Discussion

Explanation:

Let both the series a₁, a₂, a₃, ... and b₁, b₂, b₃, ... be in arithmetic progression such that the common differences of both the series are prime numbers. If a₅ = b₉, a₁₉ = b₁₉ and b₂ = 0, then a₁₁ equal?

Let the common difference of a_n and b_n be p and q respectively, where both p and q are prime numbers.

Given,
a₅ = b₉,   ...(1)
a₁₉ = b₁₉   ...(2)

(2) - (1)
⇒ a₁₉ - a₅ = b₁₉ - b₉ 
⇒ 14p = 10q
⇒ p/q = 5/7

Since p and q are prime numbers, they must be equal to 5 and 7 respectively.

Now, b₂ = 0
⇒ b₉ = b₂ + 7q = 0 + 49 = 49

From (1):
a₅ = b₉ = 49

Now, a₁₁ = a₅ + 6p = 49 + 30 = 79

Hence, option (c).

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