cat 2023 Question 20

Question

Let both the series $a_1, a_2, a_3, \ldots$ and $b_1, b_2, b_3 \ldots$ be in arithmetic progression such that the common differences of both the series are prime numbers. If $a_5=b_9, a_{19}=b_{19}$ and $b_2=0$ then $a_{11}$ equals

Accepted Answer

Let the first term of both series be $ a_1 $, and $ b_1 $, respectively, and the common difference be $ d_1 $, and $ d_2 $, respectively.

It is given that $ a_5 = b_9 $, which implies $ a_1 + 4d_1 = b_1 + 8d_2 $

=> $ a_1 - b_1 = 8d_2 - 4d_1 $ ..... Eq(1)

Similarly, it is known that $ a_{19} = b_{19} $, which implies $ a_1 + 18d_1 = b_1 + 18d_2 $

=> $ a_1 - b_1 = 18d_2 - 18d_1 $ ...... Eq(2)

Equating (1) and (2), we get:

=> $ 18d_2 - 18d_1 = 8d_2 - 4d_1 $

=> $ 10d_2 = 14d_1 $

=> $ 5d_2 = 7d_1 $

Since, $ d_1, d_2 $ are the prime numbers, which implies $ d_1 = 5, d_2 = 7 $.

It is also known that $ b_2 = 0 $, which implies $ b_1 + d_2 = 0 \Rightarrow b_1 = -d_2 = -7 $

Putting the value of $ b_1, d_1 $, and, $ d_2 $ in Eq(1), we get:

$ a_1 = 8d_2 - 4d_1 + b_1 = 56 - 20 - 7 = 29 $

Hence, $ a_{11} = a_1 + 10d_1 = 29 + 10 \cdot 5 = 29 + 50 = 79 $

The correct option is A

Question 20.

Question Explanation