Problems combining progressions

AS · P1 · 14 min

The longest series questions on Paper 1 put an arithmetic progression and a geometric progression side by side: "the first, second and fifth terms of an arithmetic progression are the first three terms of a geometric progression", or "scheme A increases by a fixed amount, scheme B by a fixed percentage". Nothing new is needed beyond the arithmetic and geometric formulas and the sum to infinity. What these questions test is translating each sentence into an equation, keeping the two progressions' letters apart, and solving the resulting simultaneous equations. This note gathers the standard patterns.

Telling them apart

Key result

	Arithmetic progression	Geometric progression
Rule	add a common difference $d$	multiply by a common ratio $r$
Recognise	"increases by the same amount"	"increases by the same percentage" or "by a factor"
Test for $p, q, s$ consecutive	$q - p = s - q$ , i.e. $2q = p + s$	$\dfrac{q}{p} = \dfrac{s}{q}$ , i.e. $q^2 = ps$
$n$ th term	$a + (n - 1)d$	$ar^{n-1}$
Sum of $n$ terms	$\tfrac{1}{2}n\left[2a + (n - 1)d\right]$	$\dfrac{a(1 - r^n)}{1 - r}$
Long-run behaviour	grows (or falls) steadily, sum unbounded	if $

The middle term tests are the key tools. Three numbers in AP have the middle one as the mean of the outer two; three numbers in GP have the middle one squared equal to the product of the outer two.

A sequence can be both: a constant sequence $5, 5, 5, \ldots$ is an AP with $d = 0$ and a GP with $r = 1$ . Questions exclude this by saying " $d \neq 0$ " or "the terms are all different", and that condition is usually what lets you divide by $d$ .

Pattern 1: terms of an AP that form a GP

Write the named AP terms with $a$ and $d$ , then apply the GP condition (middle squared equals product of outer terms). The $a^2$ terms always cancel, leaving an equation you can factorise with $d$ as a common factor. Since $d \neq 0$ , divide by $d$ to link $a$ and $d$ ; the common ratio then follows.

AP terms forming a GP

Write each named AP term as $a + (k - 1)d$ .
Apply $(\text{middle})^2 = (\text{first})(\text{third})$ .
Expand; cancel $a^2$ ; factorise out $d$ ; use $d \neq 0$ to get a relation such as $d = 2a$ .
Find the common ratio: second GP term divided by first, simplified using the relation.
Use any given numbers to find $a$ , $d$ and $r$ , then answer the remaining parts with the appropriate formula for each progression.

Pattern 2: shared terms

When two progressions share a first term, or some terms are equal, write one equation per fact. Use different letters for the two progressions ( $a$ , $d$ for the AP; $a$ , $r$ or $b$ , $r$ for the GP) and only use the same letter where the question says the values are equal.

Pattern 3: comparing growth

An AP grows by the same amount each step; a GP with $r > 1$ grows by the same proportion, so it eventually overtakes any AP, even one that starts ahead or grows faster at first. To find when, compare the $n$ th terms (or sums) by trial, since logarithms are not on Paper 1.

y = 20000 + 1500(x - 1) y = 20000(1.06)^(x - 1)

Two salary schemes starting at $20\,000$ : an annual increase of $1500$ (straight line) and an annual increase of $6\%$ (curve). The percentage scheme is behind at first and overtakes in year $10$ .

Worked examples

Three numbers in both progressions

The numbers $p$ , $5$ , $q$ are consecutive terms of an arithmetic progression, and $p$ , $4$ , $q$ are consecutive terms of a geometric progression. Find the possible values of $p$ and $q$ .

Solution

AP: $5$ is the mean of $p$ and $q$ , so $p + q = 10$ .

GP: $4^2 = pq$ , so $pq = 16$ .

Substitute $q = 10 - p$ : $p(10 - p) = 16$ , so $p^2 - 10p + 16 = 0$ and $(p - 2)(p - 8) = 0$ .

p = 2,\ q = 8 \qquad\text{or}\qquad p = 8,\ q = 2

Check: $2, 5, 8$ has difference $3$ , and $2, 4, 8$ has ratio $2$ .

The first, second and fifth terms

An arithmetic progression has first term $3$ and common difference $d$ , where $d \neq 0$ . The first, second and fifth terms of the AP are the first three terms of a geometric progression.

(a) Find $d$ and the common ratio of the GP.

(b) Find the sum of the first $10$ terms of the AP and the sum of the first $5$ terms of the GP.

Solution

(a) The terms are $3$ , $3 + d$ , $3 + 4d$ . GP condition:

\begin{aligned} (3 + d)^2 &= 3(3 + 4d) \\ 9 + 6d + d^2 &= 9 + 12d \\ d^2 - 6d &= 0 \\ d(d - 6) &= 0 \end{aligned}

$d \neq 0$ , so $d = 6$ . The GP is $3, 9, 27, \ldots$ with $r = 3$ .

(b)

S_{10}^{\text{AP}} = \tfrac{10}{2}\left[2(3) + 9(6)\right] = 5 \times 60 = 300, \qquad S_5^{\text{GP}} = \frac{3(3^5 - 1)}{3 - 1} = \frac{3 \times 242}{2} = 363

Shared first term, matching later terms

An arithmetic progression and a geometric progression both have first term $9$ . The second terms of the two progressions are equal, and the third term of the GP equals the fourth term of the AP. Given that neither progression is constant, find the common difference and the common ratio.

Solution

Let the AP have common difference $d$ and the GP common ratio $r$ .

9 + d = 9r, \qquad 9r^2 = 9 + 3d

From the first, $d = 9r - 9$ . Substituting:

9r^2 = 9 + 27r - 27 \quad\Rightarrow\quad 9r^2 - 27r + 18 = 0 \quad\Rightarrow\quad r^2 - 3r + 2 = 0 \quad\Rightarrow\quad (r - 1)(r - 2) = 0

$r = 1$ gives $d = 0$ , a constant progression, which is excluded. So $r = 2$ and $d = 9$ .

Check: AP $9, 18, 27, 36$ ; GP $9, 18, 36$ . Second terms both $18$ , and the GP's third term equals the AP's fourth term, $36$ .

Comparing two salary schemes

A company offers two salary schemes, each starting at $20\,000$ dollars in the first year.

Scheme A: the salary increases by $1500$ dollars each year.

Scheme B: the salary increases by $6\%$ of the previous year's salary each year.

(a) Find the first year in which the Scheme B salary is greater than the Scheme A salary.

(b) Find the total earned under each scheme over the first $10$ years, to the nearest dollar.

Solution

(a) In year $n$ : Scheme A pays $20\,000 + 1500(n - 1)$ and Scheme B pays $20\,000 \times 1.06^{n-1}$ . By trial:

Year $n$	Scheme A	Scheme B
$9$	$32\,000$	$31\,877$
$10$	$33\,500$	$33\,790$

Scheme B is first greater in year $10$ .

(b)

S_{10}^{A} = \tfrac{10}{2}\left[2(20\,000) + 9(1500)\right] = 5 \times 53\,500 = 267\,500

S_{10}^{B} = \frac{20\,000(1.06^{10} - 1)}{0.06} = 263\,616

Even though Scheme B pays more in year $10$ , Scheme A has paid more in total over the ten years, because it was ahead in each of the first nine.

A sum to infinity and an AP sum

A geometric progression has first term $54$ and second term $18$ . An arithmetic progression has first term $1$ and common difference $2$ . Find the number of terms of the AP whose sum equals the sum to infinity of the GP.

Solution

GP: $r = \tfrac{18}{54} = \tfrac{1}{3}$ , and $|r| < 1$ , so $S_\infty = \dfrac{54}{1 - \frac{1}{3}} = 81$ .

AP: $S_n = \tfrac{n}{2}\left[2 + 2(n - 1)\right] = n^2$ .

$n^2 = 81$ , and $n$ must be a positive integer, so $n = 9$ .

The first, fifth and eighth terms

The first, second and third terms of a geometric progression are the first, fifth and eighth terms respectively of an arithmetic progression. The first term of each progression is $64$ and the common difference of the AP is $d$ , where $d \neq 0$ .

(a) Find $d$ and the common ratio of the GP.

(b) Find the sum to infinity of the GP.

Solution

(a) The AP terms are $64$ , $64 + 4d$ , $64 + 7d$ . GP condition:

\begin{aligned} (64 + 4d)^2 &= 64(64 + 7d) \\ 4096 + 512d + 16d^2 &= 4096 + 448d \\ 16d^2 + 64d &= 0 \\ 16d(d + 4) &= 0 \end{aligned}

$d \neq 0$ , so $d = -4$ . The GP terms are $64$ , $48$ , $36$ , so $r = \tfrac{48}{64} = \tfrac{3}{4}$ .

(b) $|r| < 1$ , so $S_\infty = \dfrac{64}{1 - \frac{3}{4}} = 256$ .

(c)

S_n = \tfrac{n}{2}\left[128 + (n - 1)(-4)\right] = \tfrac{n}{2}(132 - 4n) = n(66 - 2n)

For positive $n$ , $S_n < 0$ when $66 - 2n < 0$ , i.e. $n > 33$ . The least value is $n = 34$ . (Check: $S_{33} = 0$ and $S_{34} = 34 \times (-2) = -68$ .)

Watch out

Using the same letter for both progressions. If the AP and GP have different first terms, call them $a$ and $b$ . Using $a$ for both silently assumes they are equal.

Dividing by $d$ without saying why. In $d(d - 6) = 0$ , state " $d \neq 0$ " before discarding $d = 0$ . It is usually the reason the question told you $d \neq 0$ .

Mixing up the conditions. AP: $2q = p + s$ . GP: $q^2 = ps$ . Swapping them is a common slip under time pressure.

Comparing terms when sums are asked (or the reverse). "Which scheme pays more in year $n$ " compares $n$ th terms; "total earned" compares sums.

Rejecting a valid solution. A negative ratio, or a negative common difference, is fine unless the question rules it out.

Exam tip

Write a separate equation for each sentence of the question before solving anything. This is what earns the first method marks, and it makes long questions manageable.
Label which progression each formula belongs to, such as $S_{10}^{\text{AP}}$ and $S_5^{\text{GP}}$ .
Number of terms is a positive integer. If solving gives $n = 9$ or $n = -9$ , reject the negative one; if $n$ is not an integer, re-check the setup or interpret it (for example "least $n$ ").
Trial for comparisons. When asked for the first year one scheme overtakes another, show the values for the two years either side, as in the table above.
Exact or rounded. Money is usually given to the nearest dollar; ratios and differences should be exact fractions where possible.

Summary

AP: add $d$ ; three terms $p, q, s$ satisfy $2q = p + s$ . GP: multiply by $r$ ; three terms satisfy $q^2 = ps$ .
"Same amount" means AP; "same percentage" or "same factor" means GP.
AP terms forming a GP: write them with $a$ and $d$ , apply the GP condition, cancel $a^2$ , factorise, use $d \neq 0$ .
Shared or matching terms: one equation per fact, different letters for each progression unless told equal.
A GP with $r > 1$ eventually overtakes any AP; find when by trial and show values either side.
Combine with $S_\infty = \dfrac{a}{1 - r}$ when the GP converges.

Practice questions

Question

State whether each sequence is an AP, a GP, or neither: (a) $5, 9, 13, \ldots$ (b) $5, 10, 20, \ldots$ (c) $1, 4, 9, \ldots$ (d) $3, -3, 3, \ldots$
The numbers $2$ , $x$ , $y$ are consecutive terms of an arithmetic progression, and $x$ , $y$ , $9$ are consecutive terms of a geometric progression. Find the possible values of $x$ and $y$ .
The second, third and sixth terms of an arithmetic progression with $d \neq 0$ are consecutive terms of a geometric progression. Show that $d = -2a$ , where $a$ is the first term of the AP, and find the common ratio of the GP.
The first, third and ninth terms of an arithmetic progression with first term $2$ and common difference $d \neq 0$ are the first three terms of a geometric progression. (a) Find $d$ and the common ratio. (b) Find the sum of the first $20$ terms of the AP. (c) Find the least $n$ for which the sum of the first $n$ terms of the GP exceeds $1000$ .
Two salary schemes start at $30\,000$ dollars. Scheme A increases by $2000$ dollars each year; Scheme B increases by $5\%$ each year. Find the first year in which Scheme B pays more than Scheme A, and the total paid by each scheme over the first $15$ years.
A geometric progression has first term $40$ and common ratio $\tfrac{1}{2}$ . An arithmetic progression has first term $5$ and common difference $10$ . Find the number of terms of the AP whose sum is equal to the sum to infinity of the GP.
An arithmetic progression and a geometric progression both have first term $12$ . The second term of the GP equals the fourth term of the AP, and the third term of the GP equals the sixth term of the AP. The progressions are not constant. (a) Find the common ratio of the GP and the common difference of the AP. (b) Find the sum to infinity of the GP. (c) Find the value of $n$ for which the sum of the first $n$ terms of the AP is zero.
The three numbers $a$ , $b$ , $c$ are consecutive terms of a GP with $a + b + c = 26$ and $abc = 216$ . Find $b$ , and hence find the possible values of the common ratio.

Answers

(a) AP, $d = 4$ . (b) GP, $r = 2$ . (c) Neither: differences $3, 5$ ; ratios $4, \tfrac{9}{4}$ . (d) GP, $r = -1$ .
AP: $2x = 2 + y$ , so $y = 2x - 2$ . GP: $y^2 = 9x$ . Then $(2x - 2)^2 = 9x$ , so $4x^2 - 17x + 4 = 0$ and $(4x - 1)(x - 4) = 0$ . $x = 4$ , $y = 6$ (AP $2, 4, 6$ ; GP $4, 6, 9$ ), or $x = \tfrac{1}{4}$ , $y = -\tfrac{3}{2}$ (AP $2, \tfrac{1}{4}, -\tfrac{3}{2}$ ; GP $\tfrac{1}{4}, -\tfrac{3}{2}, 9$ with $r = -6$ ).
Terms $a + d$ , $a + 2d$ , $a + 5d$ . $(a + 2d)^2 = (a + d)(a + 5d)$ gives $a^2 + 4ad + 4d^2 = a^2 + 6ad + 5d^2$ , so $0 = 2ad + d^2 = d(2a + d)$ . Since $d \neq 0$ , $d = -2a$ . Ratio $= \dfrac{a + 2d}{a + d} = \dfrac{a - 4a}{a - 2a} = \dfrac{-3a}{-a} = 3$ .
(a) $(2 + 2d)^2 = 2(2 + 8d)$ gives $4 + 8d + 4d^2 = 4 + 16d$ , so $4d^2 - 8d = 0$ , $4d(d - 2) = 0$ , $d = 2$ . GP: $2, 6, 18$ , so $r = 3$ . (b) $S_{20} = 10\left[4 + 19(2)\right] = 420$ . (c) $S_n = \dfrac{2(3^n - 1)}{2} = 3^n - 1 > 1000$ , so $3^n > 1001$ . $3^6 = 729$ and $3^7 = 2187$ , so $n = 7$ .
Year $n$ : A pays $30\,000 + 2000(n - 1)$ , B pays $30\,000 \times 1.05^{n-1}$ . Year $13$ : A $54\,000$ , B $53\,876$ . Year $14$ : A $56\,000$ , B $56\,569$ . So year $14$ . Totals: $S_{15}^{A} = \tfrac{15}{2}\left[60\,000 + 14(2000)\right] = 660\,000$ ; $S_{15}^{B} = \dfrac{30\,000(1.05^{15} - 1)}{0.05} = 647\,357$ (to the nearest dollar).
$S_\infty = \dfrac{40}{1 - \frac{1}{2}} = 80$ . AP: $S_n = \tfrac{n}{2}\left[10 + 10(n - 1)\right] = 5n^2$ . $5n^2 = 80$ , so $n = 4$ .
(a) $12r = 12 + 3d$ and $12r^2 = 12 + 5d$ . From the first, $d = 4r - 4$ . Then $12r^2 = 12 + 20r - 20$ , so $3r^2 - 5r + 2 = 0$ , $(3r - 2)(r - 1) = 0$ . $r = 1$ gives $d = 0$ (constant), so $r = \tfrac{2}{3}$ and $d = -\tfrac{4}{3}$ . (b) $S_\infty = \dfrac{12}{1 - \frac{2}{3}} = 36$ . (c) $S_n = \tfrac{n}{2}\left[24 - \tfrac{4}{3}(n - 1)\right] = 0$ with $n > 0$ gives $\tfrac{4}{3}(n - 1) = 24$ , so $n = 19$ .
In a GP, $b^2 = ac$ , so $abc = b^3 = 216$ and $b = 6$ . With $a = \tfrac{6}{r}$ and $c = 6r$ : $\tfrac{6}{r} + 6 + 6r = 26$ , so $6r^2 - 20r + 6 = 0$ , $3r^2 - 10r + 3 = 0$ , $(3r - 1)(r - 3) = 0$ . $r = 3$ (terms $2, 6, 18$ ) or $r = \tfrac{1}{3}$ (terms $18, 6, 2$ ).

Telling them apart

Pattern 1: terms of an AP that form a GP

Pattern 2: shared terms

Pattern 3: comparing growth

Worked examples

Practice questions

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