Question

Let both the series $a_1, a_2, a_3, \dots$ and $b_1, b_2, b_3, \dots$ 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

D. Let $d_1$ and $d_2$ be the common differences of series $a_n$ and $b_n$ respectively, where $d_1, d_2$ are prime numbers. Given $b_2 = 0 \implies b_n = (n-2)d_2$. Thus, $b_9 = 7d_2$ and $b_{19} = 17d_2$. We are given $a_5 = b_9 = 7d_2$ and $a_{19} = b_{19} = 17d_2$. In series $a_n$: $$a_{19} - a_5 = 14d_1 \implies 17d_2 - 7d_2 = 14d_1 \implies 10d_2 = 14d_1 \implies 5d_2 = 7d_1$$ Since $d_1, d_2$ are prime numbers, $d_1 = 5$ and $d_2 = 7$. Now, $a_{11}$ is the midpoint of $a_5$ and $a_{19}$: $$a_{11} = \frac{a_5 + a_{19}}{2} = \frac{7d_2 + 17d_2}{2} = 12d_2 = 12 	imes 7 = 84$$

Term of arithmetic progression with prime common differences

Explanation

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