Sequences and Series

Sequences and Series

A sequence is an ordered list of numbers. A series is the sum of the terms of a sequence.

Geometric Series Convergence

Adjust the parameters in the graph above to explore the relationships between variables.

Notation

• $u_n$ or $a_n$: the $n$-th term of a sequence
• $u_1$: the first term
• $d$: common difference (arithmetic)
• $r$: common ratio (geometric)
• $n$: number of terms
• $S_n$: sum of the first $n$ terms

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Arithmetic Sequences

An arithmetic sequence has a constant difference between consecutive terms:

$$u_{n+1} - u_n = d$$

General Term

$$u_n = u_1 + (n-1)d$$

Sum of the First $n$ Terms

$$S_n = \frac{n}{2}(u_1 + u_n) = \frac{n}{2}[2u_1 + (n-1)d]$$

Example

Find the 20th term and the sum of the first 20 terms of the sequence $3, 7, 11, 15, \ldots$.

$u_1 = 3$, $d = 4$.

$$u_{20} = 3 + 19 \times 4 = 79$$$$S_{20} = \frac{20}{2}(3 + 79) = 10 \times 82 = 820$$

Properties of Arithmetic Sequences

The arithmetic mean of $a$ and $b$ is $\dfrac{a+b}{2}$.

If $a, b, c$ are consecutive terms of an arithmetic sequence, then $2b = a + c$.

Example

The 5th term of an arithmetic sequence is $17$ and the 12th term is $38$. Find the first term and the common difference.

$$u_5 = u_1 + 4d = 17 \quad \mathrm{(1)}$$$$u_{12} = u_1 + 11d = 38 \quad \mathrm{(2)}$$

(2) $-$ (1): $7d = 21 \implies d = 3$.

From (1): $u_1 + 12 = 17 \implies u_1 = 5$.

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Geometric Sequences

A geometric sequence has a constant ratio between consecutive terms:

$$\frac{u_{n+1}}{u_n} = r$$

General Term

$$u_n = u_1 r^{n-1}$$

Sum of the First $n$ Terms

$$S_n = \frac{u_1(r^n - 1)}{r - 1} = \frac{u_1(1 - r^n)}{1 - r} \quad (r \neq 1)$$

When $r = 1$: $S_n = nu_1$.

Example

Find the 8th term and the sum of the first 8 terms of $2, 6, 18, 54, \ldots$.

$u_1 = 2$, $r = 3$.

$$u_8 = 2 \times 3^7 = 2 \times 2187 = 4374$$$$S_8 = \frac{2(3^8 - 1)}{3 - 1} = \frac{2(6561 - 1)}{2} = 6560$$

Geometric Mean

The geometric mean of $a$ and $b$ is $\sqrt{ab}$ (for positive $a, b$).

If $a, b, c$ are consecutive terms of a geometric sequence, then $b^2 = ac$.

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Convergence of Geometric Series

Sum to Infinity

If $|r| \lt 1$, the geometric series converges and the sum to infinity is:

$$S_{\infty} = \frac{u_1}{1 - r}$$

If $|r| \ge 1$, the series diverges (the sum to infinity does not exist).

Example

Find the sum to infinity of $1 + \dfrac{1}{2} + \dfrac{1}{4} + \dfrac{1}{8} + \cdots$.

$u_1 = 1$, $r = \dfrac{1}{2}$. Since $|r| \lt 1$:

$$S_{\infty} = \frac{1}{1 - \frac{1}{2}} = 2$$

Example

Express $0.\dot{7}$ (recurring decimal) as a fraction.

$$0.\dot{7} = 0.7777\ldots = \frac{7}{10} + \frac{7}{100} + \frac{7}{1000} + \cdots$$

This is a geometric series with $u_1 = \dfrac{7}{10}$ and $r = \dfrac{1}{10}$.

$$S_{\infty} = \frac{\frac{7}{10}}{1 - \frac{1}{10}} = \frac{\frac{7}{10}}{\frac{9}{10}} = \frac{7}{9}$$

Example

Express $0.\dot{2}\dot{7}$ (i.e., $0.272727\ldots$) as a fraction.

$$0.\dot{2}\dot{7} = \frac{27}{100} + \frac{27}{10000} + \cdots$$

$u_1 = \dfrac{27}{100}$, $r = \dfrac{1}{100}$.

$$S_{\infty} = \frac{\frac{27}{100}}{1 - \frac{1}{100}} = \frac{\frac{27}{100}}{\frac{99}{100}} = \frac{27}{99} = \frac{3}{11}$$

Conditions for Convergence

| Condition | Behaviour | | --------- | ----------------------------- | ------ | ------------------------------- | | $| r | \lt 1$ | Converges to $\dfrac{u_1}{1-r}$ | | $r = 1$ | Diverges (grows linearly) | | | | $r = -1$ | Oscillates, does not converge | | | | $| r | \gt 1$ | Diverges (grows exponentially) |

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Sigma Notation

Definition

$$\sum_{i=1}^{n} u_i = u_1 + u_2 + u_3 + \cdots + u_n$$

Properties

1. $\displaystyle\sum_{i=1}^{n} k = kn$
2. $\displaystyle\sum_{i=1}^{n} k u_i = k\sum_{i=1}^{n} u_i$
3. $\displaystyle\sum_{i=1}^{n} (u_i \pm v_i) = \sum_{i=1}^{n} u_i \pm \sum_{i=1}^{n} v_i$

Useful Sigma Summations

$$\sum_{i=1}^{n} i = \frac{n(n+1)}{2}$$ $$\sum_{i=1}^{n} i^2 = \frac{n(n+1)(2n+1)}{6}$$ $$\sum_{i=1}^{n} i^3 = \left[\frac{n(n+1)}{2}\right]^2$$

Example

Evaluate $\displaystyle\sum_{k=1}^{10} (3k - 1)$.

$$\sum_{k=1}^{10}(3k-1) = 3\sum_{k=1}^{10}k - \sum_{k=1}^{10}1 = 3 \cdot \frac{10 \times 11}{2} - 10 = 165 - 10 = 155$$

Example

Evaluate $\displaystyle\sum_{r=1}^{n} r(r+1)$.

$$\sum_{r=1}^{n}(r^2 + r) = \sum_{r=1}^{n}r^2 + \sum_{r=1}^{n}r = \frac{n(n+1)(2n+1)}{6} + \frac{n(n+1)}{2}$$$$= \frac{n(n+1)}{6}\big[(2n+1) + 3\big] = \frac{n(n+1)(2n+4)}{6} = \frac{n(n+1)(n+2)}{3}$$

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Applications

Compound Interest

If a principal $P$ is invested at a rate of $r$ per period, compounded for $n$ periods:

$$A = P(1 + r)^n$$

Population Growth

If a population $P_0$ grows at a rate of $r\%$ per year:

$$P_n = P_0\left(1 + \frac{r}{100}\right)^n$$

Depreciation

For depreciation at rate $r\%$ per year:

$$V_n = V_0\left(1 - \frac{r}{100}\right)^n$$

Example

USD 5000 is invested at 6% per year, compounded annually. Find the value after 10 years.

$$A = 5000(1.06)^{10} \approx 5000 \times 1.7908 = 8954.24$$

The investment is worth approximately USD 8954.24.

Example

A car bought for USD 25000 depreciates at 15% per year. Find its value after 5 years.

$$V_5 = 25000(0.85)^5 \approx 25000 \times 0.4437 = 11092.50$$

The car is worth approximately USD 11092.50 after 5 years.

Loan Repayments

For a loan of $L$ with monthly repayment $R$ at monthly interest rate $i$ over $n$ months:

$$L = \frac{R[1 - (1+i)^{-n}]}{i}$$

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The Binomial Theorem

Expansion of $(a + b)^n$

For positive integer $n$:

$$(a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k$$

where the binomial coefficient is:

$$\binom{n}{k} = \frac{n!}{k!(n-k)!} = {}^nC_k$$

Special Cases

$(1 + x)^n = \sum_{k=0}^{n} \binom{n}{k} x^k = 1 + nx + \frac{n(n-1)}{2!}x^2 + \frac{n(n-1)(n-2)}{3!}x^3 + \cdots$

Pascal's Triangle

Each entry is the sum of the two entries above it:

\begin`\{array}``\{cccccccc}` & & & 1 & & & \\ & & 1 & & 1 & & \\ & 1 & & 2 & & 1 & \\ 1 & & 3 & & 3 & & 1 \\ & 1 & & 4 & & 6 & & 4 & & 1 \end`\{array}`

Row $n$ (starting from row 0) gives the coefficients of $(a + b)^n$.

Example

Expand $(2x - 3)^4$.

Using the binomial theorem:

$$(2x - 3)^4 = \sum_{k=0}^{4}\binom{4}{k}(2x)^{4-k}(-3)^k$$$$= 1 \cdot (2x)^4 \cdot 1 + 4 \cdot (2x)^3 \cdot (-3) + 6 \cdot (2x)^2 \cdot 9 + 4 \cdot (2x) \cdot (-27) + 1 \cdot 81$$$$= 16x^4 - 96x^3 + 216x^2 - 216x + 81$$

Finding Specific Terms

To find the term containing $x^r$ in the expansion of $(a + bx)^n$:

The general term is $\dbinom{n}{k}a^{n-k}b^k x^k$.

Set $k = r$ to find the coefficient of $x^r$.

Example

Find the coefficient of $x^3$ in the expansion of $(2 + 3x)^7$.

The general term is $\dbinom{7}{k} 2^{7-k}(3x)^k$.

For $x^3$: $k = 3$.

Coefficient $= \dbinom{7}{3} \cdot 2^4 \cdot 3^3 = 35 \times 16 \times 27 = 15120$.

Binomial Expansion for Negative or Fractional Powers

For $|x| \lt 1$ and $n \in \mathbb{'\{'}Q{'\}'}$:

$$(1 + x)^n = 1 + nx + \frac{n(n-1)}{2!}x^2 + \frac{n(n-1)(n-2)}{3!}x^3 + \cdots$$

This is an infinite series that converges for $|x| \lt 1$.

Example

Find the expansion of $(1 + x)^{-2}$ up to the term in $x^3$.

$$(1+x)^{-2} = 1 + (-2)x + \frac{(-2)(-3)}{2}x^2 + \frac{(-2)(-3)(-4)}{6}x^3$$$$= 1 - 2x + 3x^2 - 4x^3$$

Example

Find the expansion of $\dfrac{1}{\sqrt{1+x}}$ up to the term in $x^2$.

$$(1+x)^{-1/2} = 1 + \left(-\frac{1}{2}\right)x + \frac{\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)}{2}x^2$$$$= 1 - \frac{x}{2} + \frac{3x^2}{8}$$

Validity of Expansion

The expansion $(1+x)^n$ for non-integer $n$ converges when $|x| \lt 1$.

To use this for expressions like $(2 + 3x)^{-1/2}$, factor out the constant:

$$(2+3x)^{-1/2} = 2^{-1/2}\left(1 + \frac{3x}{2}\right)^{-1/2}$$

This converges when $\left|\dfrac{3x}{2}\right| \lt 1$, i.e., $|x| \lt \dfrac{2}{3}$.

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Mixed Arithmetic-Geometric Sequences

Recurrence Relations

A sequence defined by a recurrence relation gives each term in terms of previous terms.

$$u_{n+1} = au_n + b$$

Solving First-Order Linear Recurrence Relations

For $u_{n+1} = ru_n + d$ with $u_1$ given:

If $r \neq 1$, the solution is:

$$u_n = r^{n-1}u_1 + d\frac{r^{n-1} - 1}{r - 1}$$

Example

A sequence is defined by $u_{n+1} = 3u_n + 2$ with $u_1 = 1$. Find $u_n$.

$r = 3$, $d = 2$.

$$u_n = 3^{n-1} \cdot 1 + 2 \cdot \frac{3^{n-1} - 1}{3 - 1} = 3^{n-1} + 3^{n-1} - 1 = 2 \cdot 3^{n-1} - 1$$

Verify: $u_1 = 2 \cdot 1 - 1 = 1$. $u_2 = 2 \cdot 3 - 1 = 5$. Check: $u_2 = 3(1) + 2 = 5$.

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IB Exam-Style Questions

Question 1 (Paper 1 style)

An arithmetic sequence has first term $5$ and common difference $3$. A geometric sequence has first term $3$ and common ratio $2$. Find the value of $n$ for which the $n$-th terms are equal.

$$u_n = 5 + (n-1) \times 3 = 3n + 2$$ $$v_n = 3 \times 2^{n-1}$$ $$3n + 2 = 3 \times 2^{n-1}$$

Testing values:

$n = 1$: $5 \neq 3$.

$n = 2$: $8 \neq 6$.

$n = 3$: $11 \neq 12$.

$n = 4$: $14 \neq 24$.

$n = 5$: $17 \neq 48$.

The sequences do not have equal terms for small $n$. The geometric sequence grows much faster. The only possible solution is $n = 3$ approximately (left $= 11$, right $= 12$), so there is no exact integer solution.

Question 2 (Paper 2 style)

The sum of the first three terms of a geometric sequence is $52$. The sum of the first six terms is $4732$. Find the common ratio and the first term.

$$S_3 = \frac{u_1(r^3 - 1)}{r - 1} = 52 \quad \mathrm{(1)}$$ $$S_6 = \frac{u_1(r^6 - 1)}{r - 1} = 4732 \quad \mathrm{(2)}$$

Dividing (2) by (1):

$$\frac{r^6 - 1}{r^3 - 1} = \frac{4732}{52} = 91$$ $$r^3 + 1 = 91 \implies r^3 = 90$$

Since $90 = 9 \times 10$ is not a perfect cube, this suggests the ratio might not be exact. Let us reconsider with $r^3 - 1 \neq 0$:

$$\frac{(r^3 - 1)(r^3 + 1)}{r^3 - 1} = r^3 + 1 = 91$$ $$r^3 = 90, \quad r = \sqrt[3]{90}$$

From (1): $u_1 = \dfrac{52(r-1)}{r^3 - 1} = \dfrac{52(r-1)}{89}$.

Question 3 (Paper 1 style)

Expand $(1 - 2x)^5$ in ascending powers of $x$ up to and including the term in $x^3$.

$$(1-2x)^5 = 1 + 5(-2x) + 10(-2x)^2 + 10(-2x)^3 + 5(-2x)^4 + (-2x)^5$$ $$= 1 - 10x + 40x^2 - 80x^3 + 80x^4 - 32x^5$$

Question 4 (Paper 2 style)

The first three terms of a geometric sequence are $\sin\theta$, $\cos\theta$, and $\dfrac{1}{\sqrt{3}}$, where $0 \lt \theta \lt \dfrac{\pi}{2}$.

Find the value of $\theta$.

Common ratio: $r = \dfrac{\cos\theta}{\sin\theta} = \cot\theta$.

Also: $r = \dfrac{1/\sqrt{3}}{\cos\theta}$.

$$\cot\theta = \frac{1}{\sqrt{3}\cos\theta}$$ $$\frac{\cos\theta}{\sin\theta} = \frac{1}{\sqrt{3}\cos\theta}$$ $$\cos^2\theta = \frac{\sin\theta}{\sqrt{3}}$$

Since $\cos^2\theta = 1 - \sin^2\theta$:

$$1 - \sin^2\theta = \frac{\sin\theta}{\sqrt{3}}$$ $$\sqrt{3}\sin^2\theta + \sin\theta - \sqrt{3} = 0$$

Let $u = \sin\theta$:

$$\sqrt{3}u^2 + u - \sqrt{3} = 0$$ $$u = \frac{-1 \pm \sqrt{1 + 12}}{2\sqrt{3}} = \frac{-1 \pm \sqrt{13}}{2\sqrt{3}}$$

Since $0 \lt \theta \lt \dfrac{\pi}{2}$, $\sin\theta \gt 0$:

$$\sin\theta = \frac{-1 + \sqrt{13}}{2\sqrt{3}}$$

Question 5 (Paper 1 style)

Evaluate $\displaystyle\sum_{k=1}^{50} (2k + 1)$.

$$\sum_{k=1}^{50}(2k+1) = 2\sum_{k=1}^{50}k + \sum_{k=1}^{50}1 = 2 \cdot \frac{50 \times 51}{2} + 50 = 2550 + 50 = 2600$$

Question 6 (Paper 2 style)

The series $1 + 2x + 3x^2 + 4x^3 + \cdots$ can be written as $\displaystyle\sum_{n=1}^{\infty} nx^{n-1}$.

(a) Find the sum to infinity in terms of $x$ for $|x| \lt 1$.

We know $\displaystyle\sum_{n=0}^{\infty}x^n = \frac{1}{1-x}$ for $|x| \lt 1$.

Differentiating both sides with respect to $x$:

$$\sum_{n=1}^{\infty}nx^{n-1} = \frac{1}{(1-x)^2}$$

(b) Hence find the sum to infinity of $\displaystyle\sum_{n=1}^{\infty}\frac{n}{3^n}$.

Setting $x = \dfrac{1}{3}$:

$$\sum_{n=1}^{\infty}n\left(\frac{1}{3}\right)^{n-1} = \frac{1}{(1 - 1/3)^2} = \frac{1}{(2/3)^2} = \frac{9}{4}$$

Therefore:

$$\sum_{n=1}^{\infty}\frac{n}{3^n} = \frac{1}{3}\sum_{n=1}^{\infty}n\left(\frac{1}{3}\right)^{n-1} = \frac{1}{3} \cdot \frac{9}{4} = \frac{3}{4}$$

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Summary

| Topic | Key Formula | | ---------------------- | ------------------------------------------------------------- | --- | ------ | | Arithmetic $n$-th term | $u_n = u_1 + (n-1)d$ | | | | Arithmetic sum | $S_n = \dfrac{n}{2}[2u_1 + (n-1)d]$ | | | | Geometric $n$-th term | $u_n = u_1 r^{n-1}$ | | | | Geometric sum | $S_n = \dfrac{u_1(1-r^n)}{1-r}$ | | | | Sum to infinity | $S_{\infty} = \dfrac{u_1}{1-r}$ for $| r | \lt 1$ | | Binomial theorem | $(a+b)^n = \displaystyle\sum_{k=0}^{n}\binom{n}{k}a^{n-k}b^k$ | | | | Sigma of $i$ | $\displaystyle\sum_{i=1}^{n}i = \dfrac{n(n+1)}{2}$ | | | | Sigma of $i^2$ | $\displaystyle\sum_{i=1}^{n}i^2 = \dfrac{n(n+1)(2n+1)}{6}$ | | |

Exam Strategy

For binomial expansion questions, always check the validity condition when $n$ is not a positive integer. For geometric series, always verify that $|r| \lt 1$ before computing $S_{\infty}$. In Paper 2, show all steps of sigma notation manipulations.

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Mathematical Induction

Principle

Mathematical induction is a proof technique used to prove statements about natural numbers.

Steps

1. Base case: Prove the statement is true for $n = 1$ (or the starting value).
2. Inductive hypothesis: Assume the statement is true for $n = k$.
3. Inductive step: Prove that if it is true for $n = k$, then it is true for $n = k + 1$.
4. Conclusion: By the principle of mathematical induction, the statement is true for all $n \ge 1$.

Example

Prove that $\displaystyle\sum_{i=1}^{n} i^2 = \frac{n(n+1)(2n+1)}{6}$ for all $n \ge 1$.

Base case ($n = 1$): $\displaystyle\sum_{i=1}^{1} i^2 = 1$ and $\dfrac{1 \times 2 \times 3}{6} = 1$. True.

Inductive hypothesis: Assume $\displaystyle\sum_{i=1}^{k} i^2 = \frac{k(k+1)(2k+1)}{6}$.

Inductive step: Show for $n = k + 1$:

$$\sum_{i=1}^{k+1} i^2 = \sum_{i=1}^{k} i^2 + (k+1)^2 = \frac{k(k+1)(2k+1)}{6} + (k+1)^2$$$$= \frac{k(k+1)(2k+1) + 6(k+1)^2}{6} = \frac{(k+1)[k(2k+1) + 6(k+1)]}{6}$$$$= \frac{(k+1)(2k^2 + k + 6k + 6)}{6} = \frac{(k+1)(2k^2 + 7k + 6)}{6}$$$$= \frac{(k+1)(k+2)(2k+3)}{6} = \frac{(k+1)((k+1)+1)(2(k+1)+1)}{6}$$

This is the required formula with $n = k + 1$. The statement is true by induction.

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Additional Sigma Notation Problems

Changing the Index

Sometimes it is useful to change the index of summation:

$$\sum_{i=1}^{n} a_i = \sum_{j=0}^{n-1} a_{j+1} = \sum_{k=2}^{n+1} a_{k-1}$$

Telescoping Series

A telescoping series is one where many terms cancel:

$$\sum_{k=1}^{n} \frac{1}{k(k+1)} = \sum_{k=1}^{n}\left(\frac{1}{k} - \frac{1}{k+1}\right) = 1 - \frac{1}{n+1} = \frac{n}{n+1}$$

Example

Evaluate $\displaystyle\sum_{k=1}^{n} \frac{1}{k(k+2)}$.

Using partial fractions:

$$\frac{1}{k(k+2)} = \frac{1}{2}\left(\frac{1}{k} - \frac{1}{k+2}\right)$$$$\sum_{k=1}^{n} \frac{1}{k(k+2)} = \frac{1}{2}\left(1 + \frac{1}{2} - \frac{1}{n+1} - \frac{1}{n+2}\right) = \frac{1}{2}\left(\frac{3}{2} - \frac{2n+3}{(n+1)(n+2)}\right)$$

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Arithmetic-Geometric Mean Inequality

Statement

For positive real numbers $a$ and $b$:

\frac{a + b}{2} \ge \sqrt`\{ab}`

Equality holds if and only if $a = b$.

Generalisation

For positive real numbers $a_1, a_2, \ldots, a_n$:

$$\frac{a_1 + a_2 + \cdots + a_n}{n} \ge \sqrt[n]{a_1 a_2 \cdots a_n}$$

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Additional Exam-Style Questions

Question 7 (Paper 2 style)

Use the binomial theorem to expand $(1 + x)^4$ and hence evaluate $1.01^4$ to 5 decimal places.

$$(1+x)^4 = 1 + 4x + 6x^2 + 4x^3 + x^4$$

Set $x = 0.01$:

$$1.01^4 = 1 + 4(0.01) + 6(0.0001) + 4(0.000001) + 0.00000001$$ $$= 1 + 0.04 + 0.0006 + 0.000004 + 0.00000001 = 1.04060401$$

To 5 decimal places: $1.04060$.

Question 8 (Paper 2 style)

The sum of the first $n$ terms of a sequence is $S_n = n^2 + 2n$. Find an expression for the $n$-th term $u_n$.

$$u_n = S_n - S_{n-1} = (n^2 + 2n) - ((n-1)^2 + 2(n-1))$$ $$= n^2 + 2n - (n^2 - 2n + 1 + 2n - 2) = n^2 + 2n - n^2 + 1 = 2n + 1$$

Check: $u_1 = S_1 = 3 = 2(1) + 1$. Correct.

This is an arithmetic sequence with first term 3 and common difference 2.

Question 9 (Paper 2 style)

Find the coefficient of $x^4$ in the expansion of $(2 - 3x)^7$.

The general term is $\dbinom{7}{k}2^{7-k}(-3x)^k$.

For $x^4$: $k = 4$.

Coefficient $= \dbinom{7}{4} \cdot 2^3 \cdot (-3)^4 = 35 \times 8 \times 81 = 22680$.

Question 10 (Paper 1 style)

A geometric series has first term $a$ and common ratio $r$. The sum of the first three terms is $28$ and the sum to infinity is $32$. Find $a$ and $r$.

$$a + ar + ar^2 = 28 \quad \mathrm{(1)}$$ $$\frac{a}{1-r} = 32 \implies a = 32(1-r) \quad \mathrm{(2)}$$

Substituting (2) into (1):

$$32(1-r)(1 + r + r^2) = 28$$ $$(1-r)(1+r+r^2) = 1 - r^3 = \frac{28}{32} = \frac{7}{8}$$ $$r^3 = 1 - \frac{7}{8} = \frac{1}{8} \implies r = \frac{1}{2}$$ $$a = 32\left(1 - \frac{1}{2}\right) = 16$$

Verify: $16 + 8 + 4 = 28$. And $S_\infty = 16/(1 - 1/2) = 32$. Correct.

For the A-Level treatment of this topic, see Sequences and Series.

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