Question

Let \(a_n=46+8n\) and \(b_n=98+4n\) be two sequences for natural numbers \(n ≤ 100\) . Then, the sum of all terms common to both the sequences is

• A. 14900
• B. 15000
• C. 14798
• D. 14602

Answer 1

The Correct Option is A

To find the sum of all terms common to both sequences \( a_n = 46 + 8n \) and \( b_n = 98 + 4n \) for natural numbers \( n \leq 100 \), we need to solve for values where both sequences yield the same term.

The condition becomes: \[ 46 + 8n = 98 + 4m \] 

Simplifying: \[ 8n - 4m = 52 \quad \Rightarrow \quad 4n - 2m = 26 \quad \Rightarrow \quad 2n - m = 13 \quad \Rightarrow \quad m = 2n - 13 \]

Since \( 1 \leq n \leq 100 \), find valid \( n \) for which \( m = 2n - 13 \) is also a natural number: \[ 1 \leq 2n - 13 \leq 100 \quad \Rightarrow \quad 14 \leq 2n \leq 113 \quad \Rightarrow \quad 7 \leq n \leq 56 \]

So, valid \( n \) range: \( n = 7 \) to \( 56 \)

Common terms: \( a_n = 46 + 8n \), for \( n = 7 \) to \( 56 \)

First term: \( a_7 = 46 + 8 \cdot 7 = 102 \)

Last term: \( a_{56} = 46 + 8 \cdot 56 = 494 \)

Number of terms: \( 56 - 7 + 1 = 50 \)

Sum of arithmetic sequence: \[ S = \frac{50}{2}(102 + 494) = 25 \cdot 596 = 14900 \]

Therefore, the sum of all terms common to both sequences is \( \boxed{14900} \).