<p>A geometric series has first term $a$ and common ratio $r$, where $r<0$. The sum to infinity of the series is $18$ and its third term is $\frac{8}{3}$.</p>
<ol class="abc-list"><li><div class="list-item-paragraph">Show that $27{{r}^{3}}-27{{r}^{2}}+4=0$.</div><p style="text-align: right;">[3]</p></li><li><div class="list-item-paragraph">Find the values of $r$ and $a$.</div><p style="text-align: right;">[1]</p></li></ol>
<p>The $n$ th term of the series is ${{u}_{n}}$.</p>
<ol class="abc-list" start="3"><li><div class="list-item-paragraph">It is given that</div><div class="list-item-paragraph" style="text-align: center">$80\sum\limits_{r=k+1}^{\infty }{{{u}_{r}}}=\sum\limits_{r=1}^{k}{{{u}_{r}}}$.</div><div class="list-item-paragraph">Find the value of $k$.</div><p style="text-align: right;">[3]</p></li></ol>

A geometric series has first term $a$ and common ratio $r$, where $r<0$. The sum to infinity of the series is $18$ and its third term is $\frac{8}{3}$.

The $n$ th term of the series is ${{u}_{n}}$.

$80\sum\limits_{r=k+1}^{\infty }{{{u}_{r}}}=\sum\limits_{r=1}^{k}{{{u}_{r}}}$.

Techniques of Differentiation - General

Techniques of Differentiation

H2 Math

International Baccalaureate Higher Level Mathematics

IB HL Math

N2024 P1 Q5

Given: $S_\infty = 18$ and $\frac{S_3}{3} = ar^2$

From GP sum to infinity: $18 = \frac{a}{1-r}$

From GC: $r = -\frac{1}{3}$ or $\frac{2}{3}$

(Reject $r = \frac{2}{3}$ based on context)

$a = 18\left(1 - \left(-\frac{1}{3}\right)\right) = 18 \times \frac{4}{3} = 24$

$\therefore a = 24, \quad r = -\frac{1}{3}$

Given: $80\sum_{r=k+1}^{\infty} u_r = \sum_{r=1}^{k} u_r$

Note: $\sum_{r=k+1}^{\infty} u_r = S_\infty - S_k$

Using $S_k = \frac{a(1-r^k)}{1-r}$ with $a = 24, r = -\frac{1}{3}$:

$\frac{80}{81}(18) = \frac{24\left(1-\left(-\frac{1}{3}\right)^k\right)}{1-\left(-\frac{1}{3}\right)}$

$\frac{160}{9} = 18\left[1-\left(-\frac{1}{3}\right)^k\right]$

$\frac{80}{81} = 1 - \left(-\frac{1}{3}\right)^k$

$\left(-\frac{1}{3}\right)^k = \frac{1}{81}$

Taking logarithm: $k \ln\left(\frac{1}{3}\right) = \ln\left(\frac{1}{81}\right)$

$k = \frac{\ln\left(\frac{1}{81}\right)}{\ln\left(\frac{1}{3}\right)} = 4$

N2024 P1 Q5

N2024 P1 Q5

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