For a given A.P.,S_{n} = 5n^{2} + 8n. Hence T | vedclass

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(B) Given the sum of the first $n$ terms of an $A.P.$ is $S_{n} = 5n^{2} + 8n$.We know that the $n^{th}$ term $T_{n}$ is given by the formula $T_{n} =

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$10n + 3$

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(B) Given the sum of the first $n$ terms of an $A.P.$ is $S_{n} = 5n^{2} + 8n$.We know that the $n^{th}$ term $T_{n}$ is given by the formula $T_{n} = S_{n} - S_{n-1}$.First,find $S_{n-1}$ by substituting $(n-1)$ for $n$ in the expression for $S_{n}$:$S_{n-1} = 5(n-1)^{2} + 8(n-1)$$S_{n-1} = 5(n^{2} - 2n + 1) + 8n - 8$$S_{n-1} = 5n^{2} - 10n + 5 + 8n - 8$$S_{n-1} = 5n^{2} - 2n - 3$Now,calculate $T_{n}$:$T_{n} = (5n^{2} + 8n) - (5n^{2} - 2n - 3)$$T_{n} = 5n^{2} + 8n - 5n^{2} + 2n + 3$$T_{n} = 10n + 3$.

For a given $A.P.$,$S_{n} = 5n^{2} + 8n$. Hence $T_{n} = \ldots \ldots \ldots \ldots$

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