The terms of a geometric progression ${{u}_{1}},{{u}_{2}},{{u}_{3}},{{u}_{4}},...$ are such that the

Question

The terms of a geometric progression ${{u}_{1}},{{u}_{2}},{{u}_{3}},{{u}_{4}},...$ are such that the sum to infinity is $81$ and the sum of the first $4$ terms is $80$. If ${{u}_{1}}>100\,\,$and $\,\mathrm{n}\ge \mathrm{3}$, find ${{u}_{n}}$ in terms of $n$, [5] show that ${{u}_{3}}+{{u}_{4}}+{{u}_{5}}+...+{{u}_{n}}=\,p\left[ 1-{{\left( -\frac{1}{3} \right)}^{n+q}} \right]$, where the integers $p$ and $q$ are to be found. [3]

Accepted Answer

(i) $u_n = 108\left(-\frac{1}{3}\right)^{n-1}$ (ii) $9\left[1 - \left(-\frac{1}{3}\right)^{n-2}\right]$

2009 VJC J1 MYE Q10

2009 VJC J1 MYE Q10

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