A Math

A Math (IGCSE)

E Math

H1 Math

H2 Math

IB AA HL Math

IB AA SL Math

IB AI HL Math

IB AI SL Math

IP Math

PSLE

<p>The sum of the first $n$ terms, where&nbsp;$n\geq1$, of a geometric series is given by \[S_n=\frac54\left[1-\left(-\frac13\right)^n\right].\]</p>
<ol class="abc-list"><li data-list-item-id="hci-q04-item01" data-list-item-path="hci-q04-item01"><div class="list-item-paragraph">Write down the value of the sum to infinity of the geometric series and find the least value of $n$ such that the sum of the first $n$ terms and the sum to infinity differ by less than $10^{-3}$.<span class="inline-marks">[4]</span></div></li><li data-list-item-id="hci-q04-item02" data-list-item-path="hci-q04-item02"><div class="list-item-paragraph">Given that&nbsp;$\displaystyle u_k=k\sum_{n=1}^{k}S_n$, find an expression for&nbsp;$u_k$ in terms of $k$ where&nbsp;$k\in\mathbb{Z}^+$.<span class="inline-marks">[4]</span></div></li></ol>

The sum of the first $n$ terms, where $n\geq1$, of a geometric series is given by \[S_n=\frac54\left[1-\left(-\frac13\right)^n\right].\]

Write down the value of the sum to infinity of the geometric series and find the least value of $n$ such that the sum of the first $n$ terms and the sum to infinity differ by less than $10^{-3}$.

Given that $\displaystyle u_k=k\sum_{n=1}^{k}S_n$, find an expression for $u_k$ in terms of $k$ where $k\in\mathbb{Z}^+$.

Sum to infinity

Sigma Notation

2026 HCI P1 Q4

Sequence and Series

<p>Kai and Lina play a sequence of rounds. Kai wins the first round with probability $\dfrac{3}{5}$. For every round after the first, the conditional probability that the winner of the preceding round also wins the next round is $\dfrac{2}{3}$. The match ends as soon as a player wins two consecutive rounds.</p>
<ol class="i-list"><li data-list-item-id="gp-q2-i" data-list-item-path="gp-q2-i"><div class="list-item-paragraph">The first four possible winning sequences for Kai are $KK$, $LKK$, $KLKK$ and $LKLKK$, where&nbsp;$K$ denotes a round won by Kai and $L$ denotes a round won by Lina. Find the probability of each sequence.<span class="inline-marks">[2]</span></div></li><li data-list-item-id="gp-q2-ii" data-list-item-path="gp-q2-ii"><div class="list-item-paragraph">Hence, using the sum to infinity of a geometric progression, find the probability that Kai wins the match.<span class="inline-marks">[4]</span></div></li><li data-list-item-id="gp-q2-iii" data-list-item-path="gp-q2-iii"><div class="list-item-paragraph">Given that Lina wins the first round, find the probability that Kai wins the match.<span class="inline-marks">[2]</span></div></li></ol>

Kai and Lina play a sequence of rounds. Kai wins the first round with probability $\dfrac{3}{5}$. For every round after the first, the conditional probability that the winner of the preceding round also wins the next round is $\dfrac{2}{3}$. The match ends as soon as a player wins two consecutive rounds.

Conditional Probability

Problems with Infinite Outcomes

Probability of a Sequence

Geometric Progression

Probability

<p>Ava and Ben take turns throwing a dart at a target. The probability that Ava hits the target on any throw is $\dfrac{2}{5}$, while the probability that Ben hits it on any throw is $\dfrac{1}{3}$. All throws are independent. A player wins as soon as that player hits the target. Ava throws first.</p>
<ol class="i-list"><li data-list-item-id="gp-q1-i" data-list-item-path="gp-q1-i"><div class="list-item-paragraph">Show that the probability that both players miss in one pair of consecutive throws, one by Ava followed by one by Ben, is $\dfrac{2}{5}$.<span class="inline-marks">[1]</span></div></li><li data-list-item-id="gp-q1-ii" data-list-item-path="gp-q1-ii"><div class="list-item-paragraph">Find the probability that Ava wins the game.<span class="inline-marks">[3]</span></div></li></ol>
<p>In a second game, a biased coin is used to choose who throws first. Ben is twice as likely as Ava to throw first. The same hit probabilities and rules apply.</p>
<ol class="i-list" start="3"><li data-list-item-id="gp-q1-iii" data-list-item-path="gp-q1-iii"><div class="list-item-paragraph">Find the probability that Ava wins the second game.<span class="inline-marks">[3]</span></div></li></ol>

Ava and Ben take turns throwing a dart at a target. The probability that Ava hits the target on any throw is $\dfrac{2}{5}$, while the probability that Ben hits it on any throw is $\dfrac{1}{3}$. All throws are independent. A player wins as soon as that player hits the target. Ava throws first.

In a second game, a biased coin is used to choose who throws first. Ben is twice as likely as Ava to throw first. The same hit probabilities and rules apply.

<p>An arithmetic series has positive first term $a$ and common difference $d$. The first, fifth and second terms of the arithmetic series are the first three consecutive terms of a geometric series respectively. It is also given that the terms of the arithmetic series are decreasing.</p>
<ol class="abc-list"><li><div class="list-item-content"><span class="list-item-text">Show that&nbsp;$d = -\frac{7}{16}a$.</span><span class="inline-marks">[3]</span></div></li><li><div class="list-item-content"><span class="list-item-text">Give a reason why the geometric series converges.</span><span class="inline-marks">[2]</span></div></li></ol>
<p>It is given that the sum to infinity of the geometric series is $\frac{32}{7}$.</p>
<ol class="abc-list" start="3"><li><div class="list-item-content"><span class="list-item-text">Find the sum of the first twenty terms of the arithmetic series.</span><span class="inline-marks">[2]</span></div></li><li><div class="list-item-content"><span class="list-item-text">Given that the sum of all the terms after the $n$th term of the geometric series is greater than $2$% of the sum of the first $n$ terms of the geometric series, find the possible values of $n$.</span><span class="inline-marks">[4]</span></div></li></ol>

An arithmetic series has positive first term $a$ and common difference $d$. The first, fifth and second terms of the arithmetic series are the first three consecutive terms of a geometric series respectively. It is also given that the terms of the arithmetic series are decreasing.

Give a reason why the geometric series converges.

It is given that the sum to infinity of the geometric series is $\frac{32}{7}$.

Find the sum of the first twenty terms of the arithmetic series.

Given that the sum of all the terms after the $n$th term of the geometric series is greater than 2% of the sum of the first $n$ terms of the geometric series, find the possible values of $n$.

Problems involving inequalities

Arithmetic Progression

Inter-sequence Relationships

2025 RI Promo Q11

<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 ${{u}_{n}}$.

Manipulating Summation Limits

N2024 P1 Q5

<ol class="abc-list"><li><div class="list-item-paragraph">The sum, ${{S}_{n}}$, of the first $n$ terms of an arithmetic progression is given by ${{S}_{n}}=pn+q{{n}^{2}}$.</div><div class="list-item-paragraph">Given also that ${{S}_{3}}=6$ and ${{S}_{5}}=11$,</div><ol class="i-list"><li><div class="list-item-paragraph">Find the values of $p$ and $q$.</div><p style="text-align: right;">[3]</p></li><li><div class="list-item-paragraph">Deduce, or find otherwise, an expression for the $n$-th term and the value of the common difference.</div><p style="text-align: right;">[3]</p></li></ol></li><li><div class="list-item-paragraph">Find the set of values of $\theta $ lying in the interval $-\frac{1}{2}\pi <\theta <\frac{1}{2}\pi $ such that the sum to infinity of the geometric series $1+\sin \theta +{{\sin }^{2}}\theta +\ldots $ is greater than $2$.</div><p style="text-align: right;">[4]</p></li></ol>

$u_n = S_n - S_{n-1} $

N1984 P1 Q1

<p>Dendrologists are specialised scientists who study trees and woody plants. Their work is diverse and can encompass various activities related to the identification, classification, biology, and ecology of trees. A group of dendrologists are studying the growth of $2$ species of trees, codenamed Tree Vee and Tree Jay.</p>
<p>In the 1st year, the height of Tree Vee and Tree Jay are both&nbsp;$H$ cm. In the 2nd year, Tree Vee’s height increases by $s$ cm and subsequently, the increase in height every year is $10\%$ less than the previous year’s increase.</p>
<ol class="abc-list"><li data-list-item-id="item-1760748353838-ob8qt2cy4" data-list-item-path="item-1760748353838-ob8qt2cy4"><div class="list-item-paragraph">Show that the height of Tree Vee in the 4th year is given by $\left( H+2.71s \right)$ cm.<span class="inline-marks">[1]</span></div></li><li data-list-item-id="item-1760748354498-ovilxc21a" data-list-item-path="item-1760748354498-ovilxc21a"><div class="list-item-paragraph">Show that the height of Tree Vee in the ${{n}^{\mathrm{th}}}$ year is given by $\left[ H+10s\left( 1-{{0.9}^{n-1}} \right) \right]$ cm.<span class="inline-marks">[3]</span></div></li><li data-list-item-id="item-1760748355053-ow04nd4vr" data-list-item-path="item-1760748355053-ow04nd4vr"><div class="list-item-paragraph">Hence, write down in terms of $H$ and $s$, the theoretical maximum height (in cm) of Tree Vee.<span class="inline-marks">[1]</span></div></li><li data-list-item-id="item-1760748355998-4ovla1o38" data-list-item-path="item-1760748355998-4ovla1o38"><div class="list-item-paragraph">In the 2nd year, Tree Jay’s height increases by $t$ cm and subsequently, the increase in height every year is $0.5$ cm less than the previous year’s increase. Show that the height of Tree Jay in the ${{10}^{\mathrm{th}}}$ year is given by $\left( H-18+9t \right)$ cm.<span class="inline-marks">[2]</span></div></li></ol>
<p>It is now given that&nbsp;$t=20$.</p>
<ol class="abc-list" start="5"><li data-list-item-id="item-1760748358571-xba0beb6d" data-list-item-path="item-1760748358571-xba0beb6d"><div class="list-item-paragraph">After the ${{10}^{\mathrm{th}}}$ year, Tree Jay’s height increases at a constant rate of $7$ cm per year. Express Tree Jay’s height (in cm) in the ${{n}^{\mathrm{th}}}$ year (where&nbsp;$n\ge 11$) in terms of $H$ and $n$.<span class="inline-marks">[2]</span></div></li><li data-list-item-id="item-1760748359973-dlg2kmjoa" data-list-item-path="item-1760748359973-dlg2kmjoa"><div class="list-item-paragraph">It is further given that&nbsp;$s=30$, and the 1st year is the year 2024. Find the years in which the heights of Tree Vee and Tree Jay are within 7 cm of each other, after 2034.<span class="inline-marks">[3]</span></div></li></ol>

Dendrologists are specialised scientists who study trees and woody plants. Their work is diverse and can encompass various activities related to the identification, classification, biology, and ecology of trees. A group of dendrologists are studying the growth of $2$ species of trees, codenamed Tree Vee and Tree Jay.

In the 1st year, the height of Tree Vee and Tree Jay are both $H$ cm. In the 2nd year, Tree Vee’s height increases by $s$ cm and subsequently, the increase in height every year is $10\%$ less than the previous year’s increase.

2024 VJC P1 Q11

<p>An arithmetic progression $A$ has first term $a$ and common difference $d$, where&nbsp;$a$ and $d$ are non-zero. A geometric progression $G$ has first term $b$ and common ratio $r$.</p>
<ol class="i-list"><li data-list-item-id="item-1760748272194-rl5l306mh" data-list-item-path="item-1760748272194-rl5l306mh"><div class="list-item-paragraph">The first, third and eleventh terms of $A$ are equal to the fourth, third and second term of $G$ respectively. Prove that the geometric series of $G$ is convergent.<span class="inline-marks">[4]</span></div></li><li data-list-item-id="item-1760748273236-n8oa1sewe" data-list-item-path="item-1760748273236-n8oa1sewe"><div class="list-item-paragraph">It is given instead that&nbsp;$r=\frac{1}{3}$ and the terms of another sequence $H$ is formed by squaring the terms of $G$. Find the range of values of $b$ such that the sum to infinity of $H$ exceeds the sum to infinity of $G$ by more than $\frac{3}{2}$.<span class="inline-marks">[3]</span></div></li></ol>

An arithmetic progression $A$ has first term $a$ and common difference $d$, where $a$ and $d$ are non-zero. A geometric progression $G$ has first term $b$ and common ratio $r$.

2024 TMJC P1 Q5

<ol class="abc-list"><li data-list-item-id="item-1760448834847-2mymvviyx" data-list-item-path="item-1760448834847-2mymvviyx"><div class="list-item-paragraph">Given that the sum of the first $n$ terms of a series is ${{S}_{n}}$, where&nbsp;${{S}_{n}}=3{{n}^{2}}-n$, state the value of its first term and write down an expression for&nbsp;${{S}_{n-1}}$ for&nbsp;$n\ge 2$. Hence show that the terms in the series form an arithmetic progression.<span class="inline-marks">[4]</span></div></li><li data-list-item-id="item-1760448835380-rw8oarkrn" data-list-item-path="item-1760448835380-rw8oarkrn"><div class="list-item-paragraph">The first term of an infinite geometric progression is $a$ and the common ratio is $r$, where&nbsp;$\left| r \right|&lt;1$. Given that the sum of all the terms after the $n$th term is equal to the $n$th term of the progression, show that&nbsp;$r=\frac{1}{2}$ and hence find the sum to infinity of the progression in terms of $a$.<span class="inline-marks">[4]</span></div></li><li data-list-item-id="item-1760448836093-vqo4pqusk" data-list-item-path="item-1760448836093-vqo4pqusk"><div class="list-item-paragraph">A bank has an account for investors. Compound interest is added to the account at the end of each year at a fixed rate of $3\%$ of the amount in the account at the beginning of that year.</div><div class="list-item-paragraph">Mr Ang decides to invest $\$2\mathrm{} \, 000$ on 1 January 2023 and a further $\$2\mathrm{} \, 000$ on 1 January of each subsequent year.</div><ol class="i-list"><li data-list-item-id="subitem-1760448859665-oa1uj8pcz" data-list-item-path="item-1760448836093-vqo4pqusk/subitem-1760448859665-oa1uj8pcz"><div class="list-item-paragraph">How much compound interest has his <strong>original</strong> $\$2\mathrm{} \, 000$ earned at the end of 31 December 2030?<span class="inline-marks">[2]</span></div></li><li data-list-item-id="subitem-1760448860528-gii4uhvv6" data-list-item-path="item-1760448836093-vqo4pqusk/subitem-1760448860528-gii4uhvv6"><div class="list-item-paragraph">By considering the total amount in the account at the end of the $n$ th year, find the year in which the total in the account will first exceed $\$40\mathrm{} \, 000$. Explain whether this occurs at the beginning or the end of the year.<span class="inline-marks">[6]</span></div></li></ol></li></ol>

Proving the progression

Financial Arithmetic or Iterative Pattern Recognition

2023 YIJC Promo Q12

<p>An arithmetic sequence ${{u}_{1}}$, ${{u}_{2}}$, ${{u}_{3}}$, … ${{u}_{n}}$, … is known to have a common difference of $\ln 3$.</p>
<p>The $r$ th term, , of another sequence ${{w}_{1}}$, ${{w}_{2}}$, ${{w}_{3}}$, … , is given by ${{w}_{r}}={{\mathrm{e}}^{-{{u}_{r}}}}$.</p>
<ol class="i-list"><li data-list-item-id="item-1760448781920-zpkvzqvbj" data-list-item-path="item-1760448781920-zpkvzqvbj"><div class="list-item-paragraph">Show that the sequence ${{w}_{1}}$, ${{w}_{2}}$, ${{w}_{3}}$, …, ${{w}_{n}}$ , …, is a geometric progression with common ratio $\frac{1}{3}$.<span class="inline-marks">[2]</span></div></li><li data-list-item-id="item-1760448782595-782cq85ly" data-list-item-path="item-1760448782595-782cq85ly"><div class="list-item-paragraph">Explain why $\sum\limits_{r=1}^{\infty }{{{w}_{r}}}$ converges.<span class="inline-marks">[1]</span></div></li><li data-list-item-id="item-1760448784545-qijxdxsai" data-list-item-path="item-1760448784545-qijxdxsai"><div class="list-item-paragraph">Given that&nbsp;${{u}_{1}}=\ln 3$, 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.<span class="inline-marks">[3]</span></div></li></ol>

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 ${{w}_{r}}={{\mathrm{e}}^{-{{u}_{r}}}}$.

2023 SAJC Promo Q4

<ol class="abc-list"><li data-list-item-id="item-1760448674318-sum5tm3vr" data-list-item-path="item-1760448674318-sum5tm3vr"><ol class="i-list"><li data-list-item-id="subitem-1760448675390-ndyz8magz" data-list-item-path="item-1760448674318-sum5tm3vr/subitem-1760448675390-ndyz8magz"><div class="list-item-paragraph">The sum of the first $n$  terms of a sequence ${{u}_{1}}$, ${{u}_{2}}$, ${{u}_{3}}$, … is given by ${{S}_{n}}={{\mathrm{e}}^{2}}{{n}^{2}}-3n$.</div><div class="list-item-paragraph">Show that&nbsp;${{u}_{n}}={{\mathrm{e}}^{2}}\left( 2n-1 \right)-3$.<span class="inline-marks">[2]</span></div></li><li data-list-item-id="subitem-1760448676380-m8f6zot5h" data-list-item-path="item-1760448674318-sum5tm3vr/subitem-1760448676380-m8f6zot5h"><div class="list-item-paragraph">Hence show that the sequence is an arithmetic progression.<span class="inline-marks">[2]</span></div></li><li data-list-item-id="subitem-1760448677265-vfnajhz3p" data-list-item-path="item-1760448674318-sum5tm3vr/subitem-1760448677265-vfnajhz3p"><div class="list-item-paragraph">State the first $3$ odd-numbered terms, ${{u}_{1}}$, ${{u}_{3}}$, and ${{u}_{5}}$. Hence find the least value of $n$ such that the sum of the first $n$ odd-numbered terms is more than $500$.<span class="inline-marks">[3]</span></div></li></ol></li><li data-list-item-id="item-1760448700028-hrb8cm3mg" data-list-item-path="item-1760448700028-hrb8cm3mg"><ol class="i-list"><li data-list-item-id="subitem-1760448701168-jt9wne8g9" data-list-item-path="item-1760448700028-hrb8cm3mg/subitem-1760448701168-jt9wne8g9"><div class="list-item-paragraph">Find the range of values of $x$ for which the following geometric series converges,</div><div class="list-item-paragraph" style="text-align: center">$\mathrm{e}-{{\mathrm{e}}^{\frac{1}{x}+1}}+{{\mathrm{e}}^{\frac{2}{x}+1}}-{{\mathrm{e}}^{\frac{3}{x}+1}}+...$<span class="inline-marks">[2]</span></div></li><li data-list-item-id="subitem-1760448702398-veol6qtyd" data-list-item-path="item-1760448700028-hrb8cm3mg/subitem-1760448702398-veol6qtyd"><div class="list-item-paragraph">Given that&nbsp;$x=-1$, find the sum of all the terms of this geometric series after, and including, the ${{8}^{\mathrm{th}}}$ term.<span class="inline-marks">[3]</span></div></li></ol></li></ol>

Block Sums

Subsequences of Progressions

2023 ACJC Promo Q8

<ol class="abc-list"><li data-list-item-id="item-1760418943489-zzk3vgf5z" data-list-item-path="item-1760418943489-zzk3vgf5z"><div class="list-item-paragraph">The sum of the first $n$ terms of a sequence is given by ${{S}_{n}}=\frac{{{n}^{2}}+5n}{8}$. Show that the sequence follows an arithmetic progression with common difference $d$, where&nbsp;$d$ is to be determined.</div><div class="list-item-paragraph">In a geometric progression, the first term is $100$ and its common ratio is $3d$. Find the smallest value of $k$ such that the sum of the first $k$ terms of the arithmetic progression is greater than the sum of the first $30$ terms of the geometric progression.<span class="inline-marks">[6]</span></div></li><li data-list-item-id="item-1760418944390-w7b93f6oq" data-list-item-path="item-1760418944390-w7b93f6oq"><div class="list-item-paragraph">The first and second terms of a geometric sequence are ${{u}_{1}}=a$ and ${{u}_{2}}={{a}^{2}}-a$. If all the terms of the sequence are positive, find the set of values of $a$ for which $\sum\limits_{r=1}^{\infty }{{{u}_{r}}}$ converges.<span class="inline-marks">[2]</span></div><div class="list-item-paragraph">For this sequence, it is known that the sum of all the terms after the ${{n}^{\mathrm{th}}}$ term is equal to the ${{n}^{\mathrm{th}}}$ term. Find the value of a and hence the value of $\sum\limits_{r=1}^{\infty }{{{u}_{r}}}$.<span class="inline-marks">[3]</span></div></li></ol>

2022 VJC P1 Q8

<ol class="abc-list"><li data-list-item-id="item-1760418785015-drvtt3thx" data-list-item-path="item-1760418785015-drvtt3thx"><div class="list-item-paragraph">An infinite geometric progression has first term $a$ and common ratio $r$, where&nbsp;$a$ and $r$ are non-zero. The sum of all the terms after the $n$th term of the progression is equal to twice the $n$th term. Show that the sum to infinity of the progression is three times the first term.<span class="inline-marks">[3]</span></div></li><li data-list-item-id="item-1760418785705-h0df4haho" data-list-item-path="item-1760418785705-h0df4haho"><div class="list-item-paragraph">The positive integers, starting at $~1$, are grouped into sets, as follows.</div><div class="list-item-paragraph" style="text-align: center">$\left\{ 1 \right\}$, $\left\{ 2,3 \right\}$, $\left\{ 4,5,6 \right\}$,…</div><div class="list-item-paragraph">where there are $r$ integers in the $r$th set.</div><ol class="i-list"><li data-list-item-id="subitem-1760418808813-46rddh60u" data-list-item-path="item-1760418785705-h0df4haho/subitem-1760418808813-46rddh60u"><div class="list-item-paragraph">Find, in terms of $r$, the first integer and the last integer in the $r$th set.<span class="inline-marks">[3]</span></div></li><li data-list-item-id="subitem-1760418810028-0vgiv0zq0" data-list-item-path="item-1760418785705-h0df4haho/subitem-1760418810028-0vgiv0zq0"><div class="list-item-paragraph">Prove that the sum of the integers in the $r$th set is $\frac{1}{2}r\left( 1+{{r}^{2}} \right)$.<span class="inline-marks">[2]</span></div></li></ol></li></ol>

Grouped sets

2023 YIJC P1 Q5

<p>An arithmetic series has first term $a$ and common difference $d$, where&nbsp;$a&gt;0$ and $d$ is non-zero. The first, eighth and thirteenth terms of the arithmetic series are equal to the first three terms of a geometric series respectively.</p>
<ol class="abc-list"><li><div class="list-item-paragraph">Show that&nbsp;$a=-\frac{49}{2}d$.<span class="inline-marks">[2]</span></div></li><li><div class="list-item-paragraph">Show that the geometric series is convergent and find its sum to infinity, $S$, in terms of $a$.<span class="inline-marks">[3]</span></div></li><li><div class="list-item-paragraph">Given that the sum of the first $n$ terms of the arithmetic series is $25$ times the third term of the geometric series, find the value of $n$.<span class="inline-marks">[3]</span></div></li><li><div class="list-item-paragraph">Find the least value of $k$ such that the sum of the first $k$ terms of the geometric series is more than $98\%$ of $S$.<span class="inline-marks">[3]</span></div></li></ol>

An arithmetic series has first term $a$ and common difference $d$, where $a>0$ and $d$ is non-zero. The first, eighth and thirteenth terms of the arithmetic series are equal to the first three terms of a geometric series respectively.

2022 YIJC Promo Q8

<p>The series $a+ar-a{{r}^{2}}-a{{r}^{3}}+a{{r}^{4}}+a{{r}^{5}}-a{{r}^{6}}-a{{r}^{7}}+$…, where $a&gt;0$, has its $k$th term, ${{T}_{k}}$, defined by ${{T}_{k}}=\left\{ \begin{matrix}
a{{r}^{k-1}}, \\
-a{{r}^{k-1}}, \\
\end{matrix} \right.$ $\begin{matrix}
\mathrm{if}\,k=4p-3\,\,\mathrm{or}\,\,4p-2 \\
\mathrm{if}\,k=4p-1\,\,\mathrm{or}\,\,4p\,\,\,\,\,\,\, \\
\end{matrix}$, for $p\in {{\mathbb{Z}}^{+}}$,</p>
<ol class="abc-list"><li><div class="list-item-paragraph">By rewriting the series as the sum of two geometric series, or otherwise, prove that the sum to $4n$ terms of the series is $\frac{a(1+r)(1-{{r}^{4n}})}{1+{{r}^{2}}}$.</div><p style="text-align: right;">[3]</p></li><li><div class="list-item-paragraph">State the set of values of $r$ for which the series has a sum to infinity.</div><p style="text-align: right;">[1]</p></li><li><div class="list-item-paragraph">Given that $r=-0.9$, find the least value of $n$ for the sum of $4n$ terms to be greater than $99\%$ of the sum to infinity.</div><p style="text-align: right;">[3]</p></li></ol>

The series $a+ar-a{{r}^{2}}-a{{r}^{3}}+a{{r}^{4}}+a{{r}^{5}}-a{{r}^{6}}-a{{r}^{7}}+$…, where $a&gt;0$, has its $k$th term, ${{T}_{k}}$, defined by ${{T}_{k}}=\left\{ \begin{matrix}
a{{r}^{k-1}}, \\
-a{{r}^{k-1}}, \\
\end{matrix} \right.$ $\begin{matrix}
\mathrm{if}\,k=4p-3\,\,\mathrm{or}\,\,4p-2 \\
\mathrm{if}\,k=4p-1\,\,\mathrm{or}\,\,4p\,\,\,\,\,\,\, \\
\end{matrix}$, for $p\in {{\mathbb{Z}}^{+}}$,

2023 NYJC MYE P1

<p>An arithmetic series has first term $a$ and common difference $d$, where $a$ and $d$ are non-zero. The first three terms of a geometric series are equal to the ninth, fifth and second terms respectively of the arithmetic series.</p>
<ol class="i-list"><li><div class="list-item-paragraph">Prove that the geometric series is convergent and find, in terms of $a$, the sum to infinity.</div><p style="text-align: right;">[5]</p></li><li><div class="list-item-paragraph">Given that $a>0$, find the largest value of $n$ for which the sum of the first $n$ terms of the geometric series is less than four fifths of the sum to infinity.</div><p style="text-align: right;">[3]</p></li></ol>

An arithmetic series has first term $a$ and common difference $d$, where $a$ and $d$ are non-zero. The first three terms of a geometric series are equal to the ninth, fifth and second terms respectively of the arithmetic series.

2011 IJC P1 Q7

<ol class="abc-list"><li><div class="list-item-paragraph">The sum of the first $n$ terms of a sequence, ${{S}_{n}}$ , is a quadratic polynomial in $n$. It is given that the first term and second term of the sequence is $50$ and $45$ respectively, and that the sum of the next $10$ terms is $175$. Find ${{S}_{n}}$ and hence show that this sequence is an arithmetic progression.<span class="inline-marks">[5]</span></div></li><li><div class="list-item-paragraph">At the end of 2015, the farmers of a chocolate factory had $40000$ square metres of fertile land for planting cocoa. At the end of 2016, the farmers realised that the area of fertile land had decreased by 800 square metres due to nutrient loss from the soil.</div><div class="list-item-paragraph">In each subsequent year, the decrease in the area of fertile land is $\frac{7}{8}$ of the decrease in the previous year.</div><ol class="i-list"><li><div class="list-item-paragraph">Show that the remaining area of fertile land at the end of the ${{n}^{\mathrm{th}}}$ year starting from 2016 (where 2016 is the 1st year, 2017 is the 2nd year, and so on) can be expressed as&nbsp;$a+b{{\left( \frac{7}{8} \right)}^{n}}$, where&nbsp;$a$ and $b$ are constants to be determined.<span class="inline-marks">[3]</span></div></li><li><div class="list-item-paragraph">After how many complete years, starting from 2016, will the area of fertile land first fall below $35000$ square metres? Show your working clearly.<span class="inline-marks">[3]</span></div></li><li><div class="list-item-paragraph">Find the area of fertile land available to the farmer in the long run.<span class="inline-marks">[1]</span></div></li></ol></li></ol>

2017 DHS Promo Q11

<p>The sum of the first $n$ terms of a series, ${{S}_{n}}$, is given by</p>
<p style="text-align: center">${{S}_{n}}=\frac{1}{a}\left[ 1-{{(a-1)}^{n}} \right]$, where $a$ is a constant and $a\ne 1$, $n\in {{\mathbb{Z}}^{+}}$</p>
<p>Obtain an expression for the ${{n}^{\mathrm{th}}}$ term of the series, ${{T}_{n}}$, and prove that ${{T}_{n}}$ is a geometric series.</p><p style="text-align: right;">[3]</p>
<p>If the sequence ${{T}_{n}}$ is now grouped as follows: $({{T}_{1}}),\ ({{T}_{2}},\ {{T}_{3}},\ {{T}_{4}}),\ ({{T}_{5}},\ {{T}_{6}},\ {{T}_{7}},\ {{T}_{8}},\ {{T}_{9}}),\ \ldots $ where each subsequent bracket has $2$ terms more than the previous bracket, find</p>
<ol class="i-list"><li><div class="list-item-paragraph">the total number of terms in the first $n$ brackets.</div><p style="text-align: right;">[2]</p></li><li><div class="list-item-paragraph">the middle term of the ${{11}^{\mathrm{th}}}$ bracket in terms of $a$.</div><p style="text-align: right;">[2]</p></li><li><div class="list-item-paragraph">the range of values of $a$ for the sum to infinity of the series to exist.</div><div class="list-item-paragraph">Hence, find the least value of $n$ for the sum of all the terms in the first $n$ brackets to be within $0.1\%$ of the sum to infinity of the series when $a=\frac{39}{20}$.</div><p style="text-align: right;">[4]</p></li></ol>

The sum of the first $n$ terms of a series, ${{S}_{n}}$, is given by

${{S}_{n}}=\frac{1}{a}\left[ 1-{{(a-1)}^{n}} \right]$, where $a$ is a constant and $a\ne 1$, $n\in {{\mathbb{Z}}^{+}}$

Obtain an expression for the ${{n}^{\mathrm{th}}}$ term of the series, ${{T}_{n}}$, and prove that ${{T}_{n}}$ is a geometric series.

If the sequence ${{T}_{n}}$ is now grouped as follows: $({{T}_{1}}),\ ({{T}_{2}},\ {{T}_{3}},\ {{T}_{4}}),\ ({{T}_{5}},\ {{T}_{6}},\ {{T}_{7}},\ {{T}_{8}},\ {{T}_{9}}),\ \ldots $ where each subsequent bracket has $2$ terms more than the previous bracket, find

2011 AJC P1 Q12

<ol class="abc-list" start="2"><li data-list-item-id="item-1760173813868-4qjn9vwrk" data-list-item-path="item-1760173813868-4qjn9vwrk"><div class="graph-block" data-graph-config="{"axis":false,"grid":false,"width":550,"height":143,"elements":[{"id":"01984bc9-5030-4f0d-a9cc-28d51b30d48b","type":"polygon","color":"#000000","points":[[0,0],[54,0],[27,46.765]],"fillColor":"none","pointStyle":"none","strokeWidth":1.5},{"id":"de8e0d2b-7cfc-410d-a1ee-3b06a4b7dfa6","type":"polygon","color":"#459F35","layer":2,"points":[[80,0],[134,0],[107,46.765]],"fillColor":"none","pointStyle":"none","strokeWidth":1.5},{"id":"2f1f6cab-180d-4ec2-a5df-56a7905d21fd","type":"polygon","color":"#000000","layer":3,"points":[[80,0],[98,0],[107,-15.588],[116,0],[134,0],[125,15.588],[134,31.177],[116,31.177],[107,46.765],[98,31.177],[80,31.177],[89,15.588]],"fillColor":"none","pointStyle":"none","strokeWidth":1.5},{"id":"53cc4127-a073-4978-b635-48985f08216f","type":"polygon","color":"#459F35","layer":2,"points":[[160,0],[178,0],[187,-15.588],[196,0],[214,0],[205,15.588],[214,31.177],[196,31.177],[187,46.765],[178,31.177],[160,31.177],[169,15.588]],"fillColor":"none","pointStyle":"none","strokeWidth":1.5},{"id":"aea8644f-c5b1-4ba8-8704-99cc3a32b639","type":"polygon","color":"#000000","layer":3,"points":[[160,0],[166,0],[169,-5.196],[172,0],[178,0],[181,-5.196],[178,-10.392],[184,-10.392],[187,-15.588],[190,-10.392],[196,-10.392],[193,-5.196],[196,0],[202,0],[205,-5.196],[208,0],[214,0],[211,5.196],[214,10.392],[208,10.392],[205,15.588],[208,20.785],[214,20.785],[211,25.981],[214,31.177],[208,31.177],[205,36.373],[202,31.177],[196,31.177],[193,36.373],[196,41.569],[190,41.569],[187,46.765],[184,41.569],[178,41.569],[181,36.373],[178,31.177],[172,31.177],[169,36.373],[166,31.177],[160,31.177],[163,25.981],[160,20.785],[166,20.785],[169,15.588],[166,10.392],[160,10.392],[163,5.196]],"fillColor":"none","pointStyle":"none","strokeWidth":1.5},{"id":"97de92b0-9101-4b85-a37d-3f2fe05d9444","type":"polygon","color":"#459F35","layer":2,"points":[[240,0],[246,0],[249,-5.196],[252,0],[258,0],[261,-5.196],[258,-10.392],[264,-10.392],[267,-15.588],[270,-10.392],[276,-10.392],[273,-5.196],[276,0],[282,0],[285,-5.196],[288,0],[294,0],[291,5.196],[294,10.392],[288,10.392],[285,15.588],[288,20.785],[294,20.785],[291,25.981],[294,31.177],[288,31.177],[285,36.373],[282,31.177],[276,31.177],[273,36.373],[276,41.569],[270,41.569],[267,46.765],[264,41.569],[258,41.569],[261,36.373],[258,31.177],[252,31.177],[249,36.373],[246,31.177],[240,31.177],[243,25.981],[240,20.785],[246,20.785],[249,15.588],[246,10.392],[240,10.392],[243,5.196]],"fillColor":"none","pointStyle":"none","strokeWidth":1.5},{"id":"cb0d9066-a704-4914-b74c-cc1731054067","type":"polygon","color":"#000000","layer":3,"points":[[240,0],[242,0],[243,-1.732],[244,0],[246,0],[247,-1.732],[246,-3.464],[248,-3.464],[249,-5.196],[250,-3.464],[252,-3.464],[251,-1.732],[252,0],[254,0],[255,-1.732],[256,0],[258,0],[259,-1.732],[258,-3.464],[260,-3.464],[261,-5.196],[260,-6.928],[258,-6.928],[259,-8.66],[258,-10.392],[260,-10.392],[261,-12.124],[262,-10.392],[264,-10.392],[265,-12.124],[264,-13.856],[266,-13.856],[267,-15.588],[268,-13.856],[270,-13.856],[269,-12.124],[270,-10.392],[272,-10.392],[273,-12.124],[274,-10.392],[276,-10.392],[275,-8.66],[276,-6.928],[274,-6.928],[273,-5.196],[274,-3.464],[276,-3.464],[275,-1.732],[276,0],[278,0],[279,-1.732],[280,0],[282,0],[283,-1.732],[282,-3.464],[284,-3.464],[285,-5.196],[286,-3.464],[288,-3.464],[287,-1.732],[288,0],[290,0],[291,-1.732],[292,0],[294,0],[293,1.732],[294,3.464],[292,3.464],[291,5.196],[292,6.928],[294,6.928],[293,8.66],[294,10.392],[292,10.392],[291,12.124],[290,10.392],[288,10.392],[287,12.124],[288,13.856],[286,13.856],[285,15.588],[286,17.321],[288,17.321],[287,19.053],[288,20.785],[290,20.785],[291,19.053],[292,20.785],[294,20.785],[293,22.517],[294,24.249],[292,24.249],[291,25.981],[292,27.713],[294,27.713],[293,29.445],[294,31.177],[292,31.177],[291,32.909],[290,31.177],[288,31.177],[287,32.909],[288,34.641],[286,34.641],[285,36.373],[284,34.641],[282,34.641],[283,32.909],[282,31.177],[280,31.177],[279,32.909],[278,31.177],[276,31.177],[275,32.909],[276,34.641],[274,34.641],[273,36.373],[274,38.105],[276,38.105],[275,39.837],[276,41.569],[274,41.569],[273,43.301],[272,41.569],[270,41.569],[269,43.301],[270,45.033],[268,45.033],[267,46.765],[266,45.033],[264,45.033],[265,43.301],[264,41.569],[262,41.569],[261,43.301],[260,41.569],[258,41.569],[259,39.837],[258,38.105],[260,38.105],[261,36.373],[260,34.641],[258,34.641],[259,32.909],[258,31.177],[256,31.177],[255,32.909],[254,31.177],[252,31.177],[251,32.909],[252,34.641],[250,34.641],[249,36.373],[248,34.641],[246,34.641],[247,32.909],[246,31.177],[244,31.177],[243,32.909],[242,31.177],[240,31.177],[241,29.445],[240,27.713],[242,27.713],[243,25.981],[242,24.249],[240,24.249],[241,22.517],[240,20.785],[242,20.785],[243,19.053],[244,20.785],[246,20.785],[247,19.053],[246,17.321],[248,17.321],[249,15.588],[248,13.856],[246,13.856],[247,12.124],[246,10.392],[244,10.392],[243,12.124],[242,10.392],[240,10.392],[241,8.66],[240,6.928],[242,6.928],[243,5.196],[242,3.464],[240,3.464],[241,1.732]],"fillColor":"none","pointStyle":"none","strokeWidth":1.5},{"x":16.82,"y":-25,"id":"34e5a88f-607a-436c-8e3a-3557afb4f8af","type":"text","color":"#000000","label":"\\(\\mathbf{Fig.\\ 1}\\)"},{"x":98,"y":-25,"id":"40f4a6b7-36d5-40f6-986b-b68e7232e9ee","type":"text","color":"#000000","label":"\\(\\mathbf{Fig.\\ 2}\\)"},{"x":178,"y":-25,"id":"0a0ed51b-fe51-445d-9c98-fcbcecfc28f7","type":"text","color":"#000000","label":"\\(\\mathbf{Fig.\\ 3}\\)"},{"x":258,"y":-25,"id":"88b78f92-94de-44aa-be22-3063042dfd4c","type":"text","color":"#000000","label":"\\(\\mathbf{Fig.\\ 4}\\)"}],"xAxisLabel":"\\(x\\)","yAxisLabel":"\\(y\\)","boundingBox":[-8.4076,50.5492,303.5889,-30.6323],"originLabel":"\\(O\\)","gridSettings":{"majorColor":"#94a3b8","majorStepX":1,"majorStepY":1,"minorColor":"#cbd5e1","majorOpacity":0.45,"minorOpacity":0.45,"minorElementsX":4,"minorElementsY":4,"majorStrokeWidth":0.8,"minorStrokeWidth":0.5,"showMajorGridLines":true,"showMinorGridLines":true},"showAxisLabels":false,"showNavigation":false,"keepAspectRatio":true,"xTickLabelEvery":1,"yTickLabelEvery":1,"xAxisLabelOffset":[2,-11],"yAxisLabelOffset":[-14,5],"originLabelOffset":[-5,-2]}" data-width="550" data-height="143" style="width:550px;height:143px;max-width:100%;display:block;margin:0 auto;"></div><div class="list-item-paragraph"><p>Snowflakes can be constructed by starting with an equilateral triangle (Fig. $1$), then repeatedly altering each line segment of the resulting polygon as follows:</p></div><div class="list-item-paragraph" style="margin-left: 3rem"><p>$1.$ Divide each outer line segments into three segments of equal length.</p></div><div class="list-item-paragraph" style="margin-left: 3rem"><p>$2.$ Add an equilateral triangle that has the middle segment from step $1$ as its base.</p></div><div class="list-item-paragraph" style="margin-left: 3rem"><p>$3.$ Remove the line segment that is the base of the triangle from step $2$.</p></div><div class="list-item-paragraph" style="margin-left: 3rem"><p>$4.$ Repeat the above steps for a number of iterations, $n$.</p></div><div class="list-item-paragraph"><p>The $1$st, $2$nd and $3$rd iteration produces the snowflakes in Fig. $2$, Fig. $3$ and Fig. $4$ respectively.</p></div><ol class="i-list"><li data-list-item-id="subitem-1760173916843-lcp2lyh2g" data-list-item-path="item-1760173813868-4qjn9vwrk/subitem-1760173916843-lcp2lyh2g"><div class="list-item-paragraph">If&nbsp;${{a}_{0}}$ denotes the area of the original triangle and the area of each new triangle added in the ${{n}^{th}}$ iteration is denoted by ${{a}_{n}}$, show that</div><div class="list-item-paragraph" style="text-align: center">${{a}_{n}}=\frac{1}{9}{{a}_{n-1}}$ for all positive integers $n$.<span class="inline-marks">[2]</span></div></li><li data-list-item-id="subitem-1760173934499-q0bb8y1qp" data-list-item-path="item-1760173813868-4qjn9vwrk/subitem-1760173934499-q0bb8y1qp"><div class="list-item-paragraph"><p>Write down the number of sides in the polygons in Fig.$1$, Fig. $2$ and Fig. $3$, and deduce, with clear explanations, that the number of new triangles added in the ${{n}^{th}}$ iteration is ${{T}_{n}}=k\left( {{4}^{n}} \right)$ where&nbsp;$k$ is a constant to be determined.</p><span class="inline-marks">[2]</span></div></li><li data-list-item-id="subitem-1760173941871-845h0bf3u" data-list-item-path="item-1760173813868-4qjn9vwrk/subitem-1760173941871-845h0bf3u"><div class="list-item-paragraph">Find the total area of triangles added in the ${{n}^{th}}$ iteration, ${{A}_{n}}$in terms of ${{a}_{0}}$ and $n$.<span class="inline-marks">[2]</span></div></li><li data-list-item-id="subitem-1760173953489-ffk11k6yl" data-list-item-path="item-1760173813868-4qjn9vwrk/subitem-1760173953489-ffk11k6yl"><div class="list-item-paragraph">Show that the total area of the snowflake produced after the ${{n}^{th}}$ iteration is ${{a}_{0}}\left[ \frac{8}{5}-\frac{3}{5}{{\left( \frac{4}{9} \right)}^{n}} \right]$ $\mathrm{unit}{{\mathrm{s}}^{2}}$.<span class="inline-marks">[3]</span></div><div class="list-item-paragraph">The snowflake formed when the above steps are followed indefinitely is called the Koch snowflake.</div></li><li data-list-item-id="subitem-1760174001548-7ootpzwbg" data-list-item-path="item-1760173813868-4qjn9vwrk/subitem-1760174001548-7ootpzwbg"><div class="list-item-paragraph"><p>Determine the least number of iterations needed for the area of the snowflake to exceed $99\%$ of the area of a Koch snowflake.</p><span class="inline-marks">[3]</span></div></li></ol></li></ol>

2019 ASRJC P1 Q11

<p>A shrimp farmer cultivates marine shrimps in tanks. There are $4000$ marine shrimps in a tank at the start of January $2022$. At the start of each month, the farmer will harvest $500$ marine shrimps for sale. At the end of each month, the number of marine shrimps in the tank will increase by $20%$ from the number at the start of the month after the shrimps are harvested for sale.</p>
<ol class="abc-list"><li><ol class="i-list"><li><div class="list-item-paragraph">Find the number of marine shrimps in the tank at the end of February $2022$.<span class="inline-marks">[2]</span></div></li><li><div class="list-item-paragraph">Show that the number of marine shrimps in the tank at the end of $n$ months after the start of January $2022$ is $1000{{\left( 1.2 \right)}^{n}}+3000$.<span class="inline-marks">[4]</span></div></li><li><div class="list-item-paragraph">The tank will be too congested for marine shrimps to grow well when there are more than $8000$ marine shrimps in the tank. Find when will the tank first become too congested for the marine shrimps to grow well.<span class="inline-marks">[2]</span></div></li></ol></li><li><div class="list-item-paragraph">Salt water is added to the tank at the start of each month to ensure that the marine shrimps are growing well. In January $2022$, $25$m$^{3}$ of salt water is added to the tank. In each subsequent month, the amount of salt water added to the tank is $r%$ of the amount of salt water added in the previous month.</div><ol class="i-list"><li><div class="list-item-paragraph">Assuming $r=55$, find the amount of salt water added to the tank on September $2022$.<span class="inline-marks">[2]</span></div></li><li><div class="list-item-paragraph">The capacity of the tank is $60$m$^{3}$. Find the range of values of $r$ such that the theoretical maximum total amount of salt water added to the tank will never exceed the capacity of the tank.<span class="inline-marks">[3]</span></div></li></ol></li></ol>

A shrimp farmer cultivates marine shrimps in tanks. There are $4000$ marine shrimps in a tank at the start of January $2022$. At the start of each month, the farmer will harvest $500$ marine shrimps for sale. At the end of each month, the number of marine shrimps in the tank will increase by $20%$ from the number at the start of the month after the shrimps are harvested for sale.

Math Question Bank