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

An arithmetic progression (AP) is a sequence where each term is found by adding a constant value (called the common difference) to the previous term.

For example: 3, 7, 11, 15, ... (common difference = 4)

The general formula for the nth term of an AP is: $a_n = a_1 + (n-1)d$ where $a_1$ is the first term and $d$ is the common difference.

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Let's define our two arithmetic progressions:

Series 1: $a_1, a_2, a_3, ...$ with common difference $q$ (where $q$ is prime)

Series 2: $b_1, b_2, b_3, ...$ with common difference $p$ (where $p$ is prime)

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Given: $b_2 = 0$

Using the AP formula: $b_2 = b_1 + (2-1)p = b_1 + p$

Since $b_2 = 0$:

$b_1 + p = 0$

$b_1 = -p$

So our b-series looks like: $-p, 0, p, 2p, 3p, ...$

Now we can find any term:

$b_9 = b_1 + (9-1)p = -p + 8p = 7p$

$b_{19} = b_1 + (19-1)p = -p + 18p = 17p$

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We're told:

$a_5 = b_9 = 7p$

$a_{19} = b_{19} = 17p$

For the a-series, using the AP formula:

$a_5 = a_1 + 4q = 7p$ ... (equation 1)

$a_{19} = a_1 + 18q = 17p$ ... (equation 2)

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Subtracting equation 1 from equation 2:

$(a_1 + 18q) - (a_1 + 4q) = 17p - 7p$

The $a_1$ terms cancel out:

$18q - 4q = 10p$

$14q = 10p$

Dividing both sides by 2:

$7q = 5p$

This means: $\tfrac{q}{p} = \tfrac{5}{7}$

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Here's the crucial part: Since both $p$ and $q$ are prime numbers, and we need $7q = 5p$, the only way this equation can be satisfied is if:

$q = 5$ (prime)

$p = 7$ (prime)

Because if $7q = 5p$, then $q$ must be divisible by 5 and $p$ must be divisible by 7. Since $p$ and $q$ are primes (only divisible by 1 and themselves), we must have $q = 5$ and $p = 7$.

Let's verify: $7(5) = 5(7)$

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Now we can substitute $q = 5$ into equation 1:

$a_1 + 4(5) = 7(7)$

$a_1 + 20 = 49$

$a_1 = 29$

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Finally, we can find $a_{11}$:

$a_{11} = a_1 + (11-1)q$

$a_{11} = 29 + 10(5)$

$a_{11} = 29 + 50$

$a_{11} = 79$

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Therefore, $a_{11} = 79$.