CAT 2018 Question Paper Shift 2 Solved Paper

Question : 75 of 100

Marks: +1, -0

Let

t_1, t_2

,… be real numbers such that

t_1 + t_2

+ … +

t_n

2n^2

+ 9n + 13, for every positive integer n ≥ 2.If

t_k

= 103, then k equals

Solution:

t_{1}+t_{2}+. . .+t_{n}=2 n^{2}+9 n+13 →(1)

t_{1}+t_{2}+. . .+t n-1=2(n-1)^{2}+9(n-1)+13 →(2)

From (2) - (1), we get

t_{n}=(2 n^{2}+9 n+13)-(2(n-1)^{2}

+9(n-1)+13)=4 n+7

Given

t_{k}=103 ⇒ 4 k+7=103 ⇒ k=24

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