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Given Information:

• First term of the AP ($a$) = 5
• Common difference ($d$) = 1.5
• Total number of terms ($n$) = 22

We are solving for:

1. The sum of the first 22 terms ($S_{22}$).
2. The last term ($a_{22}$).

Solution:

1. Formula for the sum of an AP:

$S_n = \frac{n}{2} \times [2a + (n-1)d]$ Substitute the given values: $S_{22} = \frac{22}{2} \times [2(5) + (22-1)(1.5)]$ $S_{22} = 11 \times [10 + 21 \times 1.5]$ $S_{22} = 11 \times [10 + 31.5] = 11 \times 41.5 = 456.5$

So, the sum of the first 22 terms is: $S_{22} = 456.5$

2. Formula for the last term:

$a_n = a + (n-1)d$ Substitute the values: $a_{22} = 5 + (22-1)(1.5)$ $a_{22} = 5 + 31.5 = 36.5$

So, the last term is: $a_{22} = 36.5$

Final Answers:

1. Sum of the first 22 terms ($S_{22}$) = 456.5
2. Last term ($a_{22}$) = 36.5

Would you like further clarification or details on the process?

Related Questions:

1. What is the formula for an arithmetic progression?
2. How would the sum change if the common ratio were negative?
3. How do you derive the formula for the sum of an arithmetic series?
4. What is the significance of the common difference in an AP?
5. How would the last term change for 50 terms instead of 22?

Tip:

Always double-check your substitutions in the formulas for accuracy during exams or problem-solving.