Solution:

We are solving two parts related to an arithmetic sequence where:

1. The sum of the first 10 terms, $S_{10}$, is 220.
2. The sum of the first 15 terms, $S_{15}$, is 180.

General Formula for the Sum of an Arithmetic Sequence

The sum of the first $n$ terms of an arithmetic sequence is given by: $S_n = \frac{n}{2} [2a + (n-1)d]$ where:

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

Part (a) Find $S_{20}$, the sum of the first 20 terms.

Step 1: Use the information to find $a$ and $d$.

For $S_{10}$: $220 = \frac{10}{2} [2a + (10-1)d]$ $220 = 5 [2a + 9d]$ 44 = 2a + 9d \tag{1}

For $S_{15}$: $180 = \frac{15}{2} [2a + (15-1)d]$ $180 = 7.5 [2a + 14d]$ 24 = 2a + 14d \tag{2}

Step 2: Solve for $a$ and $d$ using (1) and (2).

Subtract (1) from (2): $(2a + 14d) - (2a + 9d) = 24 - 44$ $5d = -20 \implies d = -4$

Substitute $d = -4$ into (1): $44 = 2a + 9(-4)$ $44 = 2a - 36 \implies 2a = 80 \implies a = 40$

Thus:

• $a = 40$
• $d = -4$

Step 3: Find $S_{20}$.

Use the formula for $S_n$ with $n = 20$: $S_{20} = \frac{20}{2} [2a + (20-1)d]$ $S_{20} = 10 [2(40) + 19(-4)]$ $S_{20} = 10 [80 - 76]$ $S_{20} = 10 \times 4 = 40$

So, the sum of the first 20 terms is: $\boxed{40}$

Part (b) Find $k$ if $S_k = 216$.

We are given $S_k = 216$ and need to find $k$. Using the formula: $216 = \frac{k}{2} [2a + (k-1)d]$ Substitute $a = 40$ and $d = -4$: $216 = \frac{k}{2} [2(40) + (k-1)(-4)]$ $216 = \frac{k}{2} [80 - 4(k-1)]$ $216 = \frac{k}{2} [80 - 4k + 4]$ $216 = \frac{k}{2} [84 - 4k]$ $432 = k (84 - 4k)$ $432 = 84k - 4k^2$ Rearrange into standard quadratic form: $4k^2 - 84k + 432 = 0$ Simplify: $k^2 - 21k + 108 = 0$

Step 1: Solve the quadratic equation.

Factorize: $k^2 - 21k + 108 = (k - 12)(k - 9)$ Thus: $k = 12 \quad \text{or} \quad k = 9$

Step 2: Verify the values.

• If $k = 12$: [ S_{12} = \frac{12}{2} [2(40) + (12-1)(-4)] = 6[80 - 44] = 6 \times 36 = 216 ]

• If $k = 9$: [ S_{9} = \frac{9}{2} [2(40) + (9-1)(-4)] = 4.5[80 - 32] = 4.5 \times 48 = 216 ]

Both values work, so: $\boxed{k = 12 \, \text{or} \, k = 9}$

Final Answers:

(a) The sum of the first 20 terms is $\boxed{40}$.
(b) The value of $k$ is $\boxed{9 \, \text{or} \, 12}$.

Would you like further clarifications or details on this?

Related Questions:

1. How does the common difference affect the behavior of an arithmetic sequence?
2. Derive the sum formula for an arithmetic sequence from the general term.
3. Solve a similar problem where $S_{10} = 300$ and $S_{15} = 500$.
4. What happens if the common difference is positive instead of negative?
5. How can quadratic equations help in solving summation problems?

Tip:

Always check your quadratic solutions by substituting them back into the original sum equation.