An arithmetic sequence ${{u}_{1}}$, ${{u}_{2}}$, ${{u}_{3}}$, … ${{u}_{n}}$, … is known to h

Question

An arithmetic sequence ${{u}_{1}}$, ${{u}_{2}}$, ${{u}_{3}}$, … ${{u}_{n}}$, … is known to have a common difference of $\ln 3$. The $r$ th term, , of another sequence ${{w}_{1}}$, ${{w}_{2}}$, ${{w}_{3}}$, … , is given by ${{w}_{r}}={{\mathrm{e}}^{-{{u}_{r}}}}$. Show that the sequence ${{w}_{1}}$, ${{w}_{2}}$, ${{w}_{3}}$, …, ${{w}_{n}}$ , …, is a geometric progression with common ratio $\frac{1}{3}$.[2] Explain why $\sum\limits_{r=1}^{\infty }{{{w}_{r}}}$ converges.[1] Given that ${{u}_{1}}=\ln 3$, find the smallest possible value of $n$ such that the sum of the first $n$ terms of the geometric progression given by ${{w}_{1}}$, ${{w}_{2}}$, ${{w}_{3}}$, … is within $0.5\%$ of the sum to infinity of the geometric series.[3]

Accepted Answer

(i) Since $\frac{{{w}_{n}}}{{{w}_{n-1}}}=\frac{1}{3}$ is a constant (independent of $n$), the sequence of terms given by ${{w}_{n}}$, $n\in {{\mathbb{Z}}^{+}}$ is a geometric progression with a common ratio of $\frac{1}{3}$. (ii) $\frac{{{w}_{n}}}{{{w}_{n-1}}}=\frac{1}{3}$ is the common ratio of the geometric progression (given). $\left| r \right|=\frac{1}{3}$ $\therefore \left| r \right|<1$ Hence, $\sum\limits_{r=1}^{\infty }{{{w}_{r}}}$ converges. (iii) Smallest possible value of $n=5$.

2023 SAJC Promo Q4

2023 SAJC Promo Q4

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