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(C) The $n^{th}$ term of a sequence can be calculated using the formula $T_n = S_n - S_{n-1}$ for $n > 1$.Given $S_n = 4n^2 + 6n$.Then $S_{n-1} = 4(n-

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$122$

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(C) The $n^{th}$ term of a sequence can be calculated using the formula $T_n = S_n - S_{n-1}$ for $n > 1$.Given $S_n = 4n^2 + 6n$.Then $S_{n-1} = 4(n-1)^2 + 6(n-1) = 4(n^2 - 2n + 1) + 6n - 6 = 4n^2 - 8n + 4 + 6n - 6 = 4n^2 - 2n - 2$.Now,$T_n = (4n^2 + 6n) - (4n^2 - 2n - 2) = 4n^2 + 6n - 4n^2 + 2n + 2 = 8n + 2$.To find the $15^{th}$ term $(T_{15})$,substitute $n = 15$ into the expression for $T_n$.$T_{15} = 8(15) + 2 = 120 + 2 = 122$.

Consider a sequence whose sum of first $n$ terms is given by $S_n = 4n^2 + 6n$,where $n \in N$. Find the $15^{th}$ term $(T_{15})$ of this sequence.

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