CAT 2023 Slot 2 · Quant · Question 20

Write the two progressions as {a}_{n}={a}_{1}+\left( n-1 \right)d and {b}_{n}={b}_{1}+\left( n-1 \right)e, where d and e are prime.

The condition {a}_{5}={b}_{9} gives {a}_{1}+4d={b}_{1}+8e.

The condition {a}_{19}={b}_{19} gives {a}_{1}+18d={b}_{1}+18e.

Subtracting the first from the second, the starting terms cancel: 14d=10e.

Dividing by 2, 7d=5e.

Since 7 and 5 are coprime, d must be a multiple of 5 and e a multiple of 7.

Both are prime, so the only possibility is d=5 and e=7.

From {b}_{2}=0, {b}_{1}+e=0, so {b}_{1}=-7.

Substituting into {a}_{1}+18d={b}_{1}+18e: {a}_{1}+90=-7+126=119.

So {a}_{1}=29.

Then {a}_{11}={a}_{1}+10d=29+50=79 — the first option.

Shortcut. Subtracting the two equalities is what makes this quick: both unknown first terms vanish and 7d=5e remains. The primality condition then leaves exactly one pair, with no searching at all.