14} = a + 13d
\]
Calculate the difference:
\[
a_{18} - a_{14} = (a + 17d) - (a + 13d)
\]
\[
a_{18} - a_{14} = a + 17d - a - 13d = 4d
\]
Given \( d = 5 \):
\[
a_{18} - a_{14} = 4(5) = 20
\]
Answer: (b) 20

---

#### 10. What is the common difference of an AP in which \( a_{18} - a_{14} = 32 \)?
Using the general term formula:
\[
a_{18} = a + 17d
\]
\[
a_{14} = a + 13d
\]
Calculate the difference:
\[
a_{18} - a_{14} = (a + 17d) - (a + 13d)
\]
\[
a_{18} - a_{14} = a + 17d - a - 13d = 4d
\]
Given \( a_{18} - a_{14} = 32 \):
\[
4d = 32
\]
\[
d = 8
\]
Answer: (d) 8

---

#### 11. Two APs have the same common difference. The first term of one of these is \(-1\) and that of the other is \(-8\). The difference between their 4th terms is:
Let the common difference be \( d \). The 4th term of the first AP is:
\[
a_{4,1} = -1 + 3d
\]
The 4th term of the second AP is:
\[
a_{4,2} = -8 + 3d
\]
Calculate the difference:
\[
a_{4,1} - a_{4,2} = (-1 + 3d) - (-8 + 3d)
\]
\[
a_{4,1} - a_{4,2} = -1 + 3d + 8 - 3d = 7
\]
Answer: (c) 7

---

#### 12. If 7 times the 7th term of an AP is equal to 11 times its 11th term, then its 18th term will be:
Let the first term be \( a \) and the common difference be \( d \). The 7th term is:
\[
a_7 = a + 6d
\]
The 11th term is:
\[
a_{11} = a + 10d
\]
Given:
\[
7(a_7) = 11(a_{11})
\]
\[
7(a + 6d) = 11(a + 10d)
\]
Expand and simplify:
\[
7a + 42d = 11a + 110d
\]
\[
7a - 11a = 110d - 42d
\]
\[
-4a = 68d
\]
\[
a = -\frac{68d}{4} = -17d
\]
Now, find the 18th term:
\[
a_{18} = a + 17d
\]
Substitute \( a = -17d \):
\[
a_{18} = -17d + 17d = 0
\]
Answer: (d) 0

---

#### 13. The 4th term from the end of the AP: \(-11, -8, -5, \ldots, 49\) is:
First, find the total number of terms \( n \) in the AP. The first term \( a = -11 \), the last term \( l = 49 \), and the common difference \( d = 3 \). Using the general term formula:
\[
l = a + (n-1)d
\]
\[
49 = -11 + (n-1)(3)
\]
\[
49 = -11 + 3(n-1)
\]
\[
49 = -11 + 3n - 3
\]
\[
49 = 3n - 14
\]
\[
63 = 3n
\]
\[
n = 21
\]
The 4th term from the end is the \((n-3)\)-th term:
\[
a_{18} = a + 17d
\]
Substitute \( a = -11 \) and \( d = 3 \):
\[
a_{18} = -11 + 17(3) = -11 + 51 = 40
\]
Answer: (b) 40

---

#### 14. The famous mathematician associated with finding the sum of the first 100 natural numbers is:
The mathematician is Gauss.
Answer: (c) Gauss

---

#### 15. If the first term of an AP is \(-5\) and the common difference is \(2\), then the sum of the first 6 terms is:
Using the sum formula:
\[
S_n = \frac{n}{2} \left[ 2a + (n-1)d \right]
\]
Substitute \( a = -5 \), \( d = 2 \), and \( n = 6 \):
\[
S_6 = \frac{6}{2} \left[ 2(-5) + (6-1)(2) \right]
\]
\[
S_6 = 3 \left[ -10 + 5(2) \right]
\]
\[
S_6 = 3 \left[ -10 + 10 \right]
\]
\[
S_6 = 3(0) = 0
\]
Answer: (a) 0

---

#### 16. The sum of the first 16 terms of the AP: \( 10, 6, 2, \ldots \) is:
First, identify the first term \( a \) and the common difference \( d \):
\[
a = 10
\]
\[
d = a_2 - a_1 = 6 - 10 = -4
\]
Using the sum formula:
\[
S_n = \frac{n}{2} \left[ 2a + (n-1)d \right]
\]
Substitute \( a = 10 \), \( d = -4 \), and \( n = 16 \):
\[
S_{16} = \frac{16}{2} \left[ 2(10) + (16-1)(-4) \right]
\]
\[
S_{16} = 8 \left[ 20 + 15(-4) \right]
\]
\[
S_{16} = 8 \left[ 20 - 60 \right]
\]
\[
S_{16} = 8(-40) = -320
\]
Answer: (a) -320

---

#### 17. In an AP if \( a = 1 \), \( a_n = 20 \), and \( S_n = 399 \), then \( n \) is:
Using the general term formula:
\[
a_n = a + (n-1)d
\]
Substitute \( a = 1 \) and \( a_n = 20 \):
\[
20 = 1 + (n-1)d
\]
\[
19 = (n-1)d \quad \text{(1)}
\]
Using the sum formula:
\[
S_n = \frac{n}{2} \left[ 2a + (n-1)d \right]
\]
Substitute \( a = 1 \), \( S_n = 399 \):
\[
399 = \frac{n}{2} \left[ 2(1) + (n-1)d \right]
\]
\[
399 = \frac{n}{2} \left[ 2 + (n-1)d \right] \quad \text{(2)}
\]
From equation (1):
\[
d = \frac{19}{n-1}
\]
Substitute \( d \) into equation (2):
\[
399 = \frac{n}{2} \left[ 2 + (n-1) \left( \frac{19}{n-1} \right) \right]
\]
\[
399 = \frac{n}{2} \left[ 2 + 19 \right]
\]
\[
399 = \frac{n}{2} \cdot 21
\]
\[
399 = \frac{21n}{2}
\]
\[
798 = 21n
\]
\[
n = \frac{798}{21} = 38
\]
Answer: (c) 38

---

#### 18. The sum of the first five multiples of 3 is:
The first five multiples of 3 are: \( 3, 6, 9, 12, 15 \). This is an AP with:
\[
a = 3, \quad d = 3, \quad n = 5
\]
Using the sum formula:
\[
S_n = \frac{n}{2} \left[ 2a + (n-1)d \right]
\]
Substitute \( a = 3 \), \( d = 3 \), and \( n = 5 \):
\[
S_5 = \frac{5}{2} \left[ 2(3) + (5-1)(3) \right]
\]
\[
S_5 = \frac{5}{2} \left[ 6 + 4(3) \right]
\]
\[
S_5 = \frac{5}{2} \left[ 6 + 12 \right]
\]
\[
S_5 = \frac{5}{2} \cdot 18 = 5 \cdot 9 = 45
\]
Answer: (a) 45

---

Final Answers:

\[
\boxed{
\begin{array}{ll}
1. & (d) 28 \\
2. & (b) 3.5 \\
3. & (c) \text{an A.P. with } d = 4 \\
4. & (b) 20 \\
5. & (d) -2, -4, -6, -8 \\
6. & (b) 137 \\
7. & (b) 33 \\
8. & \text{None of the options} \\
9. & (b) 20 \\
10. & (d) 8 \\
11. & (c) 7 \\
12. & (d) 0 \\
13. & (b) 40 \\
14. & (c) Gauss \\
15. & (a) 0 \\
16. & (a) -320 \\
17. & (c) 38 \\
18. & (a) 45 \\
\end{array}
}
\]