Question

If the first term of an A.P. is 3 and the sum of its first four terms is equal to one-fifth of the sum of the next four terms, then the sum of the first 20 terms is equal to:

• A. \( -1200 \)
• B. \( -1080 \)
• C. \( -1020 \)
• D. \( -120 \)

Hint:

For arithmetic progressions, use sum formulas effectively to simplify and solve equations.

Answer 1

The Correct Option is B

To address this problem, we will proceed in stages:

Stage 1: Comprehending the Arithmetic Progression (A.P.)

The initial term of the A.P. is established as \(a_1 = 3\). Our objective is to determine the common difference, denoted by \(d\), and subsequently, the sum of the first 20 terms.

Stage 2: Evaluating the Provided Conditions

It is stated that the sum of the initial four terms equals one-fifth of the sum of the subsequent four terms.

The sum of the first four terms, \(S_4\), is calculated as:

\(S_4 = \frac{4}{2}\left(2a_1 + 3d\right)\)

Substituting \(a_1 = 3\):

\(S_4 = 2(6 + 3d) = 12 + 6d\)

The subsequent four terms encompass the 5th through the 8th terms. Their sum, \(S_{5-8}\), is:

\(S_{5-8} = \frac{4}{2}\left(2(a_1 + 4d) + 3d\right)\)

This simplifies to:

\(S_{5-8} = 2\left(6 + 11d\right) = 12 + 22d\)

As per the problem statement, \(S_4 = \frac{1}{5}S_{5-8}\):

\(12 + 6d = \frac{1}{5}(12 + 22d)\)

Solving this equation:

Multiply all terms by 5 to remove the fraction:

\(5(12 + 6d) = 12 + 22d\)

\(60 + 30d = 12 + 22d\)

Rearrange to isolate \(d\):

\(8d = -48\)

\(d = -6\)

Stage 3: Ascertaining the Sum of the First 20 Terms

The formula for the sum of the first \(n\) terms of an A.P. is:

\(S_n = \frac{n}{2}(2a_1 + (n-1)d)\)

For the first 20 terms:

\(S_{20} = \frac{20}{2}(2 \times 3 + 19 \times (-6))\)

\(S_{20} = 10(6 - 114)\)

\(S_{20} = 10(-108)\)

\(S_{20} = -1080\)

Consequently, the sum of the first 20 terms is \(-1080\), which aligns with the provided correct option.