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

The Correct Option is A

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, \(a_{11} = 79\).