Question

What is the sum of the first 100 terms which are common to both the progressions \(17, 21, 25, \ldots\) and \(16, 21, 26, \ldots\)?

• A. 100000
• B. 101100
• C. 111000
• D. 100110

Hint:

Common terms of two APs also form an AP, whose common difference is the LCM of the two original common differences.

Answer 1

The Correct Option is B

Step 1: Write the general term of each progression.
The first progression \(17, 21, 25, \ldots\) has first term \(17\) and common difference \(4\), so its \(n\)-th term is \(4n+13\).
The second progression \(16, 21, 26, \ldots\) has first term \(16\) and common difference \(5\), so its \(m\)-th term is \(5m+11\).

Step 2: Find the pattern of common terms.
Listing out a few terms of each: the first progression gives \(17, 21, 25, 29, 33, 37, 41, \ldots\) and the second gives \(16, 21, 26, 31, 36, 41, \ldots\). The number \(21\) appears in both, and so does \(41\). The common terms themselves form a new arithmetic progression.

Step 3: Find the common difference of the new progression.
Whenever a number belongs to both progressions, the next shared number appears after stepping forward by the least common multiple of the two common differences, since that is the smallest jump that keeps both progressions in sync. Here \(\text{lcm}(4,5) = 20\). So the common terms form their own arithmetic progression with common difference \(20\).

Step 4: Identify the first common term.
Scanning the lists above, the smallest number appearing in both is \(21\). So the progression of common terms is \(21, 41, 61, 81, 101, \ldots\), with first term \(a=21\) and common difference \(d=20\).

Step 5: Sum the first 100 terms of this new progression.
Using the arithmetic series sum formula \(S_n = \dfrac{n}{2}\big[2a+(n-1)d\big]\) with \(n=100\), \(a=21\), \(d=20\):
\[ S_{100} = \frac{100}{2}\big[2(21)+(99)(20)\big] \]
\[ S_{100} = 50\big[42+1980\big] \]
\[ S_{100} = 50 \times 2022 = 101100 \]

Final Answer:
The sum of the first 100 common terms is 101100. \[ \boxed{101100} \]