The question asks about an arithmetic progression (AP) and gives the following information:

• The sum of the first 7 terms ($S_7$) is 49.
• The sum of the first 17 terms ($S_{17}$) is 289.

We are required to find the sum of the first 20 terms ($S_{20}$).

Formula for the sum of terms in an AP:

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

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

Step 1: Use the sum formula for $S_7$ and $S_{17}$

From the first equation: $S_7 = \frac{7}{2} \times (2a + 6d) = 49$ This simplifies to: $7(2a + 6d) = 98 \implies 2a + 6d = 14 \quad \text{(Equation 1)}$

For the second equation: $S_{17} = \frac{17}{2} \times (2a + 16d) = 289$ This simplifies to: $17(2a + 16d) = 578 \implies 2a + 16d = 34 \quad \text{(Equation 2)}$

Step 2: Solve for $a$ and $d$

Subtract Equation 1 from Equation 2: $(2a + 16d) - (2a + 6d) = 34 - 14$ $10d = 20 \implies d = 2$

Substitute $d = 2$ into Equation 1: $2a + 6(2) = 14 \implies 2a + 12 = 14 \implies 2a = 2 \implies a = 1$

Step 3: Find the sum of the first 20 terms

Now that we know $a = 1$ and $d = 2$, we can use the sum formula to find $S_{20}$: $S_{20} = \frac{20}{2} \times (2(1) + (20-1)(2)) = 10 \times (2 + 38) = 10 \times 40 = 400$

Final Answer:

The sum of the first 20 terms is 400, so the correct option is A.

Would you like a deeper explanation on any part? Here are some follow-up questions you might find helpful:

1. How can I derive the formula for the sum of an arithmetic progression?
2. Why do we subtract the two equations to find the common difference $d$?
3. How would the problem change if the common difference $d$ was negative?
4. What are the general characteristics of an arithmetic progression?
5. How would you solve for the $n$-th term directly in this AP?

Tip: Always double-check the order of terms in your formulas to avoid simple algebraic mistakes!