1. Find the seventeenth term of the sequence $$ 2 ; 5 ; 8 ; \ldots $$ 2. How many terms are there in the sequence $$ 3 ; 7 ; 11 ; \ldots ; 147 $$ ? 3. Which term of the arithmetic sequence $$ 10 ; 7 ; \ldots $$ will be -5 ? 4. In an arithmetic sequence the seventh term is 30 and the fourth term is 18 . Find the first term and the common difference. 5. Find the twentieth term of an arithmetic sequence if the fifth te is 11 and the twelfth term is 25 . 6. The fourth term of an arithmetic sequence is 14 and the sixtee term is 50 . Which terin has a value of 77 ?

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### **1. Find the seventeenth term of the sequence $$ 2 ; 5 ; 8 ; \ldots $$**

**Solution:**

This is an arithmetic sequence where:
- **First term (a)** = 2
- **Common difference (d)** = 5 - 2 = 3

The $$ n $$-th term of an arithmetic sequence is given by:
$$ T_n = a + (n-1) \times d $$

For the 17th term:
$$ T_{17} = 2 + (17-1) \times 3 $$
$$ T_{17} = 2 + 16 \times 3 $$
$$ T_{17} = 2 + 48 $$
$$ T_{17} = 50 $$

**Answer:** The seventeenth term is **50**.

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### **2. How many terms are there in the sequence $$ 3 ; 7 ; 11 ; \ldots ; 147 $$?**

**Solution:**

This is an arithmetic sequence where:
- **First term (a)** = 3
- **Common difference (d)** = 7 - 3 = 4
- **Last term (L)** = 147

The $$ n $$-th term is:
$$ L = a + (n-1) \times d $$
$$ 147 = 3 + (n-1) \times 4 $$
$$ 147 - 3 = (n-1) \times 4 $$
$$ 144 = (n-1) \times 4 $$
$$ n-1 = \frac{144}{4} $$
$$ n-1 = 36 $$
$$ n = 37 $$

**Answer:** There are **37 terms** in the sequence.

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### **3. Which term of the arithmetic sequence $$ 10 ; 7 ; \ldots $$ will be -5?**

**Solution:**

This is an arithmetic sequence where:
- **First term (a)** = 10
- **Common difference (d)** = 7 - 10 = -3

We need to find $$ n $$ such that $$ T_n = -5 $$.

Using the formula:
$$ T_n = a + (n-1) \times d $$
$$ -5 = 10 + (n-1) \times (-3) $$
$$ -5 - 10 = -3(n-1) $$
$$ -15 = -3(n-1) $$
$$ 5 = n - 1 $$
$$ n = 6 $$

**Answer:** The term -5 is the **6th term** of the sequence.

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### **4. In an arithmetic sequence the seventh term is 30 and the fourth term is 18. Find the first term and the common difference.**

**Solution:**

Let:
- **First term (a)** = ?
- **Common difference (d)** = ?

Given:
- $$ T_7 = a + 6d = 30 $$  (1)
- $$ T_4 = a + 3d = 18 $$  (2)

Subtract equation (2) from equation (1):
$$ (a + 6d) - (a + 3d) = 30 - 18 $$
$$ 3d = 12 $$
$$ d = 4 $$

Now, substitute $$ d = 4 $$ into equation (2):
$$ a + 3 \times 4 = 18 $$
$$ a + 12 = 18 $$
$$ a = 6 $$

**Answer:** The first term is **6** and the common difference is **4**.

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### **5. Find the twentieth term of an arithmetic sequence if the fifth term is 11 and the twelfth term is 25.**

**Solution:**

Given:
- $$ T_5 = a + 4d = 11 $$  (1)
- $$ T_{12} = a + 11d = 25 $$ (2)

Subtract equation (1) from equation (2):
$$ (a + 11d) - (a + 4d) = 25 - 11 $$
$$ 7d = 14 $$
$$ d = 2 $$

Now, substitute $$ d = 2 $$ into equation (1):
$$ a + 4 \times 2 = 11 $$
$$ a + 8 = 11 $$
$$ a = 3 $$

Now, find the 20th term:
$$ T_{20} = a + 19d $$
$$ T_{20} = 3 + 19 \times 2 $$
$$ T_{20} = 3 + 38 $$
$$ T_{20} = 41 $$

**Answer:** The twentieth term is **41**.

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### **6. The fourth term of an arithmetic sequence is 14 and the sixteenth term is 50. Which term has a value of 77?**

**Solution:**

Given:
- $$ T_4 = a + 3d = 14 $$  (1)
- $$ T_{16} = a + 15d = 50 $$ (2)

Subtract equation (1) from equation (2):
$$ (a + 15d) - (a + 3d) = 50 - 14 $$
$$ 12d = 36 $$
$$ d = 3 $$

Now, substitute $$ d = 3 $$ into equation (1):
$$ a + 3 \times 3 = 14 $$
$$ a + 9 = 14 $$
$$ a = 5 $$

We need to find $$ n $$ such that $$ T_n = 77 $$:
$$ a + (n-1)d = 77 $$
$$ 5 + (n-1) \times 3 = 77 $$
$$ (n-1) \times 3 = 72 $$
$$ n-1 = 24 $$
$$ n = 25 $$

**Answer:** The term with the value **77** is the **25th term**.

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1. The seventeenth term is **50**. 2. There are **37 terms** in the sequence. 3. The term -5 is the **6th term**. 4. The first term is **6** and the common difference is **4**. 5. The twentieth term is **41**. 6. The term with the value **77** is the **25th term**.

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