CAT 2023 Slot 3 QA Question 21

The question says: We have two sequences, a_n and b_n, defined as a_n = 46 + 8n and b_n = 98 + 4n, where n is a natural number less than or equal to 100. We need to find the sum of all terms that are common to both sequences. The options are A) 15000, B) 14900, C) 14602, D) 14798.

Hmm, okay. So, both sequences are arithmetic progressions, right? Because each term increases by a constant difference. Let me confirm that.

For a_n: The first term when n=1 is 46 + 8(1) = 54. Then, n=2: 46 + 16 = 62. So, the common difference is 8. Yep, that's an arithmetic progression with first term 54 and common difference 8.

For b_n: The first term when n=1 is 98 + 4(1) = 102. Then, n=2: 98 + 8 = 106. So, common difference is 4. That's also an arithmetic progression with first term 102 and common difference 4.

Now, we need to find the terms that are common to both sequences. That means we're looking for numbers that appear in both a_n and b_n. So, essentially, we're looking for the intersection of these two sequences.

when you have two arithmetic sequences, their common terms form another arithmetic sequence. The common difference of this new sequence is the least common multiple (LCM) of the differences of the original sequences. Let me recall the formula for that.

If the first sequence has a common difference of d1 and the second has d2, then the common difference of the common terms is LCM(d1, d2). But wait, actually, it's the LCM of d1 and d2 divided by their greatest common divisor (GCD). Hmm, no, wait, that's not right. Let me think again.

Actually, the common difference of the common terms is the LCM of d1 and d2. Because the terms have to satisfy both sequences' differences. So, for example, if one sequence increases by 8 and the other by 4, the common terms will increase by LCM(8,4) which is 8. Wait, but 4 divides 8, so LCM is 8. So, the common terms will form an arithmetic progression with difference 8.

But let me make sure. Let's denote the common terms as c_k. So, c_k must satisfy c_k = a_n = 46 + 8n and c_k = b_m = 98 + 4m for some natural numbers n and m.

So, setting them equal: 46 + 8n = 98 + 4m.

Let me rearrange this equation: 8n - 4m = 98 - 46 => 8n - 4m = 52.

Simplify this equation by dividing both sides by 4: 2n - m = 13.

So, 2n - m = 13. Therefore, m = 2n - 13.

Since m must be a natural number (as per the problem statement, n and m are natural numbers ≤100), m must be ≥1. So, 2n -13 ≥1 => 2n ≥14 => n ≥7.

Also, since m must be ≤100, 2n -13 ≤100 => 2n ≤113 => n ≤56.5. Since n is a natural number, n ≤56.

So, n can range from 7 to 56.

When n=7: m=2(7)-13=14-13=1, which is valid.

When n=56: m=2(56)-13=112-13=99, which is also valid.

So, n can take integer values from 7 to 56, inclusive. That gives us the number of common terms.

Calculate how many terms that is. From 7 to 56, the number of terms is 56 -7 +1 = 50 terms.

Now, we need to find the sum of these 50 terms. Since these terms form an arithmetic progression, we can use the formula for the sum of an AP: S = (number of terms)/2 * (first term + last term).

First, let's find the first common term. When n=7: a_7 =46 +8(7)=46+56=102.

Wait, that's interesting because b_1 is also 102. So, the first common term is 102.

Now, the last common term when n=56: a_56=46 +8(56)=46 +448=494.

Yep, that's correct.

So, the common terms form an AP starting at 102, ending at 494, with common difference 8 (since the LCM of 8 and 4 is 8). Wait, but earlier I thought the common difference would be 8, but let me confirm.

Wait, the common terms are the terms that are present in both sequences. Since a_n increases by 8 each time and b_n increases by 4, the common terms will increase by LCM(8,4)=8. So, yes, the common terms form an AP with d=8.

So, now, the number of terms is 50, first term is 102, last term is 494.

Sum S = (50/2)(102 + 494) =25(596)=25*596.

Compute 25596. 25600=15000, but 596 is 4 less than 600, so 25*(600-4)=15000 -100=14900.

So, the sum is 14900, which is option B.