CAT 2023 Slot 2 QA Question & Solution
Question
Let both the series $a_{1},a_{2},a_{3}$... and $b_{1},b_{2},b_{3}$... 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
Options
Solution
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 $b_1+d_2=0\ =>\ b_1=-d_2\ =\ -7$
Putting the value of $b_1,d_1,\ \text{and, }d_2$ in Eq(1), we get:
$a_1=8d_2-4d_1+b_1=56-20-7\ =\ 29$
Hence, $a_{11}=a_1+10d_1\ =\ 29+10\cdot5\ =\ 29+50\ =\ 79$
The correct option is B