Finding Common Terms in Two Arithmetic Progressions

Q: Why is the common difference equal to LCM(d1, d2)?

A number x is common if it matches both modular congruences: x ≡ 3 pmod 4 and x ≡ 1 pmod 5. By the Chinese Remainder Theorem, the solutions are unique modulo lcm(4, 5) = 20, which means consecutive common terms must differ by exactly 20.

The 8^th common term of the series S_1 = 3 + 7 + 11 + 15 + 19 + … S_2 = 1 + 6 + 11 + 16 + 21 + … is _______.

Your answer stays private

What feels right?

No score. Commit to your first instinct. We’ll show what your mind noticed and what it missed.

Type the value - units or words beside it are fine.

✓ Source and academic review

Question type: Numerical
Exam relevance: JEE Main · Mathematics
Academic status: Answer verified against the official NTA key
Source: Previous-year question
Editorial review: 19 September 2026

Quick checks

Students also ask

Why is the common difference equal to LCM(d1, d2)?

A number $x$ is common if it matches both modular congruences: $x \equiv 3 (m o d 4)$ and $x \equiv 1 (m o d 5)$ . By the Chinese Remainder Theorem, the solutions are unique modulo $lcm (4, 5) = 20$ , which means consecutive common terms must differ by exactly 20.

Answer

Accepted answer: 151

The first common term is $11$ and the common difference is $lcm (4, 5) = 20$ , making the $8^{th}$ common term $11 + (8 - 1) \times 20 = 151$ .

Why each option works or fails

Step-by-step solution

Given: Two arithmetic progressions: $S_{1} = 3, 7, 11, 15, 19, \dots$ with first term $a_{1} = 3$ and common difference $d_{1} = 4$ . $S_{2} = 1, 6, 11, 16, 21, \dots$ with first term $a_{2} = 1$ and common difference $d_{2} = 5$ .
Goal: Find the $8^{th}$ common term of $S_{1}$ and $S_{2}$ .
Approach: The terms common to two APs form a new AP whose first term is the first common term, and whose common difference $D$ is the LCM of the common differences of the two APs, i.e., $D = lcm (d_{1}, d_{2})$ . Then use $T_{n} = A + (n - 1) D$ with $n = 8$ .
Execute: By inspecting both sequences, the first common term is $A = 11$ . The common difference of the AP of common terms is $D = lcm (4, 5) = 20$ . The $8^{th}$ common term is: $T_{8} = A + (8 - 1) D = 11 + 7 \times 20 = 11 + 140 = 151$
Verify: Check if 151 belongs to both APs: For $S_{1}$ : $\frac{151 - 3}{4} = \frac{148}{4} = 37$ , so $151$ is the $38^{th}$ term of $S_{1}$ . For $S_{2}$ : $\frac{151 - 1}{5} = \frac{150}{5} = 30$ , so $151$ is the $31^{st}$ term of $S_{2}$ . Both are integers, confirming 151 is in both series.

Finding Common Terms in Two Arithmetic Progressions

What feels right?

The first common term is $11$ and the common difference is $lcm (4, 5) = 20$ , making the $8^{th}$ common term $11 + (8 - 1) \times 20 = 151$ .

Why each option works or fails

Solution

01Given

02Goal

03Approach

04Execute

✓Verify

We think you should solve this next.

The $4^{th}$ term of GP is $500$ and its common ratio is…

The same chapter, from another paper — Sequences and Series, 2023

A vector $\vec{v}$ in the first octant is inclined to the x-axis at $60^{\circ}$ , to…

Students also ask

Finding Common Terms in Two Arithmetic Progressions

The first common term is 1111 and the common difference is lcm⁡(4,5)=20lcm(4,5)=20, making the 8th8th common term 11+(8−1)×20=15111+(8−1)×20=151.

01Given

02Goal

03Approach

04Execute

✓Verify

The 4th4th term of GP is 500500 and its common ratio is…

The same chapter, from another paper — Sequences and Series, 2023

A vector v⃗v in the first octant is inclined to the x-axis at 60∘60∘, to…

Students also ask

The first common term is $11$ and the common difference is $lcm (4, 5) = 20$ , making the $8^{th}$ common term $11 + (8 - 1) \times 20 = 151$ .

The $4^{th}$ term of GP is $500$ and its common ratio is…

A vector $\vec{v}$ in the first octant is inclined to the x-axis at $60^{\circ}$ , to…