Sequences and Series | CBSE Class 11 Maths Notes

These Mathematics notes cover sequences and their terms, finite and infinite sequences, recurrence relations, series and sigma notation, geometric progressions, general terms and finite sums, geometric means, the relationship between arithmetic and geometric means, and applications of geometric series.

What is a sequence, and how are its terms identified?

Definition: A sequence is an arrangement of numbers in a definite order according to a rule. Each number in that arrangement is a term, and its position is part of the information describing the sequence.

Write the terms as $a_1,a_2,a_3,\ldots,a_n,\ldots$, where $a_n$ means the term in position $n$, and $n$ is a natural number. The subscript indicates a position; it is not a factor multiplying the letter.

How do finite and infinite sequences differ?

A finite sequence contains a fixed, finite number of terms. An infinite sequence does not end. These descriptions concern the number of terms, rather than how large or small the individual terms are.

For an ancestor-counting pattern over ten preceding generations, the terms are $2,4,8,16,32,\ldots,1024$. The generation is the position, and the number of ancestors in that generation is the term. Restricting the pattern to ten generations makes this sequence finite.

The successive decimal quotients obtained while dividing ten by three give $3,3.3,3.33,3.333,\ldots$. Continuing this process gives an infinite sequence. Its first term is $a_1=3$, while its sixth term is $a_6=3.33333$.

A sequence can also be regarded as a function whose domain is the natural numbers or an appropriate subset of them. The input gives the position and the output gives its term. The alternative notation $a(n)$ means the same term as $a_n$.

A progression is a sequence following a specific pattern. Recognising the rule is essential: the order of the terms determines which number is first, which comes next, and which term a requested position identifies.

How can a formula or recurrence relation generate a sequence?

A general term gives a term directly from its position. For the even natural numbers, the rule is $a_n=2n$. For the odd natural numbers, it is $a_n=2n-1$. Substituting a natural number for the position produces the corresponding term.

How is direct substitution carried out?

Worked example 1. Find the first three terms when $a_n=2n+5$.

Answer: Substitute each required position separately, keeping the same rule throughout.

1. For the first position, $a_1=2(1)+5=7$.
2. For the second position, $a_2=2(2)+5=9$.
3. For the third position, $a_3=2(3)+5=11$.

The required terms, in order, are $7,9,11$.

Worked example 2. Find the twentieth term when $a_n=(n-1)(2-n)(3+n)$.

Answer: Replace every occurrence of the position by the same value.

1. Substitute the required position: $a_{20}=(20-1)(2-20)(3+20)$.
2. Evaluate all three factors: $a_{20}=19(-18)(23)$.
3. Multiply the first two factors: $a_{20}=(-342)(23)$.
4. Complete the multiplication: $a_{20}=-7866$.

What information does a recurrence require?

A recurrence relation generates a later term from earlier terms. It needs starting information as well as the rule. Unlike direct substitution into a general-term formula, this method proceeds through the terms in their given order.

The Fibonacci sequence begins with two specified terms, $a_1=a_2=1$. Its recurrence is $a_n=a_{n-2}+a_{n-1}$ for $n>2$. Each later term therefore uses both of the immediately preceding terms.

1. Start with the given values: $a_1=1,\ a_2=1$.
2. Add the first two terms: $a_3=a_1+a_2=1+1=2$.
3. Use the next pair: $a_4=a_2+a_3=1+2=3$.
4. Continue in order: $a_5=a_3+a_4=2+3=5$.
5. Repeat the same rule: $a_6=a_4+a_5=3+5=8$.

Not every sequence needs an explicit algebraic formula. A verbal description or a theoretical rule may specify how its terms are generated. Distinguish the rule defining a sequence from a pattern guessed from a few displayed terms.

What is a series, and how does sigma notation represent it?

Definition: The series associated with a sequence is the expression obtained by adding its terms in order. It is finite or infinite according as the associated sequence is finite or infinite.

For a sequence with terms $a_1,a_2,\ldots,a_n$, the associated finite series is $a_1+a_2+\cdots+a_n$. The sum of the series is the number obtained by carrying out that addition.

Thus $1+3+5+7$ is a series, whereas $16$ is its sum. Keeping this distinction clear separates the written addition from its evaluated result. A request for a corresponding series asks for the terms joined by addition signs.

How should sigma notation be read?

The symbol $\sum$, called sigma, indicates summation. In $\sum_{k=1}^{n}a_k$, the letter $k$ is the summation index: it takes successive integer values from the lower limit, one, through the upper limit, $n$.

The compact and expanded expressions are equivalent: $$\sum_{k=1}^{n}a_k=a_1+a_2+\cdots+a_n.$$ The upper limit gives the final included index. It does not mean that only the final term is added, nor does the sigma symbol itself supply a numerical answer.

Worked example 3. A sequence has $a_1=1$ and $a_n=a_{n-1}+2$ for $n\geq2$. Find its first five terms and the corresponding series.

Answer: Begin with the specified first term and use each calculated term to find the next.

1. The starting term is $a_1=1$.
2. The second term is $a_2=a_1+2=1+2=3$.
3. The third term is $a_3=a_2+2=3+2=5$.
4. The fourth term is $a_4=a_3+2=5+2=7$.
5. The fifth term is $a_5=a_4+2=7+2=9$.
6. The first-five-term series is $1+3+5+7+9$; the continuing associated series is $1+3+5+7+9+\cdots$.

The recurrence has no stated final position, so the full sequence continues. Listing its first five terms selects a finite initial part; it does not change the rule defining the whole sequence.

How can a geometric progression be recognised?

A geometric progression, abbreviated G.P., is a sequence of non-zero terms in which each term after the first bears a constant ratio to the immediately preceding term. This constant factor is the common ratio.

Let $r$ denote the common ratio and let $a$ denote the first term. For each valid position $k$, the defining relation is $$\frac{a_{k+1}}{a_k}=r.$$ The progression can therefore be written as $a,ar,ar^2,ar^3,\ldots$.

Result: Equal successive ratios identify a G.P.

To check a displayed pattern, divide a later term by its immediate predecessor. Maintain the same order in every division. A negative ratio changes signs between successive terms; a positive ratio does not require the terms to increase.

The examples show why a G.P. should be recognised by multiplication or division. Its terms may grow, decrease in magnitude, or alternate in sign. The common feature is the ratio between consecutive terms.

1. In the first sequence, the initial ratio is $r=4/2=2$.
2. The next ratio confirms the same factor: $8/4=2$.
3. In the second sequence, $r=(-1/27)/(1/9)=-1/3$.
4. In the decimal sequence, $r=0.001/0.1=0.01$.

Note: Use non-zero terms when applying the ratio definition. A zero denominator makes the quotient undefined. Also check which term comes later: reversing a quotient changes the ratio used in the progression.

Once the first term and common ratio are known, multiplication generates the following terms. The same two quantities are the starting information used in the general-term formula and in the finite-sum formula.

How is the general term of a G.P. found and used?

Result: The exponent is one less than the position

For first term $a$, common ratio $r$, and positive integer position $n$, the general-term formula is $$a_n=ar^{n-1}.$$ The first term has not yet been multiplied by the ratio; each move to the next position introduces one additional factor.

Derivation: Building the general term

1. The first term is $a_1=a=ar^0$.
2. Multiply once to obtain $a_2=a_1r=ar$.
3. Multiply again to obtain $a_3=a_2r=ar^2$.
4. The fourth term is $a_4=a_3r=ar^3$.
5. Continuing the same rule to position $n$ gives $a_n=ar^{n-1}$.

Interpretation: Count the moves from the first position, rather than the number of positions listed. This explains why the exponent is smaller than the term number by one.

Worked example 4. Find the tenth term and the general term of $5,25,125,\ldots$.

Answer: Identify the first term and ratio before substituting the required position.

1. The first term is $a=5$, and the ratio is $r=25/5=5$.
2. For the tenth term, $a_{10}=5(5)^{10-1}=5(5)^9=5^{10}$.
3. For any positive integer position, $a_n=5(5)^{n-1}=5^n$.

Worked example 5. The third term of a G.P. is $24$, and the sixth term is $192$. Find its tenth term.

Answer: Express both given terms using the same first term and common ratio.

1. The third-term information gives $ar^2=24$.
2. The sixth-term information gives $ar^5=192$.
3. Divide the second equation by the first: $r^3=192/24=8$, so the real ratio is $r=2$.
4. Substitute into the third-term equation: $a(2)^2=24$, hence $a=24/4=6$.
5. Calculate the requested term: $a_{10}=6(2)^9=6(512)=3072$.

Dividing the two term equations removes the unknown first term. The difference between their exponents corresponds to the gap between the given positions. Substitute the recovered ratio back into a given equation before calculating a new term.

How is the sum of a finite geometric series derived?

Let $S_n$ denote the sum of the first $n$ terms of a G.P. Its first term is $a$, its common ratio is $r$, and $n$ is the number of terms included. The final included term is $ar^{n-1}$.

Result: The finite-sum formula depends on the ratio

When $r=1$, every term equals the first term, so $S_n=na$. When $r\ne1$, the equivalent formulae are $$S_n=\frac{a(1-r^n)}{1-r}=\frac{a(r^n-1)}{r-1}.$$ Choose either form while keeping its numerator and denominator together.

Derivation: Multiplying and subtracting the series

1. Write the original finite sum: $$S_n=a+ar+ar^2+\cdots+ar^{n-1}.$$
2. Multiply every term by the common ratio: $$rS_n=ar+ar^2+ar^3+\cdots+ar^n.$$
3. Subtract the second expression from the first. The overlapping terms cancel: $$(1-r)S_n=a-ar^n.$$
4. Factor the right side and divide, using $r\ne1$: $$S_n=\frac{a(1-r^n)}{1-r}.$$
5. Change the signs of both numerator and denominator: $$S_n=\frac{a(r^n-1)}{r-1}.$$

Cancellation: The original first term and the additional final term in the multiplied series remain. The extra multiplication explains why the sum formula contains $r^n$, although the last original term contains $r^{n-1}$.

Worked example 6. Find the sum of the first $n$ terms and the sum of the first five terms of $1+\frac23+\frac49+\cdots$.

Answer: Here the first term is $a=1$, and the common ratio is $r=2/3$.

1. Substitute into the finite-sum formula: $$S_n=\frac{1-(2/3)^n}{1-2/3}.$$
2. Simplify the denominator: $$S_n=3\left[1-\left(\frac23\right)^n\right].$$
3. Set the number of terms to five: $$S_5=3\left(1-\frac{32}{243}\right).$$
4. Complete the arithmetic: $$S_5=3\left(\frac{211}{243}\right)=\frac{211}{81}.$$

The finite formula calculates a specified initial portion even when the displayed series continues. It does not require a value for the sum of all the terms of an infinite series.

How can an unknown position or number of terms be determined?

First decide whether the given number is an individual term or a sum. A term position comes from the general-term formula. The number of terms giving a specified total comes from the finite-sum formula. These are different equations.

How is a given term located?

Worked example 7. Which term of the G.P. $2,8,32,\ldots$ is $131072$?

Answer: Let $n$ be the position of the given term. The first term is $a=2$, and the common ratio is $r=4$.

1. Use the term formula: $131072=2(4)^{n-1}$.
2. Divide by the first term: $4^{n-1}=131072/2=65536$.
3. Express the right side with the same base: $65536=4^8$.
4. Equate exponents: $n-1=8$, giving $n=9$.

The specified number is therefore the ninth term.

How is a specified sum used?

Worked example 8. How many terms of $3,\frac32,\frac34,\ldots$ have sum $\frac{3069}{512}$?

Answer: Here $n$ is the number of included terms, with $a=3$ and $r=1/2$.

1. Substitute the given sum: $$\frac{3069}{512}=\frac{3[1-(1/2)^n]}{1-1/2}.$$
2. Simplify the coefficient: $$\frac{3069}{512}=6\left(1-\frac1{2^n}\right).$$
3. Divide by six: $$\frac{3069}{3072}=1-\frac1{2^n}.$$
4. Rearrange the equation: $$\frac1{2^n}=1-\frac{3069}{3072}=\frac3{3072}=\frac1{1024}.$$
5. Compare powers: $2^n=1024=2^{10}$, so $n=10$.

Ten terms give the required sum.

In both calculations, reduce the equation to comparable powers before reading off the exponent. Remember the different exponents in the two formulae: the position formula contains one less than the position, while the sum formula contains the number of included terms.

Note: The answer must identify what was found. “Ninth term” names a position; “ten terms” names the size of an initial portion. Neither phrase is interchangeable with the value of a term or a sum.

What are geometric means, and how are they inserted?

For two positive numbers $a$ and $b$, their geometric mean, denoted by $G$, is $$G=\sqrt{ab}.$$ This is the positive square root. The sequence $a,G,b$ forms a G.P.; for the numbers two and eight, the geometric mean is four.

How are several positive means inserted?

Let $G_1,G_2,\ldots,G_n$ denote $n$ positive geometric means inserted between the positive endpoints $a$ and $b$. Here $n$ counts the inserted numbers. Including both endpoints gives $n+2$ terms, so there are $n+1$ ratio multiplications.

1. Write the complete progression as $a,G_1,G_2,\ldots,G_n,b$.
2. Apply the term formula at the final position: $b=ar^{n+1}$.
3. Divide and take the positive root: $$r=\left(\frac ba\right)^{1/(n+1)}.$$
4. For an inserted position $k$, where $1\leq k\leq n$, obtain $$G_k=ar^k=a\left(\frac ba\right)^{k/(n+1)}.$$

The endpoint positions must be counted before solving for the ratio. Counting only the inserted numbers gives the wrong exponent. Once the ratio is found, generate every intermediate number successively and check the final endpoint.

Can a G.P. between positive endpoints have a negative ratio?

Worked example 9. Insert three numbers between $1$ and $256$ so that the resulting sequence is a G.P.

Answer: The first term is one, and there are five terms altogether.

1. The endpoint equation is $256=1\cdot r^4$.
2. Taking the real fourth roots gives $r=4$ or $r=-4$.
3. With the positive ratio, the inserted terms are $G_1=1(4)=4$, $G_2=1(4)^2=16$, and $G_3=1(4)^3=64$.
4. With the negative ratio, they are $G_1=1(-4)=-4$, $G_2=1(-4)^2=16$, and $G_3=1(-4)^3=-64$.
5. The endpoint checks are $64(4)=256$ and $(-64)(-4)=256$.

The positive geometric means are four, sixteen and sixty-four. The broader instruction to form a real G.P. also allows the alternating-sign solution. This distinction preserves the positive-root definition of a geometric mean without discarding a valid progression.

What is the relationship between arithmetic and geometric means?

For positive real numbers $a$ and $b$, let $A$ be their arithmetic mean, abbreviated A.M., and let $G$ be their geometric mean, abbreviated G.M. Their formulae are $$A=\frac{a+b}{2},\qquad G=\sqrt{ab}.$$

Result: Arithmetic mean is at least geometric mean

The relationship is $A\geq G$. It follows by rewriting the difference between the means as a square divided by a positive number. The assumption that both given numbers are positive is part of this statement.

Derivation: Comparing the two means

1. Subtract their defining expressions: $$A-G=\frac{a+b}{2}-\sqrt{ab}.$$
2. Use a common denominator: $$A-G=\frac{a+b-2\sqrt{ab}}{2}.$$
3. Recognise the square: $$A-G=\frac{(\sqrt a-\sqrt b)^2}{2}.$$
4. A real square is non-negative, hence $$A-G\geq0,\qquad A\geq G.$$

Reasoning: The comparison comes from the square in the numerator. It is a general result for the positive numbers under consideration, rather than a pattern inferred from a few calculated means.

How do the means determine the original numbers?

Worked example 10. Two positive numbers have A.M. $10$ and G.M. $8$. Find the numbers.

Answer: Let the required positive numbers be $a$ and $b$. Their means give their sum and product.

1. From the arithmetic mean, $(a+b)/2=10$, so $a+b=20$.
2. From the geometric mean, $\sqrt{ab}=8$, so $ab=64$.
3. Use the difference-square identity: $$(a-b)^2=(a+b)^2-4ab=20^2-4(64)=400-256=144.$$
4. Taking both real square roots gives $a-b=12$ or $a-b=-12$.
5. Combining the positive difference with the sum gives $2a=32$, so $a=16$ and $b=4$. The negative difference interchanges these values.
6. Check the means: $(16+4)/2=10$ and $\sqrt{16\cdot4}=\sqrt{64}=8$.

The numbers are four and sixteen, in either order. Keeping both signs when taking a square root accounts for the two possible assignments to the named variables.

How can geometric sums solve counting and repeated-digit problems?

A geometric model applies when the stated pattern repeatedly multiplies a quantity by the same factor. Before calculating, decide whether the problem asks for the quantity at one stage or the total across several stages.

How are ancestor counts added across generations?

Worked example 11. A person has two parents, four grandparents, eight great-grandparents, and so on in the stated pattern. Find the total number of ancestors during the ten preceding generations.

Answer: Use the generation counts as the terms of a G.P.

1. The first term, ratio and number of terms are $a=2$, $r=2$ and $n=10$.
2. Apply the finite-sum formula: $$S_{10}=\frac{2(2^{10}-1)}{2-1}.$$
3. Evaluate the power and subtraction: $$S_{10}=2(1024-1)=2(1023).$$
4. Complete the multiplication: $S_{10}=2046$.

The stated pattern therefore gives a total of $2046$ ancestors across those generations.

The last generation's count is a single term, whereas the requested total includes every generation listed. This is why the sum formula, rather than just the general-term formula, is used here.

How can a series that is not geometric use a G.P. formula?

Worked example 12. Find the sum of $7+77+777+7777+\cdots$ to $n$ terms, where $n$ is a positive integer.

Answer: Let $S_n$ be this sum. Rewrite each repeated-digit number using a power of ten.

1. Factor out the repeated digit: $$S_n=\frac79(9+99+999+\cdots+(10^n-1)).$$
2. Express each bracketed term as a power less one: $$S_n=\frac79[(10-1)+(10^2-1)+\cdots+(10^n-1)].$$
3. Separate the powers from the constant subtractions: $$S_n=\frac79[(10+10^2+\cdots+10^n)-n].$$
4. The power sum has first term ten and ratio ten, giving $$S_n=\frac79\left[\frac{10(10^n-1)}{10-1}-n\right].$$
5. Simplify the denominator: $$S_n=\frac79\left[\frac{10(10^n-1)}9-n\right].$$

The original repeated-digit sequence is not a G.P. The transformation exposes a geometric sum inside it. The separate subtraction must remain because one was subtracted from every power of ten, giving as many subtractions as there are terms.

How can three unknown geometric terms be found from their sum and product?

For three unknown consecutive terms, a symmetric representation can simplify a product condition. Let $m$ denote the middle term and $r$ the common ratio. The three terms can be written as $m/r,m,mr$, with both quantities non-zero.

This notation places the middle term at the centre of the calculation. The factors involving the ratio cancel in the product, leaving the cube of the middle term. The sum condition then becomes an equation for the ratio.

How do the two conditions work together?

Worked example 13. The first three terms of a real G.P. have sum $13/12$ and product $-1$. Find the terms and the common ratio.

Answer: Use the middle-term representation to handle the product before the sum.

1. Write the given conditions: $$\frac mr+m+mr=\frac{13}{12},\qquad \left(\frac mr\right)m(mr)=-1.$$
2. Cancel the ratio factors in the product: $m^3=-1$. Its real root is $m=-1$.
3. Substitute into the sum condition: $$-\frac1r-1-r=\frac{13}{12}.$$
4. Multiply by the non-zero quantity $12r$: $-12-12r-12r^2=13r$. Rearranging gives $12r^2+25r+12=0$.
5. Factor the quadratic: $(3r+4)(4r+3)=0$, giving $r=-4/3$ or $r=-3/4$.
6. For $r=-3/4$, the terms are $(-1)/(-3/4)=4/3$, $-1$, and $(-1)(-3/4)=3/4$.
7. For $r=-4/3$, the terms are $(-1)/(-4/3)=3/4$, $-1$, and $(-1)(-4/3)=4/3$.
8. Check the sum and product for either ordering: $$\frac43-1+\frac34=\frac{16-12+9}{12}=\frac{13}{12},\qquad \frac43(-1)\frac34=-1.$$

The two answers reverse the outer terms. Both satisfy the given data, but their common ratios differ. Retaining both real solutions of the quadratic is necessary to give the complete answer.

Choice of notation: Here the middle term is called $m$, leaving $a$ available for the first-term notation used earlier. Whichever letters are chosen, identify their roles before forming equations so that a middle term is not mistaken for a first term.

Glossary

• Sequence — An arrangement of numbers in a definite order according to a rule.
• Term — An individual number occupying a particular position in an ordered sequence.
• General term — The term at an arbitrary position, often expressed using that position as a variable.
• Finite sequence — A sequence containing a fixed, finite number of terms.
• Infinite sequence — A sequence that does not end and is therefore not finite.
• Recurrence relation — A rule that generates later terms using earlier terms and specified starting information.
• Series — The expression formed by adding the terms of an associated sequence.
• Sigma notation — Compact notation using the summation symbol to indicate which indexed terms are added.
• Geometric progression — A sequence of non-zero terms with a constant ratio between each term and its predecessor.
• Common ratio — The constant factor obtained by dividing a geometric progression term by its preceding term.
• Arithmetic mean — Half the sum of two numbers, giving their average value.
• Geometric mean — The positive square root of the product of two positive numbers.

Common errors and misconceptions

• Misconception: A sequence and its associated series are identical expressions. Correct: A sequence lists ordered terms; its associated series indicates their addition.
• Misconception: Every sequence requires a direct formula for its general term. Correct: Terms may instead be generated by a recurrence or described by a verbal rule.
• Misconception: The general term of a G.P. is $ar^n$. Correct: For first term $a$ and ratio $r$, it is $a_n=ar^{n-1}$.
• Misconception: A G.P. must have an increasing sequence of positive terms. Correct: Its defining feature is a constant ratio; fractional or negative ratios also occur.
• Misconception: The usual quotient formula for a geometric sum can be used when $r=1$. Correct: That would divide by zero; instead use $S_n=na$.
• Misconception: Inserting $n$ means between two endpoints gives $n$ total terms. Correct: Both endpoints are included, so there are $n+2$ terms and $n+1$ ratio multiplications.
• Misconception: The geometric mean of positive numbers may be the negative square root. Correct: Their geometric mean is the positive root, although a differently worded insertion problem may allow a negative ratio.
• Misconception: The final generation's ancestor count equals the total across all generations. Correct: The former is a term; the latter requires adding the generation counts.

Exam-style questions with model answers

Q1. Distinguish the series $1+3+5+7$ from its sum, and evaluate that sum. [2 marks]

1. The series is the indicated addition $1+3+5+7$, containing four terms in their stated order.
2. Its sum is the evaluated number: $1+3+5+7=4+5+7=9+7=16$.

Q2. A sequence has general term $a_n=2n+5$, where $n$ is its positive integer position. Find its first three terms. [3 marks]

1. For the first term, substitute one for the position in the given rule: $a_1=2(1)+5=7$.
2. For the second term, use the same rule with the next position, multiplying that position by two before adding five: $a_2=2(2)+5=9$.
3. For the third term, substitute three: $a_3=2(3)+5=11$. Thus the required terms, written in their original order, are $7,9,11$.

Q3. Which term of the G.P. $2,8,32,\ldots$ is $131072$? Show your method. [4 marks]

1. Let $n$ be the required position. The first term is $a=2$, and the common ratio is $r=8/2=4$.
2. Use the general-term formula $a_n=ar^{n-1}$, giving the equation $131072=2(4)^{n-1}$.
3. Divide both sides by two and express the result as a power of four: $4^{n-1}=65536=4^8$.
4. Equating the exponents gives $n-1=8$, hence $n=9$. Therefore the given number occupies the ninth position in the progression.

Q4. A G.P. has first term $a$, common ratio $r\ne1$, and $n$ terms, where $n$ is a positive integer. Derive its finite-sum formula. [5 marks]

1. Let $S_n$ denote the sum of the stated terms. Write them in order as $S_n=a+ar+ar^2+\cdots+ar^{n-1}$; the final exponent is one less than the number of terms.
2. Multiply the entire equality by the common ratio. This gives $rS_n=ar+ar^2+ar^3+\cdots+ar^n$, introducing one additional power at the end.
3. Subtract the multiplied equation from the original equation. All overlapping terms cancel, leaving $(1-r)S_n=a-ar^n$.
4. Factor out the first term on the right to obtain $(1-r)S_n=a(1-r^n)$. This isolates the coefficient of the required sum.
5. Since the given ratio is not one, division by $1-r$ is valid. Therefore $S_n=a(1-r^n)/(1-r)$, equivalently $S_n=a(r^n-1)/(r-1)$.

Q5. The third and sixth terms of a real G.P. are $24$ and $192$, respectively. Find its tenth term. [5 marks]

1. Let $a$ denote the first term and $r$ the common ratio. Applying the term formula at the third position gives $ar^2=24$.
2. Applying the same formula at the sixth position gives $ar^5=192$. Both equations describe the same progression and therefore use the same unknowns.
3. Divide the sixth-term equation by the third-term equation to eliminate the first term: $r^3=192/24=8$. Taking the real cube root gives $r=2$.
4. Substitute the ratio into the third-term equation: $a(2)^2=24$. Hence $4a=24$, so the first term is $a=6$.
5. Use the recovered values at the tenth position: $a_{10}=6(2)^9=6(512)=3072$. This is the required term, rather than a sum.

Q6. Insert three real numbers between $1$ and $256$ to form a G.P. Give both possible real-ratio solutions. [4 marks]

1. Let $r$ denote the common ratio. Including both endpoints gives five terms, so the last term satisfies $256=1\cdot r^4$.
2. Solving the fourth-power equation over the real numbers gives $r=4$ or $r=-4$; both must be considered.
3. With $r=4$, successive multiplication gives the inserted terms $1(4)=4$, $4(4)=16$, and $16(4)=64$.
4. With $r=-4$, they are $1(-4)=-4$, $(-4)(-4)=16$, and $16(-4)=-64$. The positive means are the first solution.

Q7. Two positive numbers have arithmetic mean $10$ and geometric mean $8$. Determine the numbers and check both means. [5 marks]

1. Let the numbers be $a$ and $b$. Their arithmetic mean gives $(a+b)/2=10$, so $a+b=20$; this determines their sum.
2. The geometric mean gives $\sqrt{ab}=8$. Squaring this positive-root equation yields $ab=64$, determining their product as well.
3. Use the identity for their squared difference: $(a-b)^2=(a+b)^2-4ab=400-256=144$. Therefore $a-b=12$ or $a-b=-12$.
4. Using the positive difference with the sum gives $2a=32$, hence $a=16$ and $b=4$. The negative difference gives the same pair in reverse order. The two signs exchange which variable names the larger number.
5. Verify both requested means directly: $(16+4)/2=10$ and $\sqrt{16\cdot4}=\sqrt{64}=8$. Thus the two positive numbers are four and sixteen.

Key takeaways

• A sequence gives numbers in a definite order, while its associated series indicates the addition of those terms.
• A general-term formula gives a term directly from its position; a recurrence generates terms from specified earlier values.
• A geometric progression has non-zero terms and a constant ratio between each later term and its immediate predecessor.
• The general term is $a_n=ar^{n-1}$, because reaching position $n$ requires one fewer multiplication by the common ratio.
• For $r\ne1$, the finite sum is $S_n=a(1-r^n)/(1-r)$; for a ratio of one, use $S_n=na$.
• Inserting $n$ geometric means between two endpoints creates $n+2$ total terms, so the endpoint equation contains the exponent $n+1$.
• For two positive numbers, the geometric mean is the positive square root of their product, and their arithmetic mean is at least as large.
• Some repeated-digit series can be rewritten using powers of ten, allowing a geometric sum formula to evaluate part of the expression.

Test yourself

What does the subscript in $a_n$ tell you?

It identifies the position of the term in the sequence, with $n$ a natural number.

How do you distinguish a series from its sum?

The series is the indicated addition; its sum is the number obtained after adding the terms.

What starting values and rule define the Fibonacci sequence used here?

The starting values are $a_1=a_2=1$; subsequent terms satisfy $a_n=a_{n-2}+a_{n-1}$ for $n>2$.

Why does the general term of a G.P. use the exponent $n-1$?

The first term is already given, so reaching position $n$ requires only $n-1$ multiplications by the common ratio.

What happens to the finite sum when the common ratio is one?

All included terms equal the first term $a$, so the sum of $n$ terms is $S_n=na$.

How many ratio multiplications connect the endpoints when $n$ geometric means are inserted?

There are $n+2$ terms including the endpoints, so moving between them requires $n+1$ ratio multiplications.

What inequality connects the two means of positive real numbers?

With $A$ denoting the arithmetic mean and $G$ the geometric mean, the relationship is $A\geq G$.

Why is the repeated-digit series rewritten before using a G.P. sum formula?

The original sequence is not geometric; rewriting its terms exposes a power-of-ten geometric series that can be summed.