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Geometric Sequence
The Core Concept
A geometric sequence is a list of numbers where each term is found by multiplying the previous term by a constant called the common ratio ($r$).
$$a_n = a_{n-1} \cdot r$$
The recursive formula defines a term based on its immediate predecessor.
$a_n$=Current term(unitless)
$a_{n-1}$=Previous term(unitless)
$r$=Common ratio(multiplier)
$r > 1$
→The sequence grows indefinitely (Growth)
$0 < r < 1$
→The sequence shrinks toward zero (Decay)
$r < 0$
→The sequence alternates in sign
Common Ratio: Found by dividing any term by its predecessor ($r = a_n / a_{n-1}$).
Recursive Formula: Defines the relationship between adjacent terms.
Initial Term: The starting value ($a_1$) which determines the scale of the entire sequence.
Zero Terms Forbidden: No term of a G.P. can be zero, since $r = a_n / a_{n-1}$ is undefined when $a_{n-1} = 0$.
Sign Pattern: When $r < 0$, terms alternate sign — even-indexed powers of $r$ are positive, odd-indexed powers are negative.
Finding Any Term
The explicit formula allows you to calculate the $n^{th}$ term directly without knowing all previous values by applying the common ratio repeatedly.
$$a_n = a_1 r^{n-1}$$
Directly links term position to its value using powers of the ratio.
$a_n$=The nth term value(unitless)
$a_1$=First term(unitless)
$r$=Common ratio(unitless)
$n$=Position index(integer)
$n = 1$
→$a_1 r^0 = a_1$ (Consistent with initial term)
Exponent: The power is $n-1$ because the first term involves zero multiplications of $r$.
Position Index: Always a positive integer representing the term's place in line.
Finding $a_1$ and $r$ from Two Terms: Divide the later term by the earlier term to eliminate $a_1$: $\frac{a_m}{a_k} = r^{m-k}$, then solve for $r$.
Proportionality: $a_n \propto r^n$ — doubling $r$ raises every term exponentially, while doubling $a_1$ simply scales the entire sequence.
Summation of Terms
A finite geometric series is the sum of a specific number of terms in a sequence, calculated using a formula that accounts for the cumulative multiplicative effect.
$$S_n = \frac{a_1(1 - r^n)}{1 - r}$$
Calculates the total of the first $n$ terms efficiently.
$S_n$=Partial sum of n terms(unitless)
$a_1$=First term(unitless)
$r$=Common ratio(unitless)
$r = 1$
→$S_n = n \cdot a_1$ (Arithmetic-like sum)
Partial Sum: The total value up to a certain point $n$.
Alternative Form: When $|r| > 1$, use $S_n = \frac{a_1(r^n - 1)}{r - 1}$ to avoid negative numerator and denominator.
Derivation Trick: Multiply $S_n$ by $r$, then subtract: $S_n - rS_n = a_1 - a_1 r^n$. This telescoping cancellation is why the formula works.
Sigma Notation: Compact way to write sums using $\sum_{k=1}^{n} a_1 r^{k-1}$.
Infinite Sums & Convergence
If $|r| < 1$, an infinite geometric series will converge to a specific limit, meaning the sum of infinitely many terms is a finite number.
$$S_\infty = \frac{a_1}{1 - r}, \text{ for } |r| < 1$$
Determines the value an infinite series approaches as $n \to \infty$.
$S_\infty$=Sum to infinity(unitless)
$a_1$=Initial term(unitless)
$r$=Common ratio(unitless)
$|r| \ge 1$
→The series diverges (Sum approaches $\pm \infty$ or oscillates)
Convergence: The behavior of a series that approaches a finite limit.
Divergence: Occurs when terms do not shrink fast enough ($|r| \ge 1$) to have a finite sum.
Oscillatory Case: When $r = -1$, the partial sums alternate between $a_1$ and $0$ — the series neither converges nor diverges to infinity.
Recurring Decimals: Every repeating decimal is an infinite G.P. For example, $0.\overline{23} = \frac{23}{100} + \frac{23}{10000} + \cdots$ with $r = \frac{1}{100}$, giving $\frac{23}{99}$.
Derivation from Finite: $S_\infty$ is obtained from $S_n$ by noting $r^n \to 0$ as $n \to \infty$ when $|r| < 1$, so $1 - r^n \to 1$.
Geometric Mean
A number $G$ is the geometric mean between two numbers $a$ and $b$ if $a, G, b$ form a G.P. This gives $G^2 = ab$, so $G = \pm\sqrt{ab}$.
$$G = \pm\sqrt{ab}$$
The geometric mean is the square root of the product of two numbers, preserving the multiplicative relationship.
$G$=Geometric mean between $a$ and $b$(unitless)
$a, b$=The two given numbers(unitless)
$a = b$
→$G = \pm a$ — the geometric mean of equal numbers is the number itself
$a$ and $b$ have opposite signs
→$ab < 0$, so $\sqrt{ab}$ is imaginary — real G.M. does not exist
Derivation: If $a, G, b$ are in G.P., then $\frac{G}{a} = \frac{b}{G}$, which gives $G^2 = ab$.
Two G.P.s Result: $G = +\sqrt{ab}$ gives ratio $r > 0$; $G = -\sqrt{ab}$ gives ratio $r < 0$. Both are valid geometric sequences.
AM ≥ GM Inequality: For positive $a, b$: $\frac{a+b}{2} \ge \sqrt{ab}$, with equality only when $a = b$.
To insert $n$ geometric means between $a$ and $b$, we form a G.P. of $n+2$ terms: $a, G_1, G_2, \ldots, G_n, b$. The common ratio is found from $b = ar^{n+1}$.
$$r = \left(\frac{b}{a}\right)^{\frac{1}{n+1}}$$
The common ratio for inserting $n$ geometric means creates equally-spaced multiplicative steps between $a$ and $b$.
$r$=Common ratio of the resulting G.P.(unitless)
$n$=Number of geometric means to insert(integer)
$a, b$=The boundary values(unitless)
$n = 1$
→$r = \sqrt{b/a}$, recovering the single geometric mean formula
Total Terms: The sequence has $n + 2$ terms (including $a$ and $b$ at the ends).
$k$-th Mean: $G_k = a \cdot r^k = a\left(\frac{b}{a}\right)^{k/(n+1)}$ for $k = 1, 2, \ldots, n$.
Product Property: The product of all $n$ geometric means equals $(\sqrt{ab})^n$, and their geometric mean equals $\sqrt{ab}$.
Application Strategy
Worked Examples
Problem Statement
Find $a_n$ if $a_4 = \frac{8}{27}$ and $a_7 = \frac{-64}{729}$ in a G.P.
Term ratio to eliminate $a_1$Cube roots with negative numbers
Strategic Toolkit
$$\frac{a_m}{a_k} = r^{m-k}$$
Step 1
Divide Terms to Find $r$
Divide $a_7$ by $a_4$ to eliminate $a_1$: $\frac{a_7}{a_4} = r^{7-4} = r^3$.
$r^3 = \frac{-64/729}{8/27} = \frac{-64}{729} \times \frac{27}{8} = \frac{-8}{27}$
Step 2
Solve for $r$
Take the cube root of both sides.
$r = \sqrt[3]{\frac{-8}{27}} = -\frac{2}{3}$
Step 3
Find $a_1$
Substitute $r$ into $a_4 = a_1 r^3$.
$a_1 \left(-\frac{2}{3}\right)^3 = \frac{8}{27} \implies a_1 \cdot \frac{-8}{27} = \frac{8}{27} \implies a_1 = -1$
Step 4
Write the General Term
Substitute $a_1 = -1$ and $r = -\frac{2}{3}$ into $a_n = a_1 r^{n-1}$.
$a_n = (-1)\left(-\frac{2}{3}\right)^{n-1} = (-1)^n \left(\frac{2}{3}\right)^{n-1}$
Problem Statement
Insert three geometric means between 2 and $\frac{1}{2}$.
n Geometric MeansFinding common ratio from boundary values
Strategic Toolkit
$$r = \left(\frac{b}{a}\right)^{1/(n+1)}$$
Step 1
Identify Parameters
We need 3 means between $a = 2$ and $b = 1/2$, so the G.P. has $n + 2 = 5$ terms.
$a_1 = 2, \quad a_5 = \frac{1}{2}, \quad n = 3 \text{ means}$
Step 2
Find the Common Ratio
Use $a_5 = a_1 r^4$ to solve for $r$.
$\frac{1}{2} = 2r^4 \implies r^4 = \frac{1}{4} \implies r^2 = \frac{1}{2} \implies r = \pm\frac{1}{\sqrt{2}}$
Step 3
Calculate the Means (taking $r = \frac{1}{\sqrt{2}}$)
Multiply successively by $r$ starting from $a_1 = 2$.
$G_1 = \sqrt{2}, \quad G_2 = 1, \quad G_3 = \frac{1}{\sqrt{2}}$
Problem Statement
Convert $2.\overline{23}$ to a common fraction.
Recurring decimals as infinite G.P.Infinite Sum Formula
Strategic Toolkit
$$S_\infty = \frac{a_1}{1 - r}$$
Step 1
Express as Infinite G.P.
Separate the integer part and write the repeating block as a series.
$2.\overline{23} = 2 + (0.23 + 0.0023 + 0.000023 + \cdots)$
Step 2
Identify $a_1$ and $r$
The repeating part has $a_1 = \frac{23}{100}$ and $r = \frac{1}{100}$.
$a_1 = 0.23, \quad r = \frac{0.0023}{0.23} = \frac{1}{100}$
Step 3
Apply Infinite Sum
Sum the repeating part and add the integer part.
$2 + \frac{23/100}{1 - 1/100} = 2 + \frac{23/100}{99/100} = 2 + \frac{23}{99} = \frac{221}{99}$
Problem Statement
Find the sum of the first 8 terms of the sequence: 3, 6, 12, 24...
Common Ratio IdentificationFinite Sum Formula
Strategic Toolkit
$$S_n = \frac{a_1(1 - r^n)}{1 - r}$$
Step 1
Identify Key Variables
Extract the first term, ratio, and number of terms from the problem.
$a_1 = 3, r = \frac{6}{3} = 2, n = 8$
Step 2
Apply Finite Sum Formula
Substitute the identified values into the summation formula.
$S_8 = \frac{3(1 - 2^8)}{1 - 2}$
Step 3
Simplify and Calculate
Evaluate the power and perform the arithmetic operations.
$S_8 = \frac{3(1 - 256)}{-1} = \frac{3(-255)}{-1} = 765$
Problem Statement
Determine if the infinite series $10 + 5 + 2.5 + ...$ converges, and if so, find its sum.
Convergence TestInfinite Sum Formula
Strategic Toolkit
$$S_\infty = \frac{a_1}{1 - r}$$
Step 1
Test for Convergence
Calculate the common ratio $r$ and check if $|r| < 1$.
$a_1 = 10, a_2 = 5 \implies r = \frac{5}{10} = 0.5$. Since $|0.5| < 1$, it converges.
Step 2
Calculate the Infinite Sum
Use the convergence formula to find the limit of the sum.
$S_\infty = \frac{10}{1 - 0.5} = \frac{10}{0.5} = 20$
Memorize formulas
nth Term Formula
The term index $n$ is the power of the ratio, offset by 1.
Formula
$$a_n = a_1 r^{n-1}$$
Common Ratio
Always 'New divided by Old'.
Formula
$$r = \frac{a_n}{a_{n-1}}$$
Finite Sum Formula
The ratio power $n$ matches the number of terms being summed.
Formula
$$S_n = \frac{a_1(1 - r^n)}{1 - r}$$
Infinite Sum Formula
The denominator $1-r$ represents the 'gap' left by the decay.
Formula
$$S_\infty = \frac{a_1}{1 - r}$$
Geometric Mean
The multiplicative midpoint between two numbers.
Formula
$$G = \pm\sqrt{ab}$$
n Geometric Means Ratio
Common ratio when inserting $n$ means between $a$ and $b$.
Formula
$$r = \left(\frac{b}{a}\right)^{\frac{1}{n+1}}$$
Review
1
If $a_4 = \frac{8}{27}$ and $a_7 = \frac{-64}{729}$ in a G.P., what is $a_1$?
Ans
$-1$
Review
Tap to reveal
Find the common ratio from two separated terms
In a geometric progression, the ratio of terms three places apart equals $r^3$. Use the given $a_7$ and $a_4$ to determine the sign and magnitude of $r$.
$$r^3 = \frac{a_7}{a_4} = \frac{-64/729}{8/27} = -\frac{8}{27}, \qquad r = -\frac{2}{3}$$
Back-substitute to get the first term
Now use $a_4 = a_1 r^3$ and solve directly for $a_1$.
$$a_1\left(-\frac{8}{27}\right) = \frac{8}{27} \;\Rightarrow\; a_1 = -1$$
2
Find the 11th term of the sequence $1+i, 2, \frac{4}{1+i}, \ldots$
Ans
$\frac{1024}{(1+i)^9}$
Review
Tap to reveal
Identify the common ratio
Divide the second term by the first term to get the G.P. ratio, then simplify it into a cleaner complex form.
$$r = \frac{2}{1+i} = 1-i$$
Use the nth-term formula
Apply $a_n = a_1 r^{n-1}$ with $a_1 = 1+i$ and $n=11$, then use $(1-i)(1+i)=2$ to simplify the power.
$$a_{11} = (1+i)(1-i)^{10} = \frac{1024}{(1+i)^9}$$
3
A ball is dropped from 40 m and rebounds $\frac{3}{4}$ of the distance it falls. What total distance does it travel before coming to rest?
Ans
280 m
Review
Tap to reveal
Write the falling and rising distances as geometric sums
The ball first falls 40 m, then rises 30 m and each later rebound is multiplied by $\frac{3}{4}$. So the downward and upward parts are both geometric series.
$$\text{Downward} = 40 + 30 + 22.5 + \cdots, \qquad \text{Upward} = 30 + 22.5 + \cdots$$
Add the two geometric totals
Compute the infinite sums separately and combine them to get the full travel distance.
$$40 + \frac{30}{1-3/4} + \frac{30}{1-3/4} = 40 + 120 + 120 = 280\ \text{m}$$
4
If $\frac{a_5}{a_3} = \frac{4}{9}$ and $a_2 = \frac{4}{9}$ in a G.P., find $a_n$.
Ans
$\left(\frac{2}{3}\right)^n$
Review
Tap to reveal
Extract the common ratio from the term ratio
For a G.P., $\frac{a_5}{a_3} = r^2$. That gives the ratio directly.
$$r^2 = \frac{4}{9} \Rightarrow r = \frac{2}{3}$$
Use the second term to fix the first term
Since $a_2 = a_1 r$, solve for $a_1$ and then write the general term.
$$a_1 = \frac{4/9}{2/3} = \frac{2}{3}, \qquad a_n = a_1 r^{n-1} = \left(\frac{2}{3}\right)^n$$
5
The A.M. of two positive numbers is 10 and their positive G.M. is 8. Find the numbers.
Ans
4 and 16
Review
Tap to reveal
Convert the means into sum and product
The arithmetic mean gives the sum, and the geometric mean gives the product of the two positive numbers.
$$a+b=20, \qquad ab=64$$
Solve the quadratic with those values
Form the monic quadratic whose roots are the numbers and factor it.
$$t^2-20t+64=0 \Rightarrow (t-4)(t-16)=0$$
6
Three real geometric means are inserted between 2 and 162. What is the sum of all five terms?
Ans
242
Review
Tap to reveal
Use the first and fifth terms to get the ratio
Three inserted means create a 5-term G.P., so the last term is $a_1 r^4$.
$$2r^4 = 162 \Rightarrow r^4 = 81 \Rightarrow r = 3$$
List the five terms and add them
Once the ratio is known, the full sequence is immediate and the sum is straightforward.
$$2, 6, 18, 54, 162 \quad \Rightarrow \quad 2+6+18+54+162=242$$
7
The vulgar fraction equivalent to the recurring decimal $1.\overline{53}$ is:
Ans
$\frac{152}{99}$
Review
Tap to reveal
Convert the repeating decimal to an equation
Let the decimal be $x = 1.5353\ldots$. Multiplying by 100 shifts one full repeating block.
$$100x = 153.5353\ldots$$
Subtract to isolate the repeat
Subtract the original number so the recurring tail cancels, then solve for $x$.
$$99x = 152 \Rightarrow x = \frac{152}{99}$$
8
Three consecutive terms of a G.P. have sum 21 and product 64. The smallest term is:
Ans
1
Review
Tap to reveal
Use the product to find the middle term
For three consecutive G.P. terms $a/r, a, ar$, the product is $a^3$.
$$a^3 = 64 \Rightarrow a = 4$$
Use the sum to find the ratio
Substitute $a=4$ into the sum and solve the resulting quadratic for $r$.
$$\frac{4}{r}+4+4r=21 \Rightarrow 4r^2-17r+4=0 \Rightarrow r=4 \text{ or } \frac14$$
Write the terms and pick the smallest
The two possible orderings give the same set of terms, so the least value is clear.
$$1, 4, 16 \Rightarrow \text{smallest} = 1$$
9
The sum of an infinite G.P. is half the sum of the squares of its terms. If the sum of its first two terms is $\frac{9}{2}$, find $a_1$.
Ans
3
Review
Tap to reveal
Translate the condition into series formulas
Write the infinite sum and the sum of squares using the standard G.P. formulas.
$$S_1=\frac{a_1}{1-r}, \qquad S_2=\frac{a_1^2}{1-r^2}$$
Use the half-sum relation to connect $a_1$ and $r$
Set $S_1 = \frac12 S_2$ and simplify; the common factor $(1-r)$ cancels.
$$\frac{a_1}{1-r}=\frac{a_1^2}{2(1-r)(1+r)} \Rightarrow a_1=2(1+r)$$
Apply the first-two-terms condition
Since the first two terms sum to $\frac92$, substitute the expression for $a_1$ and solve for $r$, then recover $a_1$.
$$a_1(1+r)=\frac92 \Rightarrow 2(1+r)^2=\frac92 \Rightarrow r=\frac12, \; a_1=3$$
10
If $a, b, c, d$ are in G.P., which of the following is also in G.P.?
Ans
$a^2 - b^2, b^2 - c^2, c^2 - d^2$
Review
Tap to reveal
Parameterize the original G.P.
Write the terms as $a, ar, ar^2, ar^3$ so the pattern is explicit.
$$b=ar, \quad c=ar^2, \quad d=ar^3$$
Factor each candidate term
Each difference of squares picks up the same factor, leaving powers of $r^2$ between successive terms.
$$a^2-b^2=a^2(1-r^2),\ b^2-c^2=a^2r^2(1-r^2),\ c^2-d^2=a^2r^4(1-r^2)$$
11
If $y = 1 + 2x + 4x^2 + 8x^3 + \ldots$ and the series converges, what is $x$ in terms of $y$?
Ans
$\frac{y-1}{2y}$
Review
Tap to reveal
Recognize the geometric series
The coefficients double each time, so the common ratio is $2x$ and the first term is 1.
$$y = 1 + 2x + (2x)^2 + \cdots = \frac{1}{1-2x}$$
Solve the sum formula for $x$
Rearrange the equation for $y$ and isolate $x$ algebraically.
$$1-2x = \frac{1}{y} \Rightarrow x = \frac{y-1}{2y}$$
12
If $\frac{1}{a}, \frac{1}{b}, \frac{1}{c}$ are in G.P., the common ratio is:
Ans
$\pm\sqrt{\frac{a}{c}}$
Review
Tap to reveal
Use the geometric-mean property
For three consecutive G.P. terms, the middle term squared equals the product of the first and third terms.
$$\left(\frac{1}{b}\right)^2 = \frac{1}{a}\cdot\frac{1}{c} \Rightarrow b^2 = ac$$
Compute the common ratio directly
The ratio between the first two terms is $\frac{1/b}{1/a} = \frac{a}{b}$, and substituting $b = \pm\sqrt{ac}$ gives the final form.
$$r = \frac{a}{b} = \pm\sqrt{\frac{a}{c}}$$
Question 1 of 12
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1
If $a_4 = \frac{8}{27}$ and $a_7 = \frac{-64}{729}$ in a G.P., what is $a_1$?