AP: First Term 3, Sum of First 4 Terms = 1/5 of Next 4 — Sum of 20 Terms

JEE Main 2025 — 23 January, Morning Shift. Previous Year Question.

Problem

If the first term of an arithmetic progression is $3$, and the sum of the first $4$ terms is equal to $\dfrac{1}{5}$th of the sum of the next $4$ terms, then the sum of the first $20$ terms is:

Key insight. “The next four terms” means terms 5 through 8 — not a vague continuation but a specific, calculable block with its own sum formula. Writing both sums (first 4, next 4) in terms of the first term and common difference turns the given ratio into a single linear equation for $d$.

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Approach

Let the AP have first term $a=3$ and common difference $d$. Write the sum of the first 4 terms and the sum of the next 4 terms (terms 5 through 8) explicitly in terms of $a$ and $d$. Use the given ratio condition to solve for $d$, then compute the sum of the first 20 terms using the standard AP sum formula.

Solution

Step 1 — Write the sum of the first 4 terms

$$S_{1-4} = \frac{4}{2}\big[2a+(4-1)d\big] = 2(2a+3d) = 4a+6d$$

With $a=3$: $S_{1-4} = 12+6d$

Step 2 — Write the sum of the next 4 terms (terms 5 to 8)

$$a_5+a_6+a_7+a_8 = (a+4d)+(a+5d)+(a+6d)+(a+7d) = 4a+22d$$

With $a=3$: $S_{5-8} = 12+22d$

Step 3 — Apply the given condition

$$S_{1-4} = \frac{1}{5}S_{5-8}$$

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

Step 4 — Solve for d

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

$$60+30d = 12+22d$$

$$8d = -48 \implies d=-6$$

Step 5 — Compute the sum of the first 20 terms

$$S_{20} = \frac{20}{2}\big[2a+(20-1)d\big] = 10(2(3)+19(-6))$$

$$= 10(6-114) = 10(-108) = -1080$$

Answer

Common mistakes

• Misinterpreting “next four terms” as terms 1 through 8 combined, rather than specifically terms 5 through 8. This changes the entire setup and leads to a completely different (wrong) equation.
• Sign errors when solving for a negative common difference. Since $d=-6$ makes later terms significantly smaller (and eventually negative), it’s worth double-checking the final substitution into the sum formula carefully, as negative numbers are easy to mishandle in $2a+(n-1)d$.

Practise next

• The first term of an AP is $5$, and the sum of its first $3$ terms is $\tfrac15$ of the sum of the next $3$ terms. Find the sum of the first $15$ terms, using the same block-sum approach.