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Question 5 :Let t₁, t₂,.....be real numbers such that t₁+ t₂ +... + t_n = 2n² + 9n + 13, for every positive integer n ≥ 2. If t_k=103, then k equals[TITA]

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Video Explanation

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Explanatory Answer

Method of solving this CAT Question from Sequence and Series

Let t₁, t₂,.....be real numbers such that t₁+ t₂ +... + t_n = 2n² + 9n + 13
If t_k = 103 we have to find the value of k.
Let us assume this as sum to n terms i.e.
S_n = t₁+ t₂ +... + t_n = 2n² + 9n + 13
Given an expression in terms of n we can also take sum to n-1 terms i.e.
S_n-1 = t₁+ t₂ +... + t_n-1 = 2(n-1)² + 9(n-1) +13
This S_n and S_n-1 can be subtracted one from the another such that,
S_n = t₁+ t₂ +... + t_n = 2n² + 9n + 13
S_n-1 = t₁+ t₂ +... + t_n-1 = 2(n-1)² + 9(n-1) +13
S_n - S_n-1 = t_n = 2n² + 9n + 13 - 2(n-1)² - 9(n-1) - 13
S_n - S_n-1 = t_n = 2n² + 9n + 13 - 2(n-1)² - 9n + 9 - 13
S_n - S_n-1 = t_n = 2[n² - (n-1)²] + 9
S_n - S_n-1 = t_n = 2[n² - n² + 2n - 1] + 9

t_n = 2[2n - 1] + 9
t_n = 4n - 2 + 9
t_n = 4n + 7
So t_k = 4k + 7 = 103
k = \\frac{103 - 7}{4})
k = 24
The 24^th term is 103

The question is "Let t₁, t₂,.....be real numbers such that t₁+ t₂ +... + t_n = 2n² + 9n + 13, for every positive integer n ≥ 2. If t_k=103, then k equals [TITA] "

Hence, the answer is 24