Ratio of Two Patterned Series (α/β) – Sequences and Series JEE Main 2026 PYQ

Nishant Kumar Gupta

Former Faculty · Allen · Aakash · Narayana

Maths mentor behind PadhoLikhoJEE and Quantum Chemistry Classes (Arrah, Bihar). Teaches the “why” behind every JEE trap — not just the trick.

Quick Summary

Question Type:

Pattern summation (AP pairing technique)

Chapter:

Sequences and Series

Difficulty:

⭐⭐⭐ Medium

Time to Solve:

90 seconds

Key Concept:

Group the 40 terms into 20 consecutive pairs; each pair sum (3+5), (8+10), … is an AP with common difference 10 — α = 2060, β = 1030, so α/β = 2

Correct Answer:

Option (1) – 2

Why:

The 20 pair-sums 8, 18, …, 198 add to 2060 while β = 4 + 9 + … + 99 = 1030 — exactly half

The Question

JEE Main 2026 (8 April – Evening Shift) – Sequences and Series

Let α = 3 + 5 + 8 + 10 + 13 + 15 + … up to 40 terms, and
β = Σ_k=1²⁰ (5k − 1).
The value of (α / β) is:

Note: the printed series alternates gaps of 2 and 3
(3, 5, 8, 10, 13, 15, …), giving 20 pairs whose sums are 8, 18, 28, …

1. (1) 2 ✓
2. (2) 7/4
3. (3) 5/3
4. (4) 9/5

Quick Answer

Correct Option: (1) – 2

Reasoning: never add 40 terms one by one — pair them and read an AP:

• Pair consecutive terms: (3+5) = 8, (8+10) = 18, (13+15) = 28, … — the k-th pair is (5k − 2) + 5k = 10k − 2
• α = sum of 20 pair-sums = 20/2 · (8 + 198) = 10 · 206 = 2060
• β = 4 + 9 + 14 + … + 99 = 20/2 · (4 + 99) = 10 · 103 = 1030
• α/β = 2060/1030 = 2 ✓

Understanding the Concept

Why Pairing Works

A sequence built from two interleaved APs — here 3, 8, 13, … (difference 5) and 5, 10, 15, …
(difference 5) — is not itself an AP, so the standard n/2·(first + last) formula does not
apply directly. But consecutive pairs of the interleaved sequence, (3+5), (8+10),
(13+15), …, rise by a constant 10, because each pair collects one element of each sub-AP.
Pair sums of a double-interleaved AP are always an AP themselves, and an AP with 20 terms is
one formula away from its total. This pairing manoeuvre converts an ugly 40-term irregular
sum into a 20-term AP sum — the entire trick of the question.

Counting Terms Correctly

“Up to 40 terms” means 40 individual numbers, i.e. 20 pairs. Misreading it as 20 pairs (40
numbers being “20 terms”) halves the work and the answer. The general audit: the k-th odd-index
term is 5k − 2 (3, 8, …, 98 for k = 1..20) and the k-th even-index term is 5k (5, 10, …, 100),
so the 40th term is 100 — a quick terminal-value check that the pairing covered everything.

The Second Series Is a Plain AP

β = Σ (5k − 1) for k = 1..20 is the AP 4, 9, 14, …, 99: first term 4, common difference 5,
last term 5·20 − 1 = 99. Its sum is 20·(4 + 99)/2 = 1030. There is nothing hidden here — the
examiners place a transparent AP beside a disguised one to test whether you can see through
the interleaving. Note the pretty structural fact: β is exactly the average of the two
sub-series totals of α (1010 + 1050 over 2), a coincidence of the chosen pattern that makes
the final ratio land on the clean integer 2.

Sanity-Scaling Before Dividing

α ≈ 40 terms averaging about 50 gives ≈ 2000; β ≈ 20 terms averaging about 50 gives ≈ 1000;
so α/β should land near 2 — and 2060/1030 = 2 exactly. Estimating magnitudes before
the precise sum is a free error-catcher: if your computed α had come out as 200 or 20000, the
estimate would flag the slip instantly. Ratio questions reward this habit doubly, because a
mistake in either numerator or denominator survives into the quotient.

Detailed Step-by-Step Solution

Final Answer

Essential Facts for This Topic

Sum Toolkit for Patterned Series

1. AP sum: S = n/2 · (first + last); last term = a + (n − 1)d
2. Interleaved double-AP: pair consecutive terms — pair sums form an AP with difference d₁ + d₂ (here 5 + 5 = 10)
3. Σ (ak + b), k = 1..n: = a·n(n+1)/2 + bn — one line for β-style series
4. Term-count audit: n terms of a paired sequence = n/2 pairs; check the last term (here the 40th term is 100)
5. Magnitude sanity check: estimate ≈ (count) × (average) before trusting a computed total
6. Splitting trick: α = (odd-position sum) + (even-position sum) = 1010 + 1050 — a second route to 2060

Common Mistakes to Avoid

Key Concept Summary

What You Must Remember

1. Pairing converts interleaved APs into one AP: pair sums 8, 18, …, 198 with difference 10
2. α = 2060 via n/2·(first + last) on the 20 pair-sums — never add 40 terms individually
3. β = 4 + 9 + … + 99 = 1030 — a transparent AP: Σ(5k − 1) = 5·210 − 20
4. α/β = 2060/1030 = 2 exactly — the clean integer confirms the pairing was done right
5. Double-route check: odd-position total 1010 + even-position total 1050 = 2060 agrees
6. Estimate first: ≈2000/≈1000 predicted the ratio ≈ 2 before any exact sum

The Golden Rule for Interleaved Series

“When a series is two APs woven together, add consecutive pairs — the pair sums form a fresh AP and one formula finishes the job.”

Frequently Asked Questions

Q1: Why do consecutive pair sums form an AP here?

A: Each pair contains one term from each of the two interleaved APs, both with common difference 5. Moving from the k-th pair to the next shifts both components up by 5, so the pair sum rises by exactly 10 every time — the defining property of an AP with common difference 10.

Q2: What are the first and last pair sums?

A: The first pair is (3, 5) with sum 8. The k-th pair is (5k − 2, 5k), so the 20th pair is (98, 100) with sum 198. The 20 pair-sums 8, 18, …, 198 give α = 20·(8 + 198)/2 = 2060.

Q3: Is there a way to find α without pairing?

A: Yes — split by position. The odd-position terms 3, 8, 13, …, 98 form an AP of 20 terms summing to 1010, and the even-position terms 5, 10, 15, …, 100 sum to 1050. Adding the two subtotals gives α = 2060, agreeing with the pairing method.

Q4: How is β computed in one line?

A: β = Σ(5k − 1) for k = 1 to 20 = 5·(1 + 2 + … + 20) − 20 = 5·210 − 20 = 1050 − 20 = 1030. Equivalently it is the AP 4, 9, …, 99 summed by the n/2·(first + last) formula.

Q5: What fast check confirms the ratio 2 is right?

A: Magnitude estimation: α has 40 terms averaging roughly 51–52 (≈ 2060) and β has 20 terms averaging about 51.5 (≈ 1030), so the ratio must be near 2 — and the exact computation confirms it is exactly 2. An answer far from 2 would instantly signal a term-count or endpoint error.

Prerequisites to Solve This Question

Before attempting this problem, you should be comfortable with:

1. AP sum formula: S = n/2 (a + l) and l = a + (n − 1)d — used twice here
2. Pattern recognition: spotting two interleaved APs inside one displayed series
3. Σ notation fluency: expanding Σ(5k − 1) into first term, last term and count
4. Term-index bookkeeping: 40 terms = 20 pairs; k-th pair = (5k − 2, 5k)
5. Estimation habit: count × average as a pre-computation magnitude check

After Solving This, You Can:

• Sum any interleaved double-AP series by pairing, in under a minute
• Handle triple-interleaved patterns (three sub-APs) via grouping into blocks
• Solve “sum of terms in odd positions” and “even positions” variants directly
• Attack Σ(ak + b) and Σk(ak + b) one-liners without writing terms out
• Use magnitude estimation to catch term-count errors before finalising an option

Study Tips for This Topic

For JEE Main:

1. Count first, sum second: settle “how many terms/pairs” before touching the formula — it is the only real trap here
2. Keep both routes warm: pairing and odd/even splitting should give identical totals; use one to check the other
3. Budget 90 seconds: pair-sum AP (30s) + β one-liner (20s) + ratio (10s)

Common JEE Variants:

• “Find the last term” — here the 40th term is 100; tests the index bookkeeping directly
• Mixed gaps: series like 1 + 3 + 7 + 9 + 13 + 15 + … (gaps 2, 4) — same pairing, pair-sum difference 6
• Three-way interleave: 1 + 4 + 5 + 8 + 9 + 12 + … — group in threes
• “How many terms until the sum crosses N?” — invert the pair-AP sum formula
• Σk(5k − 1) weighted versions — bring in Σk² = n(n+1)(2n+1)/6

Difficulty Rating & Exam Frequency

Difficulty Level: ⭐⭐⭐ (3/5) – Medium (no formula beyond AP sums, but the interleaving and term-count demand real care)

JEE Main Frequency: High — patterned-series ratios and pair-sum tricks recur across recent JEE Main sessions

JEE Advanced Frequency: Low to Medium — Advanced prefers telescoping and AGP, but interleaved counting appears inside easier slots

Topic Importance: High — summation technique questions are reliable scorers once the pairing reflex is built

Written by Nishant Kumar Gupta
Former Faculty · Allen · Aakash · Narayana — Quantum Chemistry Classes, Arrah (Bihar)
Last Updated: 2 Sep 2026 ·
Question Source: JEE Main 2026 (8 April – Evening Shift) Previous Year Question ·
Topic: Sequences and Series – Interleaved AP Pairing and Ratio