The ${{n}^{\mathrm{th}}}$ term of a sequence is ${{T}_{n}}=\ln \left( 3{{x}^{n-1}} \right)$ wher

Question

The ${{n}^{\mathrm{th}}}$ term of a sequence is ${{T}_{n}}=\ln \left( 3{{x}^{n-1}} \right)$ where $x$ is a positive constant not equal to $1$. Show that the sequence is an arithmetic progression for all positive integers $n$.[2] When $\ln 3$ is subtracted from the ${{19}^{\mathrm{th}}}$, ${{7}^{\mathrm{th}}}$ and ${{3}^{\mathrm{rd}}}$ terms of the arithmetic progression in part (i), these terms become the first three terms of a geometric progression. Find the common ratio of the geometric progression.[2] Find the range of values of $x$ for which the sum of the first $20$ terms of the arithmetic progression exceeds the sum to infinity of the geometric progression.[3]

Accepted Answer

(i) AP with common difference $\ln x$; (ii)(a) $r = \frac{1}{3}$; (ii)(b) $x > 3^{-\frac{20}{163}}$ or $x > 0.874$

2015 MI P1 Q4

2015 MI P1 Q4

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