The problem states:

• The 4th term of an arithmetic sequence is 20.
• The sum of the first 5 terms is 80.
• We are asked to find the sum of the first 11 terms.

Let's solve it step-by-step.

Step 1: Understanding the Arithmetic Sequence

For an arithmetic sequence, each term can be represented as: $a_n = a + (n-1)d$ where:

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

Step 2: Use the Information About the 4th Term

Since the 4th term is given as 20: $a_4 = a + 3d = 20$

Step 3: Use the Sum of the First 5 Terms

The sum of the first $n$ terms of an arithmetic sequence is given by: $S_n = \frac{n}{2} \times (2a + (n-1)d)$ For the first 5 terms ($S_5 = 80$): $S_5 = \frac{5}{2} \times (2a + 4d) = 80$ Expanding this equation: $5 \times (2a + 4d) = 160$ $2a + 4d = 32$

Step 4: Solve the System of Equations

Now we have two equations:

1. $a + 3d = 20$
2. $2a + 4d = 32$

Let's solve these equations simultaneously.

Solve Equation 1 for $a$:

$a = 20 - 3d$

Substitute $a$ in Equation 2:

$2(20 - 3d) + 4d = 32$ $40 - 6d + 4d = 32$ $-2d = -8$ $d = 4$

Substitute $d = 4$ back into Equation 1:

$a = 20 - 3 \times 4$ $a = 20 - 12$ $a = 8$

So, the first term $a = 8$ and the common difference $d = 4$.

Step 5: Find the Sum of the First 11 Terms

Now we use the sum formula for $S_{11}$: $S_{11} = \frac{11}{2} \times (2a + 10d)$ Substitute $a = 8$ and $d = 4$: $S_{11} = \frac{11}{2} \times (2 \times 8 + 10 \times 4)$ $= \frac{11}{2} \times (16 + 40)$ $= \frac{11}{2} \times 56$ $= 11 \times 28$ $= 308$

Answer

The sum of the first 11 terms is 308, so the correct answer is: D. 308

Would you like more details or have any questions?

Here are 5 related questions to deepen understanding:

1. How would the answer change if the 4th term were different?
2. How can we verify that our values of $a$ and $d$ satisfy the initial conditions?
3. What would be the sum of the first 15 terms of this sequence?
4. How does the common difference impact the overall sum of terms?
5. What if we only knew the sum of the first 4 terms; could we still solve for $a$ and $d$?

Tip: In arithmetic sequences, the common difference $d$ determines how quickly the sequence grows or shrinks.