Let's break down each problem step by step.

Problem 5: Arithmetic Series

Given:

• First term ($a_1$) = 2
• Last term ($a_n$) = 20
• Common difference ($d$) = 2

a) Determine the number of terms in the series

The $n$th term of an arithmetic series is given by the formula: $a_n = a_1 + (n-1) \times d$ Substitute the given values: $20 = 2 + (n-1) \times 2$

Let's solve for $n$:

$20 - 2 = (n-1) \times 2$ $18 = 2n - 2$ $20 = 2n$ $n = \frac{18}{2} + 1 = 10$

So, the number of terms in the series is 10.

b) Determine the sum of all the terms in the series

The sum $S_n$ of the first $n$ terms of an arithmetic series can be calculated by: $S_n = \frac{n}{2} \times (a_1 + a_n)$

Substitute the values: $S_{10} = \frac{10}{2} \times (2 + 20) = 5 \times 22 = 110$

So, the sum of all the terms in the series is 110.

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Problem 6: Geometric Series

Given:

• Sum of the series ($S_n$) = 1365
• Common ratio ($r$) = 4
• Number of terms ($n$) = 6

We need to find the first term ($a_1$).

The sum $S_n$ of a geometric series is given by: $S_n = a_1 \times \frac{r^n - 1}{r - 1}$

Substitute the given values: $1365 = a_1 \times \frac{4^6 - 1}{4 - 1}$

Calculate $4^6$: $4^6 = 4096$ Now, substitute and solve for $a_1$: $1365 = a_1 \times \frac{4096 - 1}{3}$ $1365 = a_1 \times \frac{4095}{3} = a_1 \times 1365$ $a_1 = \frac{1365}{1365} = 1$

So, the value of the first term is 1.

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Questions for Further Exploration:

1. What is the formula for the sum of the first $n$ terms in an arithmetic series when the common difference is negative?
2. How would the sum change if the first term of the arithmetic series was negative?
3. Can you derive the formula for the sum of a geometric series with a common ratio of less than 1?
4. How would you solve for the number of terms in a geometric series if the sum and first term are known, but the common ratio is unknown?
5. How does the sum of an infinite geometric series differ from that of a finite series?

Tip: In both arithmetic and geometric series, identifying the first term, common difference or ratio, and the number of terms early on is crucial for quickly solving related problems.