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 solve this problem, let's break it down step-by-step:

Step 1: Understanding the Arithmetic Progression (A.P.) 

The first term of the A.P. is given as \(a_1 = 3\). We need to find the common difference (denoted as \(d\)) and eventually the sum of the first 20 terms.

Step 2: Conditions given

We are told that the sum of the first four terms is equal to one-fifth of the sum of the next four terms.

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

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

Substitute \(a_1 = 3\):

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

The next four terms would be the 5th to the 8th terms. The sum \(S_{5-8}\) is:

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

Which simplifies to:

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

According to the problem, \(S_4 = \frac{1}{5}S_{5-8}\):

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

Solving this equation:

Multiply through by 5 to eliminate the fraction:

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

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

Rearrange to find \(d\):

\(8d = -48\)

\(d = -6\)

Step 3: Find the sum of the first 20 terms

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

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

For the first 20 terms:

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

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

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

\(S_{20} = -1080\)

Therefore, the sum of the first 20 terms is \(-1080\), which matches the given correct option.

Answer 2

Step 1 — Convert the relation into a simpler form

Start with the given equation: \(S_4 = \dfrac{1}{5}\,(S_8 - S_4)\). Multiply both sides by 5: \(5S_4 = S_8 - S_4\). Bring \(S_4\) terms together: \(6S_4 = S_8\).

Step 2 — Use the formula for sum of n terms of an AP

The sum of the first \(n\) terms of an AP with first term \(a\) and common difference \(d\) is \(S_n=\dfrac{n}{2}\big(2a+(n-1)d\big)\). Apply this for \(n=4\) and \(n=8\):

\(S_4=\dfrac{4}{2}\big(2a+(4-1)d\big)=2\big(2a+3d\big)\),
\(S_8=\dfrac{8}{2}\big(2a+(8-1)d\big)=4\big(2a+7d\big).\)

Step 3 — Substitute into \(6S_4 = S_8\) and solve for \(d\)

Use \(6S_4 = S_8\):

\(6\cdot\big[2(2a+3d)\big] = 4(2a+7d).\)

Plug \(a=3\):

\(6\cdot 2\big(2\cdot 3 + 3d\big) = 4\big(2\cdot 3 + 7d\big)\)  ⇒  \(12(6 + 3d) = 4(6 + 7d).\)

Simplify step-by-step:

Left: \(12\cdot 6 + 12\cdot 3d = 72 + 36d.\)
Right: \(4\cdot 6 + 4\cdot 7d = 24 + 28d.\)

Equate and isolate \(d\):

\(72 + 36d = 24 + 28d\)  ⇒  \(36d - 28d = 24 - 72\)  ⇒  \(8d = -48\)  ⇒  \(d = -6\).

Step 4 — Quick numeric sanity check (compute S₄ and S₈ directly)

Write first few terms of the AP with \(a=3\) and \(d=-6\):

Terms: \(a_1=3,\; a_2=3-6=-3,\; a_3=-9,\; a_4=-15,\; a_5=-21,\; a_6=-27,\; a_7=-33,\; a_8=-39,\dots\)

Now compute partial sums:

\(S_4 = 3 + (-3) + (-9) + (-15) = -24.\)

\(S_8 = S_4 + (-21) + (-27) + (-33) + (-39) = -24 - 120 = -144.\)

Check relation: \(6S_4 = 6(-24) = -144 = S_8\). The relation holds — \(d=-6\) is correct.

Step 5 — Compute \(S_{20}\)

Use the sum formula with \(n=20\), \(a=3\), \(d=-6\):

\(S_{20}=\dfrac{20}{2}\big(2a+(20-1)d\big)=10\big(2\cdot 3 + 19(-6)\big).\)

Evaluate inside the bracket:

\(2\cdot 3 + 19(-6) = 6 - 114 = -108.\)

Thus \(S_{20} = 10\times (-108) = \mathbf{-1080}.\)

Interpretation

The sum \(S_{20}\) is negative because the common difference is large and negative: after the first term the sequence quickly becomes negative and the negative terms dominate the sum of the first 20 terms.

Common mistakes to avoid

• Arithmetic slip when simplifying \(12(6+3d)=4(6+7d)\).
• Forgetting to substitute \(a=3\) before expanding.
• Sign errors when computing sums of negative terms — always compute stepwise or use the formula.

Final answer

\( \boxed{S_{20} = -1080} \)