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: Understanding the Concept.
A number that lies in both progressions must satisfy both general term formulas at once. Instead of listing terms and spotting a pattern, we can set the two general terms equal and study what that forces.

Step 2: Key Formula or Approach.
First progression: term $= 17 + 4(p-1) = 4p+13$ for $p = 1, 2, 3, \ldots$
Second progression: term $= 16 + 5(q-1) = 5q+11$ for $q = 1, 2, 3, \ldots$
A common term needs $4p+13 = 5q+11$, that is $4p+2 = 5q$, so $q = \dfrac{4p+2}{5}$. Since $q$ must be a whole number, $4p+2$ must divide evenly by $5$.

Step 3: Detailed Explanation.
Check $p=1,2,3,\ldots$ for when $4p+2$ is divisible by 5:
$p=1: 6$, not divisible.
$p=2: 10$, divisible by 5. This gives the term $4(2)+13=21$.
The pattern of valid $p$ repeats every 5 steps, since $4p+2 \pmod 5$ cycles with period 5. The next valid value is $p=7$, giving $4(7)+13=41$.
So valid $p$ values are $2, 7, 12, 17, \ldots$, an arithmetic sequence with common difference $5$ in $p$, which through $4p+13$ turns into a common difference of $4 \times 5 = 20$ in the actual term values. This matches the two step sizes $4$ and $5$ combining into $\text{lcm}(4,5)=20$, confirming the common terms form an AP with common difference 20 starting at 21.

Step 4: Final Answer.
The common-term sequence is $21, 41, 61, \ldots$ with first term $21$ and common difference $20$. Summing 100 terms:
\[ S_{100} = \frac{100}{2}\left[2(21) + 99(20)\right] = 50(42+1980) = 50 \times 2022 = 101100 \]
So the required sum is 101100, matching the pattern-based approach. \[ \boxed{101100} \]