CAT 2023 Slot 3 · Quant · Question 21

The first sequence runs from {a}_{1}=54 to {a}_{100}=846 in steps of 8.

The second runs from {b}_{1}=102 to {b}_{100}=498 in steps of 4.

A common term must appear in both lists, so it must lie in the overlap of the two ranges, which is from 102 to 498.

Every term of the first sequence leaves remainder 6 on division by 8, since 54=6\times 8+6.

Every term of the second leaves remainder 2 on division by 4, since 102=25\times 4+2.

A number that is 6 more than a multiple of 8 is automatically 2 more than a multiple of 4, so the first condition is the binding one.

So the common terms are the numbers of the form 8k+6 lying between 102 and 498.

The smallest is 102 itself, which appears as {a}_{7} and as {b}_{1}.

The largest is 494, since 498 leaves remainder 2 rather than 6; and 494={a}_{56}={b}_{99}.

The common terms therefore form an arithmetic progression from 102 to 494 with common difference 8.

The number of terms is \frac{494-102}{8}+1=\frac{392}{8}+1=50.

The sum is \frac{50\times \left( 102+494 \right)}{2}=25\times 596.

That is 14900 — the second option.

Shortcut. Common terms of two APs form a third AP whose step is the LCM of the two steps, here \text{LCM}\left( 8,4 \right)=8. Once the first common term 102 is spotted, the range is capped by the shorter sequence's last term 498, and the sum is a single average-times-count.