For a sequence,if S_{n} = frac{5^{n} - 2^{n}} | vedclass

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(A) Given the sum of the first $n$ terms $S_{n} = \frac{5^{n} - 2^{n}}{2^{n}} = (\frac{5}{2})^{n} - 1$. The $n^{th}$ term $T_{n}$ is given by $T_{n} =

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$\frac{375}{16}$

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(A) Given the sum of the first $n$ terms $S_{n} = \frac{5^{n} - 2^{n}}{2^{n}} = (\frac{5}{2})^{n} - 1$. The $n^{th}$ term $T_{n}$ is given by $T_{n} = S_{n} - S_{n-1}$ for $n > 1$. For $n = 4$,$T_{4} = S_{4} - S_{3}$. $S_{4} = (\frac{5}{2})^{4} - 1 = \frac{625}{16} - 1 = \frac{625 - 16}{16} = \frac{609}{16}$. $S_{3} = (\frac{5}{2})^{3} - 1 = \frac{125}{8} - 1 = \frac{125 - 8}{8} = \frac{117}{8} = \frac{234}{16}$. $T_{4} = \frac{609}{16} - \frac{234}{16} = \frac{375}{16}$.

For a sequence,if $S_{n} = \frac{5^{n} - 2^{n}}{2^{n}}$,then its fourth term is

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