Key Concepts and Formulas

• Arithmetic Progression (A.P.): A sequence where the difference between consecutive terms is constant.
  • First term: $a$
  • Common difference: $d$
  • $n$-th term: $a_n = a + (n-1)d$
• Sum of first $n$ terms of an A.P.: $S_n = \frac{n}{2} [2a + (n-1)d]$

Step-by-Step Solution

Step 1: Understand the Problem and Identify Given Information We are given an Arithmetic Progression (A.P.) with its first term $a=3$. The problem states a relationship between the sum of the first four terms ($S_4$) and the sum of the next four terms (terms 5 to 8). We need to find the sum of the first 20 terms ($S_{20}$).

Step 2: Translate the Given Condition into an Equation The condition is: "the sum of its first four terms is equal to one-fifth of the sum of the next four terms."

• The sum of the first four terms is $S_4$.
• The sum of the next four terms (5th, 6th, 7th, 8th) can be represented as the sum of the first eight terms minus the sum of the first four terms. That is, $S_8 - S_4$.

The equation is: $S_4 = \frac{1}{5} (S_8 - S_4)$

Step 3: Simplify the Equation to Relate $S_4$ and $S_8$ To make the equation easier to work with, we can simplify it: Multiply both sides by 5: $5S_4 = S_8 - S_4$ Add $S_4$ to both sides: $6S_4 = S_8$ This simplified relationship will be crucial for finding the common difference, $d$.

Step 4: Express $S_4$ and $S_8$ using the Sum Formula and Substitute into the Simplified Equation We use the formula $S_n = \frac{n}{2} [2a + (n-1)d]$ with $a=3$.

For $S_4$: $S_4 = \frac{4}{2} [2(3) + (4-1)d] = 2 [6 + 3d]$

For $S_8$: $S_8 = \frac{8}{2} [2(3) + (8-1)d] = 4 [6 + 7d]$

Now, substitute these into the equation $6S_4 = S_8$: $6 \cdot (2 [6 + 3d]) = 4 [6 + 7d]$ $12 (6 + 3d) = 4 (6 + 7d)$

Step 5: Solve the Equation for the Common Difference ($d$) We now solve the linear equation for $d$: $72 + 36d = 24 + 28d$ Subtract $28d$ from both sides: $72 + 8d = 24$ Subtract $72$ from both sides: $8d = 24 - 72$ $8d = -48$ Divide by 8: $d = \frac{-48}{8}$ $d = -6$ The common difference of the A.P. is $-6$.

Step 6: Calculate the Sum of the First 20 Terms ($S_{20}$) Now that we have the first term ($a=3$) and the common difference ($d=-6$), we can find $S_{20}$ using the sum formula: $S_{20} = \frac{20}{2} [2a + (20-1)d]$ Substitute $a=3$ and $d=-6$: $S_{20} = 10 [2(3) + (19)(-6)]$ $S_{20} = 10 [6 - 114]$ $S_{20} = 10 [-108]$ $S_{20} = -1080$

Common Mistakes & Tips

• Interpreting "Next Four Terms": Be careful to correctly interpret "sum of the next four terms" as $S_{n+k} - S_n$, not just $S_k$. In this case, it's $S_8 - S_4$.
• Algebraic Errors: Pay close attention to signs and distribution when simplifying equations involving $d$. A single sign error can lead to an incorrect value for $d$.
• Verification: After finding $d$, it's a good practice to verify it by substituting it back into the original condition ($6S_4 = S_8$) to ensure consistency. $S_4 = 2(6 + 3(-6)) = 2(6-18) = 2(-12) = -24$. $S_8 = 4(6 + 7(-6)) = 4(6-42) = 4(-36) = -144$. $6S_4 = 6(-24) = -144$, which equals $S_8$. The value $d=-6$ is correct.

Summary The problem involves an arithmetic progression where the first term is given and a condition relates the sum of the first four terms to the sum of the subsequent four terms. By translating this condition into an equation ($S_4 = \frac{1}{5}(S_8 - S_4)$) and simplifying it to $6S_4 = S_8$, we were able to substitute the general sum formula for an A.P. to solve for the common difference $d = -6$. With $a=3$ and $d=-6$, we then calculated the sum of the first 20 terms using the formula $S_{20} = \frac{20}{2}[2a + (19)d]$, resulting in $S_{20} = -1080$.

The final answer is $\boxed{-1080}$, which corresponds to option (C).