Term index in a sequence

CAT 2018 Slot 2 · Quantitative Ability · Easy · Sequences & Series

Let t1,t2,…t_1, t_2, \dots be real numbers such that t1+t2+⋯+tn=2n2+9n+13t_1 + t_2 + \dots + t_n = 2n^2 + 9n + 13, for every positive integer n≥2n \ge 2. If tk=103t_k = 103, then kk equals

Answer

24

Explanation

Let Sn=t1+t2+⋯+tn=2n2+9n+13S_n = t_1 + t_2 + \dots + t_n = 2n^2 + 9n + 13 for n≥2n \ge 2. For n≥3n \ge 3, tn=Sn−Sn−1t_n = S_n - S_{n-1}. tn=(2n2+9n+13)−(2(n−1)2+9(n−1)+13)t_n = (2n^2 + 9n + 13) - (2(n-1)^2 + 9(n-1) + 13) tn=2n2+9n+13−(2n2−4n+2+9n−9+13)t_n = 2n^2 + 9n + 13 - (2n^2 - 4n + 2 + 9n - 9 + 13) tn=4n+7t_n = 4n + 7

We are given tk=103t_k = 103: 4k+7=1034k + 7 = 103 4k=96  ⟹  k=244k = 96 \implies k = 24

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