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

• A. 79
• B. 83
• C. 84
• D. 86

Answer 1

Answer: (A)

79

Answer 2

Let the first term of both series be a1 and b1, respectively, and the common differences be d1 and d2, respectively.
It is given that $a_5 = b_9$, which implies

$a_1 + 4d_1 = b_1 + 8d_2:$
$a_1 - b_1 = 8d_2 - 4d_1$--------(1)
Similarly, it is known that $a_{19} = b_{19}$, which implies

$a_1 + 18d_1 = b_1 + 18d_2:$
$a_1 - b_1 = 18d_2 - 18d_1$----------(2)

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

$18d_2 - 18d_1 = 8d_2 - 4d_1$
$10d_2 = 14d_1$
$5d_2 = 7d_1$

Since $d_1$ and $d_2$ are prime numbers, this implies $d_1 = 5$ and $d_2 = 7$.

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

Putting the values 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,

$a_{11} = a_1 + 10d_1 = 29 + 10 \times 5 = 29 + 50 = 79$

Therefore, $a_{11} = 79$.