Let S_n denote the sum of the first n terms of an arithmetic progression. If S₁₅ = 495 and S₃₀ = 1890 , then the sum of the even-indexed terms among the first 30 terms, i.e., _ k=1 ¹⁵ a_ 2k , is equal to :

Step-by-step solution

Let the first term be a and the common difference be d . Given S₁₅ = 495 15 2 [2a + 14d] = 495 15(a + 7d) = 495 a + 7d = 33 . Given S₃₀ = 1890 30 2 [2a + 29d] = 1890 15(2a + 29d) = 1890 2a + 29d = 126 . Multiplying the first equation by 2 gives 2a + 14d = 66 . Subtracting this from the second equation yields 15d = 60 d = 4 . Substituting d = 4 into a + 7d = 33 gives a + 28 = 33 a = 5 . The required sum is _ k=1 ¹⁵ a_ 2k = a₂ + a₄ + + a₃₀ . This is an arithmetic progression of 15 terms with the first term A = a₂ = a