CAT 2023 Slot 2 Solved Paper

Let the first term of both series be a1, and b1, respectively, and the common difference be d1, and d2, respectively.
It is given that a5=b9, which implies a1+4d1=b1+8d2
⇒a1−b1=8d2−4d1.....Eq(1)
Similarly, it is known that a_ {19}=b−{19}, which implies a1+18d1=b1+18d2
⇒a1−b1=18d2−18d1.....Eq(2)
Equating (1) and (2), we get:
‌⇒18d2−18d1=8d2−4d1
‌⇒10d2=14d1
‌⇒5d2=7d1
Since, d1,d2 are the prime numbers, which implies d1=5,d2=7.
It is also known that b2=0, which implies b1+d2=0⇒b1=−d2=−7
Putting the value of b1,d1, and, d2 in Eq(1), we get:

Hence, The correct option is B