Consider two AP's $S_1$ & $S_2$ such that $S_1 : \{ \text{First term} = 1, \text{Common difference} = 5, \text{no. of terms} = 101 \}$ $S_2 : \{ \text{First term} = 9, \text{Common difference} = 7, \text{no. of terms} = 71 \}$ Find the number of common terms in both AP's.
Step 1: Understanding the Concept:
Common terms of two arithmetic progressions also form an arithmetic progression.
The common difference of the new AP is the LCM of the original common differences. Step 2: Key Formula or Approach:
Last term of \( S_1 = 1 + (101-1)5 = 501 \).
Last term of \( S_2 = 9 + (71-1)7 = 499 \).
New common difference \( D = \text{LCM}(5, 7) = 35 \). Step 3: Detailed Explanation:
First common term:
\( S_1 : 1, 6, 11, 16, 21 \dots \)
\( S_2 : 9, 16, 23 \dots \)
The first common term is \( a = 16 \).
Let there be \( n \) common terms. The \( n \)-th common term must be \( \le \min(501, 499) \).
\[ 16 + (n-1)35 \le 499 \]
\[ (n-1)35 \le 483 \implies n-1 \le \frac{483}{35} = 13.8 \]
\[ n \le 14.8 \implies n = 14 \]. Step 4: Final Answer:
The number of common terms is 14.
