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 equals:
Question:
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 equals:
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 equals:
Options
Answer: 79
Explanation:
Solution: Step 1: Use b₂ = 0 → b₁ + q = 0 → b₁ = −q Step 2: Use a₅ = b₉ → a₁ + 4p = b₁ + 8q → a₁ + 4p = −q + 8q → a₁ + 4p = 7q → a₁ = 7q − 4p Step 3: Use a₁₉ = b₁₉ → a₁ + 18p = b₁ + 18q → a₁ + 18p = −q + 18q → a₁ + 18p = 17q → a₁ = 17q − 18p Step 4: Equate a₁: 7q − 4p = 17q − 18p → 14p = 10q → 7p = 5q → p = 5, q = 7 Step 5: Find a₁ → a₁ = 7q − 4p = 49 − 20 = 29 Step 6: Find a₁₁ → a₁₁ = a₁ + 10p = 29 + 50 = 79 Answer: 79
Explanation:
Solution: Step 1: Use b₂ = 0 → b₁ + q = 0 → b₁ = −q Step 2: Use a₅ = b₉ → a₁ + 4p = b₁ + 8q → a₁ + 4p = −q + 8q → a₁ + 4p = 7q → a₁ = 7q − 4p Step 3: Use a₁₉ = b₁₉ → a₁ + 18p = b₁ + 18q → a₁ + 18p = −q + 18q → a₁ + 18p = 17q → a₁ = 17q − 18p Step 4: Equate a₁: 7q − 4p = 17q − 18p → 14p = 10q → 7p = 5q → p = 5, q = 7 Step 5: Find a₁ → a₁ = 7q − 4p = 49 − 20 = 29 Step 6: Find a₁₁ → a₁₁ = a₁ + 10p = 29 + 50 = 79 Answer: 79
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