Two sequences {t_n} and {s_n} are defined by | vedclass

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(A) Given $t_n = \log \left( \frac{5^{n+1}}{3^{n-1}} \right) = \log(5^{n+1}) - \log(3^{n-1}) = (n+1)\log 5 - (n-1)\log 3 = n(\log 5 - \log 3) + (\log

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${t_n}$ is an $A.P.$,${s_n}$ is a $G.P.$

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(A) Given $t_n = \log \left( \frac{5^{n+1}}{3^{n-1}} \right) = \log(5^{n+1}) - \log(3^{n-1}) = (n+1)\log 5 - (n-1)\log 3 = n(\log 5 - \log 3) + (\log 5 + \log 3) = n \log(5/3) + \log 15$.Since $t_n$ is of the form $an + b$,it is an Arithmetic Progression $(A.P.)$ with common difference $d = \log(5/3)$.Given $s_n = [\log(5/3)]^n$. This is of the form $ar^{n-1}$ where $a = \log(5/3)$ and $r = \log(5/3)$.Since $s_n$ is of the form $ar^n$,it is a Geometric Progression $(G.P.)$ with common ratio $r = \log(5/3)$.Thus,${t_n}$ is an $A.P.$ and ${s_n}$ is a $G.P.$

Two sequences ${t_n}$ and ${s_n}$ are defined by $t_n = \log \left( \frac{5^{n+1}}{3^{n-1}} \right)$ and $s_n = \left[ \log \left( \frac{5}{3} \right) \right]^n$. Then:

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