JEE Mains Maths · Sequences and Series

Common Terms and Sub-Progressions

Terms shared by two progressions, and terms of one progression picked out by a divisibility condition, form a new progression: find its first term and step, then count and add.

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Why this matters

Eleven PYQs, eight of them numerical-answer questions with no options to check against. Each asks for a count or a sum of the terms two progressions share, or of the terms that pass a divisibility test. Two ideas cover the page.

Concept 1 of 2: Common terms of two progressions

The terms two APs share form an AP of their own. Its step is the LCM of the two common differences. Find its first term by listing a few terms of each. Its last term is the largest one not beyond either progression's last term. For an AP and a GP, list the GP's terms (there are few) and test each against the AP.

Definition

Step of the common AP: $D=\operatorname{lcm}(d_1,d_2)$ .
First common term: list terms of both until one repeats.
Last common term: the largest $T+kD$ not beyond either last term.
AP and GP: a GP term lies in the AP $a+kd$ exactly when it leaves remainder $a$ on division by $d$ .

Common difference of the shared terms

D=\operatorname{lcm}(d_1,d_2)

Worked example

Add the terms common to

2,7,12,\dots,197

and

3,7,11,\dots,199

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2024 · 1 February 2024 · Q84Moderate

Example 1 · Sequences and Series · Common Terms and Sub-Progressions

Let

3,7,11,15,\ldots.,403

and

2,5,8,11,\ldots,404

be two arithmetic progressions. Then the sum, of the common terms in them, is equal to

Start at the first COMMON term

The shared AP starts at the first term both lists contain, which is usually neither progression's first term.

Drill 7 more on common terms of two progressions

Concept 2 of 2: Terms picked out by a divisibility condition

In an AP, the terms divisible by a number

m

recur at a fixed interval, so they form a smaller AP. To add the terms that are NOT divisible, add everything and subtract that smaller AP. Numbers in a range with a fixed remainder on division by

m

are an AP with step

m

. For a condition such as coprime to 45, remove the multiples of 3 and of 5, then add back the multiples of 15.

Definition

Required sum $=$ (sum of all terms) $-$ (sum of the excluded sub-AP).
Fixed remainder $r$ on division by $m$ : an AP with step $m$ .
Two conditions: inclusion–exclusion, since multiples of both were removed twice.

Sum of the terms that pass

\text{sum}=S_{\text{all}}-S_{\text{excluded}}

Worked example

Add all three-digit numbers that leave remainder 1 on division by 7.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2023 · 10 April 2023 · Q83Moderate

Example 2 · Sequences and Series · Common Terms and Sub-Progressions

The sum of all those terms, of the arithmetic progression

3,8,13,\ldots\ldots.373

, which are not divisible by 3 , is equal to

Add the overlap back

Removing multiples of 3 and then multiples of 5 removes the multiples of 15 twice. Add their sum back once.

Drill 2 more on terms picked out by a divisibility condition

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (2)

Common terms of two progressions
Common difference of the shared terms
$D=\operatorname{lcm}(d_1,d_2)$
Terms picked out by a divisibility condition
Sum of the terms that pass
$\text{sum}=S_{\text{all}}-S_{\text{excluded}}$

Watch out for (2)

Start at the first COMMON term
→ Common terms of two progressions
Add the overlap back
→ Terms picked out by a divisibility condition

Test yourself on Sequences and Series

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