Let's solve each part of the question step by step.

Part (a): Sum of the First 23 Terms of the AP

The given arithmetic progression (AP) is: $4, -3, -10, \ldots$

Step 1: Identify the first term ($a$) and the common difference ($d$).

• First term, $a = 4$
• Common difference, $d = -3 - 4 = -7$

Step 2: Use the formula for the sum of the first $n$ terms of an AP:

$S_n = \frac{n}{2} \left(2a + (n-1)d\right)$

For the first 23 terms:

$S_{23} = \frac{23}{2} \left(2(4) + (23-1)(-7)\right)$

Step 3: Substitute the values into the formula:

$S_{23} = \frac{23}{2} \left(8 + 22(-7)\right) = \frac{23}{2} \left(8 - 154\right) = \frac{23}{2} \times (-146) = 23 \times (-73) = -1679$

So, the sum of the first 23 terms is $-1679$.

Part (b): Sum of the First 20 and 100 Terms

The first term of the series is given as $a = 4$ and the common difference $d = 3$.

(i) Sum of the First 20 Terms:

Use the sum formula $S_n = \frac{n}{2} \left(2a + (n-1)d\right)$:

$S_{20} = \frac{20}{2} \left(2(4) + (20-1)(3)\right)$

$S_{20} = 10 \times \left(8 + 57\right) = 10 \times 65 = 650$

So, the sum of the first 20 terms is $650$.

(ii) Sum of the First 100 Terms:

$S_{100} = \frac{100}{2} \left(2(4) + (100-1)(3)\right)$

$S_{100} = 50 \times \left(8 + 297\right) = 50 \times 305 = 15250$

So, the sum of the first 100 terms is $15250$.

Part (c): Sum of the Arithmetic Series with First Term 1, Common Difference 3, and Last Term 100

To find the sum of the series, we first need to determine the number of terms ($n$).

Step 1: Find $n$ using the formula for the $n$th term:

$a_n = a + (n-1)d$

Given that the last term ($a_n$) is 100:

$100 = 1 + (n-1) \times 3$

$99 = (n-1) \times 3 \quad \Rightarrow \quad n-1 = 33 \quad \Rightarrow \quad n = 34$

Step 2: Now, find the sum using the sum formula:

$S_n = \frac{n}{2} \times (a + a_n) = \frac{34}{2} \times (1 + 100) = 17 \times 101 = 1717$

So, the sum of the series is $1717$.

Part (d): Find the First Term Given the Sum Condition

Given:

• Sum of the first 20 terms equals the sum of the first 22 terms.
• Common difference, $d = 2$.

Let the first term be $a$.

Step 1: Write the equation for the sum of the first 20 terms and the first 22 terms:

$S_{20} = \frac{20}{2} \left(2a + 19d\right) \quad \text{and} \quad S_{22} = \frac{22}{2} \left(2a + 21d\right)$

Given $S_{20} = S_{22}$:

$10 \times \left(2a + 38\right) = 11 \times \left(2a + 42\right)$

Step 2: Simplify the equation:

$20a + 380 = 22a + 462$

$462 - 380 = 22a - 20a$

$2a = 82 \quad \Rightarrow \quad a = 41$

So, the first term is $41$.

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Would you like more details or have any other questions?

Here are 5 related questions you might consider:

1. How do you find the common difference if only the first and last terms are given?
2. What is the general formula for the $n$th term of an arithmetic sequence?
3. How can the sum of an infinite arithmetic series be determined?
4. What happens to the sum of an arithmetic series if the common difference is zero?
5. Can an arithmetic series have a negative sum if all its terms are positive?

Tip: In arithmetic sequences, understanding how the sum formula derives from the sum of the first and last terms can help clarify its application across different problems.