Let S₁ be an arithmetic progression comprising 120 terms with the first few terms being 4, 7, 10, 13, and S₂ be another arithmetic progression comprising 100 terms with the first few terms being 1, 6, 11, 16, . The sum of all the terms that

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Let S₁ be an arithmetic progression comprising 120 terms with the first few terms being 4, 7, 10, 13, and S₂ be another arithmetic progression comprising 100 terms with the first few terms being 1, 6, 11, 16, . The sum of all the terms that are common to both S₁ and S₂ is

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Correct answer: 4524

For the first arithmetic progression S₁ : First term a₁ = 4 , common difference d₁ = 3 , number of terms n₁ = 120 . The last term of S₁ is T₁₂₀ = 4 + (120 - 1)3 = 4 + 119 3 = 361 . For the second arithmetic progression S₂ : First term a₂ = 1 , common difference d₂ = 5 , number of terms n₂ = 100 . The last term of S₂ is T₁₀₀ = 1 + (100 - 1)5 = 1 + 99 5 = 496 . The common terms of S₁ and S₂ form a new arithmetic progression. Writing out the first few terms of S₁ : 4, 7, 10, 13, 16, 19, Writing out the first few terms

Let S₁ be an arithmetic progression comprising 120 terms with the first few terms being 4, 7, 10, 13, and S₂ be another arithmetic progression comprising 100 terms with the first few terms being 1, 6, 11, 16, . The sum of all the terms that are common to both S₁ and S₂ is

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