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

(a) Sum of the first 23 terms of the AP $4, -3, -10, \ldots$

To find the sum of the first 23 terms of an arithmetic progression (AP), we can use the formula:

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

where:

• $S_n$ is the sum of the first $n$ terms,
• $a$ is the first term,
• $d$ is the common difference, and
• $n$ is the number of terms.

For this AP:

• $a = 4$,
• $d = -3 - 4 = -7$,
• $n = 23$.

Now, plug in the values:

$S_{23} = \frac{23}{2} \times \left(2 \times 4 + (23-1) \times (-7)\right)$ $S_{23} = \frac{23}{2} \times \left(8 + 22 \times (-7)\right)$ $S_{23} = \frac{23}{2} \times \left(8 - 154\right)$ $S_{23} = \frac{23}{2} \times (-146)$ $S_{23} = 23 \times (-73) = -1679$

So, the sum of the first 23 terms is $\boxed{-1679}$.

(b) Arithmetic series with first term 4 and common difference $d$:

(i) Sum of the first 20 terms

We use the same formula:

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

Here:

• $a = 4$,
• $d$ is the common difference,
• $n = 20$.

$S_{20} = \frac{20}{2} \times \left(2 \times 4 + (20-1) \times d\right)$ $S_{20} = 10 \times \left(8 + 19d\right)$ $S_{20} = 80 + 190d$

So, the sum of the first 20 terms is $\boxed{80 + 190d}$.

(ii) Sum of the first 100 terms

Using the same approach:

$S_{100} = \frac{100}{2} \times \left(2a + (100-1) \times d\right)$ $S_{100} = 50 \times \left(8 + 99d\right)$ $S_{100} = 400 + 4950d$

So, the sum of the first 100 terms is $\boxed{400 + 4950d}$.

(c) Sum of the arithmetic series with first term 1, common difference 3, and last term 100

For this, we use the formula:

$S_n = \frac{n}{2} \times \left(a + l\right)$

where $l$ is the last term. First, let's find $n$ using the formula for the nth term:

$l = a + (n-1) \times d$

Given $l = 100$, $a = 1$, $d = 3$:

$100 = 1 + (n-1) \times 3$ $99 = (n-1) \times 3$ $n-1 = 33 \Rightarrow n = 34$

Now, calculate the sum:

$S_{34} = \frac{34}{2} \times (1 + 100)$ $S_{34} = 17 \times 101 = 1717$

So, the sum of this series is $\boxed{1717}$.

(d) The sum of the first 20 terms is equal to the sum of the first 22 terms. If the common difference is 2, find the first term.

Let the first term be $a$. We know:

$S_{20} = S_{22}$

Using the sum formula:

$\frac{20}{2} \times \left(2a + 19 \times 2\right) = \frac{22}{2} \times \left(2a + 21 \times 2\right)$ $10 \times \left(2a + 38\right) = 11 \times \left(2a + 42\right)$ $20a + 380 = 22a + 462$ $380 - 462 = 22a - 20a$ $-82 = 2a \Rightarrow a = -41$

So, the first term is $\boxed{-41}$.

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Would you like more details or have any questions about these solutions?

Here are five related questions:

1. How do you find the sum of the first n terms of an arithmetic series when the last term is given?
2. What happens to the sum of an arithmetic series if the common difference is negative?
3. How can you derive the formula for the nth term of an arithmetic progression?
4. How does the common difference influence the sum of the series as the number of terms increases?
5. Can the sum of an arithmetic series be zero? If so, under what conditions?

Tip: In arithmetic progressions, the common difference (d) determines whether the sequence is increasing, decreasing, or constant.