Exercise 4.3 — Question 2

Topic: Geometric Sequences and Series

SLO M-11-A-23: Solve problems by analyzing geometric sequences and series up to $n$ terms.

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Key Formulas

General term (nth term) of a geometric sequence: $a_n=a_1\cdot r^{n-1}$ where $a_1$ is the first term and $r$ is the common ratio.

Sum of first $n$ terms of a geometric series ($r\ne 1$): $S_n=\frac{a_1(1-r^n)}{1-r}=\frac{a_1(r^n-1)}{r-1}$

Sum when $r=1$: $S_n=n\cdot a_1$

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Worked Example (Typical Q-2 Type)

Problem: Find the sum of the geometric series: $3+6+12+24+\cdots$ up to 8 terms.

Solution:

Identify the parameters:

• First term: $a_1=3$
• Common ratio: $r=\frac{6}{3}=2$
• Number of terms: $n=8$

Apply the sum formula (since $r=2\ne 1$): $S_8=\frac{a_1(r^n-1)}{r-1}=\frac{3(2^8-1)}{2-1}$

$S_8=\frac{3(256-1)}{1}=3\times 255=765$

Answer: $S_8=765$

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Another Common Type: Finding Missing Terms

Problem: The first term of a geometric series is $5$, the common ratio is $3$, and the sum is $605$. Find the number of terms.

Solution:

Using $S_n=\frac{a_1(r^n-1)}{r-1}$:

$605=\frac{5(3^n-1)}{3-1}=\frac{5(3^n-1)}{2}$

$1210=5(3^n-1)$

$242=3^n-1$

$3^n=243=3^5$

$\therefore n=5$

Answer: There are $5$ terms.

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Key Points to Remember

• A sequence is geometric if the ratio between consecutive terms is constant: $r=\frac{a_{n+1}}{a_n}$
• If $r>1$: the series is increasing
• If $0<r<1$: the series is decreasing
• If $r<0$: the terms alternate in sign