The sum of the first n terms of an A.P. is gi | vedclass

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(C) The $n^{th}$ term of an $A.P.$ is given by the formula $a_{n} = S_{n} - S_{n-1}$.Given $S_{n} = \frac{3n^{2}}{2} + \frac{5n}{2}$.To find the $25^{

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$76$

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(C) The $n^{th}$ term of an $A.P.$ is given by the formula $a_{n} = S_{n} - S_{n-1}$.Given $S_{n} = \frac{3n^{2}}{2} + \frac{5n}{2}$.To find the $25^{th}$ term $(a_{25})$,we use $a_{25} = S_{25} - S_{24}$.First,calculate $S_{25} = \frac{3(25)^{2}}{2} + \frac{5(25)}{2} = \frac{3(625) + 125}{2} = \frac{1875 + 125}{2} = \frac{2000}{2} = 1000$.Next,calculate $S_{24} = \frac{3(24)^{2}}{2} + \frac{5(24)}{2} = \frac{3(576) + 120}{2} = \frac{1728 + 120}{2} = \frac{1848}{2} = 924$.Therefore,$a_{25} = 1000 - 924 = 76$.

The sum of the first $n$ terms of an $A.P.$ is given by $S_{n} = \frac{3n^{2}}{2} + \frac{5n}{2}$. Find the $25^{th}$ term of the $A.P.$

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