CAT 2018 Set-2 | Question: 80

Question

Let $t_{1}, t_{2},\dots$ be a real numbers such that $t_{1}+t_{2}+\dots+t_{n}=2n^{2}+9n+13$, for every positive integers $n\geq2$.If $t_{k}=103$ , then $k$ equals

Answer 1

Given that,
Let, $ t_{1}, t_{2}, \dots$ be a real numbers.
And,

• $ t_{1} + t_{2}+ \dots + t_{n} = 2n^{2} + 9n + 13 \quad \longrightarrow (1) $
• $ t_{1} + t_{2} + \dots + t_{n-1} = 2(n-1)^{2} + 9(n-1) + 13 \quad \longrightarrow (2) $

From the equation $(1),$ subtract the equation $(2),$ we get

$ (t_{1} + t_{2} + \dots + t_{n-1} + t_{n}) – ( t_{1} + t_{2} + \dots + t_{n-1}) = 2n^{2} + 9n +13 – [ 2(n-1)^{2} + 9(n-1) +13] $

$ \Rightarrow t_{n} = 2n^{2} + 9n +13 – [ 2(n^{2}+ 1 -2n) +9n – 9 + 13] $

$ \Rightarrow t_{n} = \require{cancel} {\color{Red} {\cancel{2n^{2}}}} + {\color{Teal} {\cancel{9n}}} + {\color{Blue} {\cancel{13}}} – {\color{Red} {\cancel{2n^{2}}}}  – 2 + 4n  – {\color{Teal} {\cancel{9n}}} + 9 – {\color{Blue} {\cancel{13}}} $

$ \Rightarrow t_{n} = 4n + 7 \quad \longrightarrow (3) $

We have, $ t_{k} = 103 $

From the equation $(3),$ we get

$ t_{k} = 4k + 7 $

$ \Rightarrow 103 = 4k + 7 $

$ \Rightarrow 4k = 96 $

$ \Rightarrow \boxed {k = 24} $

Correct Answer $: 24 $