H2 Math Arithmetic and Geometric Progressions: Study Guide & Examples

JC1 · H2 Mathematics · Study guide

Use term positions, selected sums and contribution timing to set up AP/GP questions, with complete aligned working and checks.

Concepts · worked methods · graphs15 min read · practice linked below

What is the question asking you to find?

Start by distinguishing a term from a sum. A monthly deposit is one term; the total saved is a sum. Then identify the first term, the common difference or common ratio, and the number of terms.

When you are given…	Start with…
A formula for $S_n$ , but asked about the sequence	Find $T_n=S_n-S_{n-1}$
AP terms that form a GP	Write the specified terms and equate their ratios
Only even, odd or remaining terms	Write the first few selected terms
A sum close to $S_\infty$	Express the difference before using logs
Deposits, interest or withdrawals	Work through the first two months

Given the sum, find the term and prove AP

Recover the sequence from its sum

The sum of the first $n$ terms is $S_n=3n^2+2n$ . Show that the sequence is an AP and state its first term and common difference.

Method: subtract the sum of the first $n-1$ terms. For $n\geq2$ ,

\begin{aligned}T_n&=S_n-S_{n-1}\\&=3n^2+2n\\&\quad-\bigl[3(n-1)^2+2(n-1)\bigr]\\&=3n^2+2n-(3n^2-4n+1)\\&=6n-1.\end{aligned}

Check the first term separately: $T_1=S_1=5$ , which agrees with $6(1)-1$ . To prove AP, show that the difference between consecutive terms is constant:

\begin{aligned}T_{n+1}-T_n&=[6(n+1)-1]\\&\quad-(6n-1)\\&=6.\end{aligned}

Hence the sequence is an AP with first term 5 and common difference 6.

Check: the first three terms are $5,11,17$ . Their sum is 33, which agrees with $S_3$ . A few terms are a check; the general difference above is the proof.

AP terms form consecutive terms of a GP

Use the stated term positions

An AP has first term $a>0$ and non-zero common difference $d$ . Its 2nd, 6th and 8th terms are three consecutive terms of a GP. Find $d$ in terms of $a$ , and determine whether the geometric series is convergent.

The AP term formula is $T_n=a+(n-1)d$ . The three terms, in the stated order, are

a+d,\qquad a+5d,\qquad a+7d.

For consecutive non-zero GP terms, the ratios are equal:

\frac{a+5d}{a+d}=\frac{a+7d}{a+5d}.

\begin{aligned}(a+5d)^2&=(a+d)(a+7d)\\a^2+10ad+25d^2&=a^2+8ad+7d^2\\2ad+18d^2&=0\\2d(a+9d)&=0.\end{aligned}

Reject $d=0$ , since the common difference is non-zero. Thus $d=-a/9$ , and

\begin{aligned}r&=\frac{a+5d}{a+d}\\&=\frac{a-5a/9}{a-a/9}\\&=\frac{4a/9}{8a/9}\\&=\frac12.\end{aligned}

Since $|r|=1/2<1$ , the geometric series is convergent. Its first term is $a+d=8a/9>0$ .

Check: taking $a=90$ gives $d=-10$ . The specified terms are $80,40,20$ , with common ratio $1/2$ .

Variation: find the largest partial sum of the AP

For the same AP with $a=90$ and $d=-10$ ,

\begin{aligned}T_n&=90-10(n-1)\\&=100-10n.\end{aligned}

Terms 1 to 9 are positive, the 10th term is zero, and subsequent terms are negative. So the greatest partial sum occurs at both $n=9$ and $n=10$ :

\begin{aligned}S_9&=\frac92[2(90)+8(-10)]\\&=450,\\S_{10}&=S_9+T_{10}=450.\end{aligned}

Do not round a continuous maximum without checking the neighbouring integer values.

Even-numbered, odd-numbered and remaining terms

These refer to positions, not whether the values themselves are even or odd. Write the selected terms first: their first term and common difference or ratio may have changed.

Sum the first five odd-numbered GP terms

Use the GP $80,40,20,10,5,\ldots$ from above. Find the sum of its first five odd-numbered terms.

The selected terms are $T_1,T_3,T_5,T_7,T_9$ :

The new first term is 80, the new common ratio is $(1/2)^2=1/4$ , and there are five terms. Using the GP sum formula,

\begin{aligned}\text{Required sum}&=\frac{80[1-(1/4)^5]}{1-1/4}\\&=106.5625.\end{aligned}

Check: $T_9$ is the last selected term; it does not mean that there are nine terms in this sum. For the first five even-numbered terms, the first term would be 40, with the same ratio $1/4$ .

Variation: sum integers that are not divisible by a given number

Find the sum of the integers from 20 to 80 inclusive that are not divisible by 3.

Strategy: find the sum of all terms, then subtract the sum of the terms divisible by 3. Those terms are $21,24,\ldots,78$ .

\begin{aligned}21+(m-1)3&=78,\\m&=20.\end{aligned}

There are $80-20+1=61$ integers altogether. Therefore,

\begin{aligned}\text{Required sum}&=\frac{61}{2}(20+80)\\&\quad-\frac{20}{2}(21+78)\\&=3050-990\\&=2060.\end{aligned}

For any AP, the sum is $\frac{\text{number of terms}}2(\text{first term}+\text{last term})$ . Check both endpoints before counting the excluded terms.

How many terms give the required accuracy?

Within 0.1% of the sum to infinity

For the GP $80,40,20,\ldots$ , find the smallest $n$ such that the difference between $S_n$ and $S_\infty$ is less than 0.1% of $S_\infty$ .

Method: express the sum of all terms after the $n$ th term. The first of these is $T_{n+1}=80(1/2)^n$ , so

\begin{aligned}S_\infty&=\frac{80}{1-1/2}=160,\\S_\infty-S_n&=\frac{80(1/2)^n}{1-1/2}\\&=160(1/2)^n.\end{aligned}

The difference is positive here. Since $0.1\%=0.001$ ,

\begin{aligned}160(1/2)^n&<0.001(160),\\(1/2)^n&<0.001,\\n\ln(1/2)&<\ln0.001,\\n&>\frac{\ln0.001}{\ln(1/2)}\\&\approx9.966.\end{aligned}

The inequality reverses when dividing by $\ln(1/2)<0$ . Hence the smallest $n$ is 10.

Number of terms	Difference as a percentage of $S_\infty$	Less than 0.1%?
9	0.1953125%	No
10	0.09765625%	Yes

Important difference: “within 0.1” means an absolute difference of less than 0.1; it does not mean 0.1%. For the same GP, that would give

\begin{aligned}160(1/2)^n&<0.1,\\(1/2)^n&<\frac1{1600},\\n&>\frac{\ln(1/1600)}{\ln(1/2)}\\&\approx10.644.\end{aligned}

This time the smallest $n$ is 11. With a negative common ratio, use $|S_\infty-S_n|$ because the partial sums can lie on either side of the sum to infinity.

Contextual problems: deposits and interest

Compare two ways of saving over one year

One saver puts $250 aside on 1 January, then increases the amount by $10 on the first day of each subsequent month. This money earns no interest.

A second saver deposits $300 into a bank account on the first day of every month, starting on 1 January. Interest of 0.3% is added at the end of each month. Both start with no savings. Who has more on 31 December, just after the December interest is added?

No-interest savings: the monthly deposits form an AP with $a=250$ , $d=10$ , and 12 terms.

\begin{aligned}T_{12}&=250+11(10)=360,\\S_{12}&=\frac{12}{2}(250+360)\\&=3660.\end{aligned}

Bank account: let $q=1.003$ , the monthly growth factor. Work through the first two months. Let $B_n$ be the balance after month $n$ 's interest.

\begin{aligned}B_1&=300q,\\B_2&=(300q+300)q\\&=300q^2+300q.\end{aligned}

The January deposit earns 12 months of interest; the December deposit still earns one month. At the end of December,

\begin{aligned}B_{12}&=300q^{12}+300q^{11}+\cdots+300q\\&=300q+300q^2+\cdots+300q^{12}.\end{aligned}

In ascending powers of $q$ , this is a GP with first term $300q$ , ratio $q$ , and 12 terms:

\begin{aligned}B_{12}&=\frac{300q(q^{12}-1)}{q-1}\\&=\frac{300(1.003)(1.003^{12}-1)}{0.003}\\&\approx3670.98.\end{aligned}

The bank saver has $3670.98, compared with $3660: $10.98 more. The bank deposits total only $3600; the interest makes the difference.

Why does January have power 12?

Follow one $300 deposit to 31 December. Count the month-ends at which it earns interest.

The 1 January deposit earns interest at 12 month-ends, so its value on 31 December is $300 × 1.003¹² = $310.98.

Variation: money is withdrawn before interest is added

An account starts with $5000. The owner withdraws $300 at the beginning of each month, starting immediately, and the remaining money earns 0.3% interest at month-end. What is left after two months?

\begin{aligned}B_1&=(5000-300)(1.003)\\&=4714.10,\\B_2&=(4714.10-300)(1.003)\\&=4427.3423.\end{aligned}

So the balance is $4427.34. The two-month expression is

B_2=5000q^2-300q^2-300q.

The withdrawals are subtracted, and the first withdrawal affects two interest calculations. If a payment is made after interest, change the order of operations before deriving the pattern.

Before you finish

Have you found one term or a sum, as asked?
For selected terms, have you changed the first term, the difference or ratio, and the number of terms?
Before using a sum to infinity, have you shown $|r|<1$ ?
For a smallest integer, does the preceding integer fail the condition?
For money, does each power match the number of interest periods?

Continue with AP/GP practice questions and worked solutions