Question 20.

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

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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

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