AP and GP (Progressions)

SSC CGL · CHSL · CPO — method by method

3 Part37 worked examples144 practice questions

A progression is a list of numbers in which every term is made from the one before by a fixed rule. Two rules are asked most: adding the same number each time (arithmetic progression, AP) and multiplying by the same number each time (geometric progression, GP). The third, the harmonic progression, is really an AP seen upside down.

In every method we first see where the formula comes from, so you need not memorise it. The basic summing tricks (sums from 1 to $n$ , cancelling series) are in the earlier chapter Series and Simplification Tricks.

How to useRead each method’s “idea” first, then the solved examples, then try the practice questions yourself and open the solution afterwards.

Arithmetic progression (AP)

Adding the same number each time

AP terms, sums, the middle term, inserting means, condition puzzles and counting multiples.

In an AP the difference between every two neighbouring terms is the same — the common difference $d$ . The first term is $a$ . These two numbers decide the whole AP.

Terms, counting and sums

T_{n}

n

S_{n}

The $n$ th term, the number of terms, the sum of the first $n$ terms, a term from the end and story questions.

The idea

In the AP $a, a + d, a + 2 d, \dots$ the second term is $a$ plus one $d$ , the third plus two $d$ ’s. So the $n$ th term is $n - 1$ steps away: $T_{n} = a + (n - 1) d$ . (The first term needs no step, which is why it is $n - 1$ , not $n$ .) When the first and last terms are known, the number of terms is $n = \frac{last - first}{d} + 1$ — like fence posts, always one more than the gaps.

◭ Diagram

the height grows by

d = 3

each time; first+last = second+second-last = … and the average is the middle height

For the sum use Gauss’s pairing: first+last, second+second-last, … every pair adds to the same number (when one term goes up by $d$ the other goes down by $d$ ). So $S_{n} = \frac{n}{2} \times (first + last)$ . Putting $a + (n - 1) d$ for the last term gives the other form.

∑ Formula

T_{n} = a + (n - 1) d

The

n

th term and the number of terms

$a$: first term
$d$: common difference = any term − the term before it
$n$: the term number (or number of terms)
$l$: last term

∑ Formula

n = \frac{l - a}{d} + 1

The

n

th term and the number of terms

$a$: first term
$d$: common difference = any term − the term before it
$n$: the term number (or number of terms)
$l$: last term

∑ Formula

S_{n} = \frac{n}{2} [2 a + (n - 1) d] = \frac{n}{2} (a + l)

The sum of the first

n

terms

$a, d, n$: as above
$l$: the $n$ th (last) term

Spot itWhenever numbers grow or fall by equal steps (rent, savings, rows of seats, rungs) it is an AP.

Steps

Identify $a$ , $d$ and which third thing ( $n$ , $l$ or the sum) is given.
$d$ = any term − the one before (it may be negative).
For a term use $a + (n - 1) d$ ; for a count use $\frac{l - a}{d} + 1$ ; for a sum use $\frac{n}{2} (a + l)$ .
Check: the other formula should give the same answer.

Worked examples

Example 1 · Basic

Find the 25th term of the progression

8, 14, 20, 26, \dots

Solution, step by step

1The common difference $d$ = any term − the term before it = $14 - 8 = 6$ . (Using the next term 20 gives the same, so it really is an AP.)
2From the first term 8 to the 25th term we take 24 steps (the first term needs no step), each adding $6$ : $T_{25} = a + (n - 1) d = 8 + 24 \times 6$ .
3 $8 + 144 = 152$ .

Answer $152$

Example 2 · Basic

Which term of the progression

11, 17, 23, 29, \dots

is 215?

Solution, step by step

1The first term is 11 and the common difference 6. To reach 215 we must cover $215 - 11 = 204$ .
2Each step is 6, so the number of steps = $\frac{204}{6} = 34$ .
3After 34 steps we stand on the 35th term (the starting term counts too), so 215 is the 35th term.

Answer 35th

Example 3 · Standard

Find the sum:

8 + 15 + 22 + \dots + 190

Solution, step by step

1The common difference is $d = 7$ . The number of terms is $n = \frac{190 - 8}{7} + 1 = 27$ .
2When the first and last terms are known use $S_{n} = \frac{n}{2} (a + l)$ . Reason: first+last, second+second-last, … every pair has the same sum ( $8 + 190 = 198$ ), so the total = the number of terms × half of (first+last).
3 $S = \frac{27}{2} (8 + 190) = \frac{27}{2} \times 198 = 2673$ .

Answer $2673$

First the number of terms, then the sum: two stages.

Example 4 · Standard

The first row of a hall has 22 seats and every next row has 3 more seats than the one before. How many seats are there in 18 rows altogether?

Solution, step by step

1The same amount is added every time, so this is an AP: $a = 22, d = 3, n = 18$ .
2The last term is $T_{18} = 22 + 17 \times 3 = 73$ .
3 $S_{18} = \frac{18}{2} (22 + 73) = 855$ .

Answer $855$

Example 5 · Advanced

Find the 9th term from the end of

6, 13, 20, \dots, 279

Solution, step by step

1Look from the other end: the first term is 279 and the common difference is $- d = - 7$ .
2 $T = 279 + (9 - 1) \times (- 7) = 279 - 56$ .
3 $223$ . (Another way: there are 40 terms, so the 9th from the end is the 32nd from the start: $6 + 31 \times 7 = 223$ .)

Answer $223$

Practice

BasicTry it yourself first

Find the 22nd term of the progression

- 15, - 11, - 7, - 3, \dots

BasicTry it yourself first

Which term of the progression

5, 13, 21, 29, \dots

is 205?

BasicTry it yourself first

How many terms are there in

4, 13, 22, \dots, 319

BasicTry it yourself first

Find the sum of the first 24 terms of

60, 65, 70, 75, \dots

StandardTry it yourself first

Find the 18th term of the progression

120, 113, 106, 99, \dots

StandardTry it yourself first

How many terms are there in

250, 245, 240, \dots, - 100

StandardTry it yourself first

Find the sum:

12 + 17 + 22 + \dots + 167

StandardTry it yourself first

Find the sum of the first 30 terms of

25, 23, 21, 19, \dots

StandardTry it yourself first

Naseem saves ₹400 in the first month and ₹50 more than the previous month in each following month. How much does Naseem save in 15 months altogether (in ₹)?

StandardTry it yourself first

A ladder has 15 rungs. The lowest rung is 90 cm long and the topmost 48 cm, with the rungs in between changing in equal steps. What is the total length of all rungs (in cm)?

AdvancedTry it yourself first

Find the 12th term from the end of

10, 14, 18, \dots, 206

AdvancedTry it yourself first

Find the sum:

3 + 11 + 19 + \dots + 299

AdvancedTry it yourself first

A company pays an employee ₹18,000 a month in the first year and raises it by ₹750 a month every year. What will the monthly pay be in year 12 (in ₹)?

AdvancedTry it yourself first

Which term of the progression

- 20, - 17, - 14, - 11, \dots

is 100?

TrapTrap: (1) writing

T_{n} = a + n d

(it is

n - 1

); (2) forgetting the

+ 1

in the count; (3) dropping the minus sign of

d

in a falling AP; (4) putting the last term where the number of terms should go in the sum.

Middle term, smart assumptions and means

T_{n}

from

S_{n}

, inserting means

The middle term of an odd number of terms, assuming three or four terms, a term from $S_{n}$ , inserting means and averages.

The idea

The middle term = the average. If an AP has an odd number of terms there are equal numbers of terms on each side of the middle one and the ups and downs make up for each other in pairs. So the total = number of terms × the middle term, and the middle term = total ÷ number of terms. (With an even number of terms, the average of the two middle ones.)

The assumption trick: take three terms as $a - d, a, a + d$ and four as $a - 3 d, a - d, a + d, a + 3 d$ ; then $d$ cancels in the sum and $a$ comes out at once. Likewise five terms $a - 2 d, \dots, a + 2 d$ .

∑ Formula

T_{n} = S_{n} - S_{n - 1}

T_{n}

from

S_{n}

, and the difference after inserting

k

means

$S_{n}$: the sum of the first $n$ terms
$T_{n}$: the $n$ th term
$A, B$: the two numbers between which means are inserted
$k$: the number of means inserted

∑ Formula

d = \frac{B - A}{k + 1}

T_{n}

from

S_{n}

, and the difference after inserting

k

means

$S_{n}$: the sum of the first $n$ terms
$T_{n}$: the $n$ th term
$A, B$: the two numbers between which means are inserted
$k$: the number of means inserted

If $S_{n} = p n^{2} + q n$ is given then $T_{n} = S_{n} - S_{n - 1} = 2 p n + (q - p)$ : so $a = p + q$ and $d = 2 p$ . After inserting $k$ means there are $k + 2$ terms in all and $k + 1$ steps from first to last, so $d = \frac{B - A}{k + 1}$ . The average of the multiples in a range = (first multiple + last multiple) ÷ 2.

Spot itWords like “sum of consecutive numbers”, “three/four numbers in AP”, “the sum formula is given”, “insert numbers between”, “average” call for this method.

Steps

For consecutive numbers find the gap (integers 1, even/odd 2, multiples of 5 gap 5).
With an odd count: middle = total ÷ count; then reach the end terms by counting steps from the middle.
For sum-and-product questions take three or four terms symmetrically.
In $S_{n}$ questions take $S_{n} - S_{n - 1}$ .

Worked examples

Example 6 · Basic

The sum of 9 consecutive even numbers is 324. Find the largest of them.

Solution, step by step

1The 9 numbers are 2 apart, so they form an AP. The count is odd, so the middle term equals the average: $middle term = \frac{324}{9} = 36$ .
2There are 4 terms on each side of the middle one, so the largest is 4 steps (each 2) from it: $36 + 4 \times 2 = 44$ .
3The numbers run from 28 to 44; check: 9 × 36 = 324.

Answer $44$

Example 7 · Standard

Three numbers in AP have sum 36 and product 1428. Find the largest.

Solution, step by step

1Take the three terms as $a - d, a, a + d$ . With this choice $d$ cancels in the sum: $3 a = 36$ , so $a = 12$ .
2Product: $(a - d) a (a + d) = a (a^{2} - d^{2}) = 12 (144 - d^{2}) = 1428$ , so $144 - d^{2} = 119$ and $d^{2} = 25$ .
3 $d = \pm 5$ . The numbers are 7, 12, 17; the largest is $17$ .

Answer $17$

Example 8 · Standard

The sum of the first

n

terms of an AP is

S_{n} = 3 n^{2} + 5 n

. Find its 12th term.

Solution, step by step

1The $n$ th term = (sum of first $n$ ) − (sum of first $n - 1$ ): $T_{n} = S_{n} - S_{n - 1}$ .
2Replace $n$ by $(n - 1)$ in $S_{n}$ to get $S_{n - 1} = 3 (n - 1)^{2} + 5 (n - 1)$ , and subtract: $T_{n} = 6 n + 2$ .
3 $T_{1} = 8$ (first term), $T_{2} - T_{1} = 6$ (common difference); $T_{12} = 6 \times 12 + 2 = 74$ .

Answer $74$

Example 9 · Advanced

6 arithmetic means are inserted between 9 and 51 so that the whole list is an AP. Find the 4th mean.

Solution, step by step

1After inserting 6 means, 9 is the first term and 51 the last, 8 terms in all, so there are 7 steps from first to last.
2 $d = \frac{51 - 9}{7} = 6$ .
3The 4th mean is 4 steps from 9: $9 + 4 \times 6 = 33$ .

Answer $33$

Practice

BasicTry it yourself first

The sum of 13 consecutive integers is 650. Find the largest of them.

BasicTry it yourself first

The sum of 11 consecutive multiples of 5 is 825. Find the largest of them.

BasicTry it yourself first

9 arithmetic means are inserted between 4 and 64 so that the whole list is an AP. Find the 6th mean.

BasicTry it yourself first

Find the average of all multiples of 7 between 50 and 500 (both included).

StandardTry it yourself first

The sum of 9 consecutive odd numbers is 567. Find the largest of them.

StandardTry it yourself first

Three numbers in AP have sum 27 and product 585. Find the largest.

StandardTry it yourself first

The sum of the first

n

terms of an AP is

S_{n} = 2 n^{2} - n

. Find its common difference.

StandardTry it yourself first

The sum of the first

n

terms of an AP is

S_{n} = 4 n^{2} + 3 n

. Find its 15th term.

StandardTry it yourself first

Find the average of all multiples of 9 between 100 and 1000 (both included).

AdvancedTry it yourself first

Four numbers in AP add up to 40, and the product of the first and the last is 91. Find the largest.

AdvancedTry it yourself first

Four numbers in AP add up to 60, and the product of the first and the last is 189. Find the largest.

AdvancedTry it yourself first

5 arithmetic means are inserted between -10 and 50 so that the whole list is an AP. Find the 3rd mean.

AdvancedTry it yourself first

The sum of 11 consecutive even numbers is 990. Find the smallest of them.

AdvancedTry it yourself first

The sum of the first

n

terms of an AP is

S_{n} = 5 n^{2} - 2 n

. Find its first term.

TrapTrap: (1) consecutive even/odd numbers differ by 2, not 1; (2) after inserting

k

means there are

k + 2

terms, not

k

; (3) with four assumed terms neighbours are

2 d

apart; (4) forgetting to replace

n

(n - 1)

when forming

S_{n - 1}

Condition puzzles

Think before you use algebra

$S_{p} = S_{q}$ , terms at symmetric positions, $T_{p} = q, T_{q} = p$ , ratios of sums, and the greatest sum.

The idea

These questions give some terms or sums and hide a special property of APs. Understand why each property holds and every question takes two lines.

Condition	Result	Reason
$S_{p} = S_{q}$	$S_{p + q} = 0$ , $T_{p + q} = - a$	it gives $2 a + (p + q - 1) d = 0$
positions $i + j = N + 1$	$a_{i} + a_{j} = a_{1} + a_{N}$	one term gains $d$ while the other loses $d$
$T_{p} = q, T_{q} = p$	$d = - 1$ , $T_{p + q} = 0$	by subtracting
$\frac{S _{n}}{S _{n}^{'}} = \frac{c _{1} n + d _{1}}{c _{2} n + d _{2}}$	$\frac{T _{m}}{T _{m}^{'}} = \frac{c _{1} ( 2 m - 1 ) + d _{1}}{c _{2} ( 2 m - 1 ) + d _{2}}$	twice the $m$ th term = $2 a + (n - 1) d$ at $n = 2 m - 1$
AP with $d < 0$	the sum grows while terms are positive	negative terms take away

Here: $S_{p}$ = the sum of the first $p$ terms; $T_{p}$ = the $p$ th term; $a$ , $d$ = first term and common difference; $a_{i}$ = the $i$ th term; $N$ = the number of terms whose sum is wanted; $S_{n}$ , $S_{n}^{'}$ = the sums of the first $n$ terms of two APs; $c_{1}, d_{1}, c_{2}, d_{2}$ = the numeric coefficients in the ratio; $m$ = the term number whose ratio is wanted.

◭ Diagram

terms whose positions add to

N + 1

(the same-colour pairs) have the same sum

Spot itUse this method when the question says “

S_{p} = S_{q}

”, “

a_{3} + a_{8} + \dots =

”, “ratio

(5 n + 4) : (9 n + 6)

” or “how many terms add to zero”.

Steps

Identify which row of the table the condition resembles.
Apply that row’s rule (find $d$ or $a_{1} + a_{N}$ ).
Put it into the formula for what is asked ( $T$ or $S$ ).
Check the rule with a small example AP.

Worked examples

Example 10 · Basic

The 8th term of an AP is 13 and its 13th term is 8. Find its 21st term.

Solution, step by step

1 $T_{13} - T_{8} = 8 - 13 = - 5$ and this is the gap of 5 steps, so $d = \frac{- 5}{5} = - 1$ .
2From $T_{8} = a + 7 d = 13$ we get $a = 13 + 7 = 20$ .
3 $T_{21} = 20 + 20 \times (- 1) = 0$ .

Answer $0$

Example 11 · Standard

In an AP the sum of the first 9 terms equals the sum of the first 16 terms. If its first term is 36, find the 25th term.

Solution, step by step

1From $S_{9} = S_{16}$ : $\frac{9}{2} [2 a + 8 d] = \frac{16}{2} [2 a + 15 d]$ . Bringing everything to one side and factorising differences like $9^{2} - 1 6^{2}$ gives $(9 - 16) [2 a + (9 + 16 - 1) d] = 0$ .
2Since 9≠16, the factor $(9 - 16)$ is not zero, so the bracket must be: $2 a + (9 + 16 - 1) d = 0$ , i.e. $d = \frac{- 2 a}{24} = \frac{- 72}{24} = - 3$ .
3 $T_{25} = a + 24 d = 36 + 24 \times (- 3) = - 36$ . (In every such AP $T_{9 + 16} = - a$ and $S_{9 + 16} = 0$ .)

Answer $- 36$

Example 12 · Standard

For the terms of an AP,

a_{3} + a_{8} + a_{12} + a_{19} + a_{23} + a_{28} = 192

. Find the sum of its first 30 terms.

Solution, step by step

1Look at the positions in pairs: 3+28=31, 8+23=31, 12+19=31. In each pair the two terms are equally far from the first and the 30th term.
2Each such pair adds up to $a_{1} + a_{30}$ (in an AP a pair’s sum does not change when one term goes up by $d$ and the other down by $d$ ). There are 3 pairs, so $3 (a_{1} + a_{30}) = 192$ , i.e. $a_{1} + a_{30} = 64$ .
3 $S_{30} = \frac{30}{2} (a_{1} + a_{30}) = \frac{30}{2} \times 64 = 960$ .

Answer $960$

Example 13 · Advanced

The ratio of the sums of the first n terms of two APs is

(5 n + 4) : (9 n + 6)

. What is the ratio of their 7th terms?

Solution, step by step

1Since $S_{n} = \frac{n}{2} [2 a + (n - 1) d]$ , the ratio of sums is $\frac{2 a + ( n - 1 ) d}{2 a ^{'} + ( n - 1 ) d ^{'}}$ (the halves and n cancel). This $2 a + (n - 1) d$ is just twice the middle term of the n-term AP.
2The 7th term is $a + (m - 1) d$ ; twice that is $2 a + (m - 1) \cdot 2 d$ — which is $2 a + (n - 1) d$ with $n - 1 = 2 (m - 1)$ , i.e. $n = 2 m - 1 = 13$ .
3Then $\frac{5 ( 13 ) + 4}{9 ( 13 ) + 6} = \frac{69}{123} = 23 : 41$ .

Answer $23 : 41$

Example 14 · Advanced

How many terms of

30, 26, 22, 18, \dots

are needed for the sum to be zero?

Solution, step by step

1 $a = 30, d = - 4$ . Put $S_{n} = \frac{n}{2} [2 a + (n - 1) d]$ equal to 0.
2 $n$ cannot be zero, so the bracket must be zero: $2 (30) + (n - 1) (- 4) = 0$ , i.e. $n - 1 = \frac{- 60}{- 4} = 15$ .
3 $n = 16$ . (The first 8 terms are positive; later negative terms bring the sum back to 0.)

Answer $16$

Practice

BasicTry it yourself first

The 11th term of an AP is 17 and its 17th term is 11. Find its 28th term.

BasicTry it yourself first

How many terms of

40, 35, 30, 25, \dots

are needed for the sum to be zero?

BasicTry it yourself first

The 7th term of an AP is 26 and the 12th term is 51. Find its 25th term.

StandardTry it yourself first

In an AP the sum of the first 14 terms equals the sum of the first 21 terms. If its first term is 51, find the 35th term.

StandardTry it yourself first

In an AP the sum of the first 8 terms equals the sum of the first 13 terms. If its first term is 30, find the 21st term.

StandardTry it yourself first

For the terms of an AP,

a_{5} + a_{11} + a_{16} + a_{25} + a_{30} + a_{36} = 270

. Find the sum of its first 40 terms.

StandardTry it yourself first

How many terms of

26, 22, 18, 14, \dots

are needed for the sum to be zero?

StandardTry it yourself first

What is the largest sum that terms of

37, 33, 29, 25, \dots

can reach?

AdvancedTry it yourself first

For the terms of an AP,

a_{2} + a_{9} + a_{16} + a_{23} = 130

. Find the sum of its first 24 terms.

AdvancedTry it yourself first

The ratio of the sums of the first n terms of two APs is

(3 n + 8) : (5 n + 2)

. What is the ratio of their 10th terms?

AdvancedTry it yourself first

The ratio of the sums of the first n terms of two APs is

(7 n + 1) : (3 n + 5)

. What is the ratio of their 12th terms?

AdvancedTry it yourself first

What is the largest sum that terms of

47, 41, 35, 29, \dots

can reach?

TrapTrap: (1) dividing in

S_{p} = S_{q}

only because

p \neq = q

— remember that is the reason; (2) putting

m

instead of

2 m - 1

in the ratio question; (3) missing a pair when counting symmetric pairs; (4) taking the zero-sum and greatest-sum counts to be the same in a falling AP.

Counting multiples, their sums and digits

“or”, “and”, “neither … nor”

Counting multiples in a range, multiples of two numbers (LCM), their sums and the digits of page numbers.

The idea

The multiples of $k$ , $k, 2 k, 3 k, \dots$ , form an AP with difference $k$ . To count them from $L$ to $U$ : find the smallest multiple that is $L$ or more and the largest that is $U$ or less, then $n = \frac{last - first}{k} + 1$ . Or directly: $⌊ \frac{U}{k} ⌋ - ⌊ \frac{L - 1}{k} ⌋$ .

For two numbers $a$ and $b$ : “ $a$ or $b$ ” = multiples of $a$ + multiples of $b$ − the multiples of both (counted twice); numbers divisible by both are the multiples of the LCM (not of the product, unless $a, b$ are co-prime). “Neither $a$ nor $b$ ” = total − (“ $a$ or $b$ ”). Sums work the same way: $S_{a} + S_{b} - S_{LCM}$ .

∑ Formula

n (A \cup B) = n (A) + n (B) - n (A \cap B)

Counting “or” (inclusion–exclusion)

$n (A)$: the number of multiples of $a$
$n (B)$: the number of multiples of $b$
$n (A \cap B)$: the number of common (LCM) multiples

Digits of page numbers: pages 1–9 are 9 pages of 1 digit, 10–99 are 90 pages of 2 digits, 100–999 are 900 of 3 digits, then 4 digits. Total digits = the sum of these blocks; in reverse, subtract whole blocks first, then divide what is left by the digits of the last block.

Spot itUse this for “how many numbers divisible by

a

b

”, “by

a

but not

b

”, “by neither”, “the sum of all 3-digit numbers that …”, “how many digits are printed”.

Steps

For each multiplier find the first and last multiples.
Count: $\frac{last - first}{k} + 1$ ; sum: $\frac{n}{2} (first + last)$ .
With two numbers find the LCM and use inclusion–exclusion.
Finally check plausibility (a count cannot exceed the total).

Worked examples

Example 15 · Basic

How many multiples of 13 are there from 100 to 999 (both included)?

Solution, step by step

1The smallest multiple that is 100 or more: 104 (round 100 ÷ 13 up). The largest that is 999 or less: 988.
2They form an AP with difference 13: $n = \frac{988 - 104}{13} + 1 = 69$ .
3Check: $⌊ \frac{999}{13} ⌋ - ⌊ \frac{99}{13} ⌋ = 76 - 7 = 69$ — dividing directly gives the same.

Answer $69$

Example 16 · Standard

How many numbers from 100 to 500 (both included) are divisible by 6 or 8?

Solution, step by step

1Multiples of 6: $⌊ \frac{500}{6} ⌋ - ⌊ \frac{99}{6} ⌋ = 83 - 16 = 67$ ; multiples of 8: $62 - 12 = 50$ .
2Numbers divisible by both 6 and 8 are multiples of their LCM 24 ( $LCM (6, 8) = 24$ ; not the product 48 unless 6, 8 are co-prime): $20 - 4 = 16$ .
3The numbers divisible by both are counted twice in “or”, so subtract them once: $67 + 50 - 16 = 101$ .

Answer $101$

Trap: not 48 — the LCM of 6 and 8 is 24.

Example 17 · Standard

Find the sum of all numbers from 100 to 999 that are divisible by 9.

Solution, step by step

1First multiple 108, last multiple 999, common difference 9.
2Number of terms: $n = \frac{999 - 108}{9} + 1 = 100$ .
3 $S = \frac{100}{2} (108 + 999) = 55, 350$ .

Answer $55, 350$

Example 18 · Advanced

Find the sum of all numbers from 1 to 100 that are divisible by 4 or 10.

Solution, step by step

1The multiples of 4 add up to $1300$ and those of 10 to $550$ (each AP by $\frac{n}{2} (first + last)$ ).
2The multiples of the LCM 20 are in both lists, so they were added twice. Their sum is $300$ .
3 $1300 + 550 - 300 = 1550$ .

Answer $1550$

Example 19 · Advanced

A total of 966 digits were used to print the page numbers of a book. How many pages does it have?

Solution, step by step

1Pages 1 to 99 use $9 + 180 = 189$ digits, leaving 777 digits.
2After page 99 each page needs 3 digits: $\frac{777}{3} = 259$ pages.
3Total pages $99 + 259 = 358$ .

Answer $358$

Practice

BasicTry it yourself first

How many multiples of 7 are there from 50 to 500 (both included)?

BasicTry it yourself first

How many digits are used to print the page numbers of a book of 425 pages?

BasicTry it yourself first

How many numbers from 100 to 500 (both included) are divisible by both 6 and 8?

StandardTry it yourself first

How many multiples of 17 are there from 200 to 1200 (both included)?

StandardTry it yourself first

How many numbers from 150 to 450 (both included) are divisible by 4 or 6?

StandardTry it yourself first

How many numbers from 1 to 200 (both included) are divisible by neither 4 nor 10?

StandardTry it yourself first

Find the sum of all numbers from 200 to 900 that are divisible by 7.

StandardTry it yourself first

How many digits are used to print the page numbers of a book of 1250 pages?

AdvancedTry it yourself first

How many numbers from 100 to 500 (both included) are divisible by 6 but not by 8?

AdvancedTry it yourself first

How many numbers from 50 to 900 (both included) are divisible by 9 or 12?

AdvancedTry it yourself first

Find the sum of all numbers from 100 to 999 that are divisible by 13.

AdvancedTry it yourself first

Find the sum of all numbers from 50 to 400 that are divisible by 6 or 9.

AdvancedTry it yourself first

A total of 3101 digits were used to print the page numbers of a book. How many pages does it have?

TrapTrap: (1) the LCM, not the product, for “and/both”; (2) taking a first multiple below

L

or a last one above

U

; (3) forgetting the total

U - L + 1

in “neither”; (4) forgetting the 189 digits before page 100.

Geometric progression (GP)

Multiplying by the same number each time

GP terms, sums, infinite sums, recurring decimals, uses, and AM-GM-HM with harmonic progressions.

In a GP the ratio of every two neighbouring terms is the same — the common ratio $r$ . The first term is $a$ . Where an AP takes “steps”, a GP takes “leaps” (multiplications), so GP terms grow (or shrink) very fast.

GP terms and sums

T_{n} = a r^{n - 1}

and

S_{n}

The $n$ th term, the number of terms, the sum, assuming three terms and inserting geometric means.

The idea

In the GP $a, a r, a r^{2}, \dots$ the $n$ th term comes from multiplying by $r$ a total of $n - 1$ times (the first term is already there): $T_{n} = a r^{n - 1}$ . The sum formula comes from a neat trick: multiply $S_{n} = a + a r + \dots + a r^{n - 1}$ by $r$ to get $r S_{n} = a r + a r^{2} + \dots + a r^{n}$ ; subtract, and all the middle terms cancel, leaving $S_{n} - r S_{n} = a - a r^{n}$ .

∑ Formula

T_{n} = a r^{n - 1}

The term and the sum of a GP

$a$: first term
$r$: common ratio = any term ÷ the term before it
$n$: term number or number of terms

∑ Formula

S_{n} = \frac{a ( r ^{n} - 1 )}{r - 1} = \frac{a ( 1 - r ^{n} )}{1 - r} (r \neq = 1)

The term and the sum of a GP

$a$: first term
$r$: common ratio = any term ÷ the term before it
$n$: term number or number of terms

Use the first form when $r > 1$ and the second when $r < 1$ — they are equal, only the denominator stays positive. With negative $r$ the signs of the terms alternate ( $3, - 6, 12, \dots$ ). To assume three terms take $\frac{a}{r}, a, a r$ ( $r$ cancels in the product). Inserting geometric means: after $k$ means there are $k + 1$ leaps, so $r^{k + 1} = \frac{B}{A}$ .

Spot itWhenever numbers change by the same factor each time (doubling, halving, 10% growth) it is a GP.

Steps

$r$ = any term ÷ the one before (check with two pairs).
When two terms are given their quotient is $r^{gap}$ .
For a term $a r^{n - 1}$ ; for the sum the suitable form of the formula.
If the number of terms is asked, match $a r^{n - 1} = l$ against powers of $r$ .

Worked examples

Example 20 · Basic

Find the 8th term of the geometric progression (GP)

5, 15, 45, 135, \dots

Solution, step by step

1The common ratio $r$ = any term ÷ the term before it = $\frac{15}{5} = 3$ . (The next pair gives the same: $\frac{45}{15} = 3$ .)
2From the first term to the 8th we multiply by $r$ 7 times: $T_{8} = a r^{n - 1} = 5 \times 3^{7}$ .
3 $3^{7} = 2187$ , so $T_{8} = 10, 935$ .

Answer $10, 935$

Example 21 · Standard

The 3rd term of a GP is 18 and the 6th term is 486 (all terms positive). Find its 9th term.

Solution, step by step

1From the 3rd to the 6th term we multiply by $r$ 3 times, so $r^{3} = \frac{T _{6}}{T _{3}} = 27$ .
2For positive terms $r = 3$ (check: $3^{3} = 27$ ).
3The 9th term is 3 steps after the 6th: $T_{9} = 486 \times 3^{3} = 13, 122$ .

Answer $13, 122$

Example 22 · Standard

Find the sum of the first 6 terms of the GP

729, 243, 81, 27, \dots

Solution, step by step

1 $a = 729, r = \frac{1}{3}, n = 6$ . $r$ is less than 1, so it is convenient to use $S_{n} = \frac{a ( 1 - r ^{n} )}{1 - r}$ (the denominator stays positive).
2 $r^{6} = (\frac{1}{3})^{6} = \frac{1}{729}$ .
3 $S_{6} = \frac{729 ( 1 - \frac{1}{729} )}{1 - ( \frac{1}{3} )} = 1092$ .

Answer $1092$

For

r < 1

the form

\frac{a ( 1 - r ^{n} )}{1 - r}

is more convenient.

Example 23 · Advanced

Three positive numbers are in GP. Their product is 216 and their sum is 26. Find the largest.

Solution, step by step

1Take the three as $\frac{a}{r}, a, a r$ . In the product $r$ cancels: $a^{3} = 216$ , so $a = 6$ .
2Sum: $\frac{6}{r} + 6 + 6 r = 26$ . Multiply by $r$ : $6 r^{2} + 6 r + 6 = 26 r$ , i.e. $6 r^{2} - 20 r + 6 = 0$ .
3Quadratic formula: $r = \frac{20 \pm 2 0 ^{2} - 4 ( 6 ) ( 6 )}{2 \times 6} = \frac{20 \pm 16}{12}$ , i.e. $r = 3$ or $r = \frac{1}{3}$ .
4Both values give the same three numbers (in reverse order): 2, 6, 18. The largest is $18$ .

Answer $18$

Practice

BasicTry it yourself first

Find the 9th term of the geometric progression (GP)

3, - 6, 12, - 24, \dots

BasicTry it yourself first

Find the sum of the first 7 terms of the GP

4, 12, 36, 108, \dots

BasicTry it yourself first

How many terms does the GP

2, 6, 18, \dots, 1458

have, and what is the sum of them all? (Give the sum.)

BasicTry it yourself first

Find the 7th term of the geometric progression (GP)

4096, 1024, 256, 64, \dots

StandardTry it yourself first

The 4th term of a GP is 40 and the 7th term is 320 (all terms positive). Find its 10th term.

StandardTry it yourself first

Find the sum of the first 8 terms of the GP

5, - 10, 20, - 40, \dots

StandardTry it yourself first

How many terms does the GP

5, 10, 20, \dots, 2560

have, and what is the sum of them all? (Give the sum.)

StandardTry it yourself first

Insert 3 geometric means between 3 and 768 (all positive). Find the 2nd mean.

StandardTry it yourself first

Find the sum of the first 12 terms of the GP

1, 2, 4, 8, \dots

AdvancedTry it yourself first

Three positive numbers are in GP. Their product is 64 and their sum is 14. Find the largest.

AdvancedTry it yourself first

Three positive numbers are in GP. Their product is 729 and their sum is 39. Find the largest.

AdvancedTry it yourself first

Insert 5 geometric means between 2 and 1458 (all positive). Find the 4th mean.

AdvancedTry it yourself first

The 2nd term of a GP is 21 and the 4th term is 189 (all terms positive). Find its 7th term.

AdvancedTry it yourself first

Insert 5 geometric means between 4 and 256 (all positive). Find the 3rd mean.

TrapTrap: (1) writing

T_{n} = a r^{n}

(it is

n - 1

); (2) dropping the sign of

r

; (3) taking a negative root of

r^{q - p}

when the terms are stated positive; (4) putting

r^{n - 1}

for

r^{n}

S_{n}

Infinite GP and recurring decimals

S_{\infty} = \frac{a}{1 - r}

The infinite sum, finding $a$ or $r$ , converting recurring decimals to fractions and the series of squares.

The idea

Take a square. Colour half, then half of what is left, then half of that … each piece is smaller, but the total coloured area creeps towards the whole square. This is $\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \dots = 1$ . For a general GP too: if $- 1 < r < 1$ then $r^{n}$ goes towards zero and $S_{n} = \frac{a ( 1 - r ^{n} )}{1 - r}$ becomes $\frac{a}{1 - r}$ .

◭ Diagram

the leftover piece halves every time; the sum creeps up to 1

◭ Diagram

24, 8, \frac{8}{3}, \dots

: the partial sums

S_{1}, S_{2}, \dots

move towards the infinite sum 36

∑ Formula

S_{\infty} = \frac{a}{1 - r}, ∣ r ∣ < 1

The sum of an infinite GP

$a$: first term
$r$: common ratio (between $- 1$ and $1$ )
$S_{\infty}$: the sum of all the terms

Recurring decimals: $0. \overline{36} = 0.36 + 0.0036 + 0.000036 + \dots$ is a GP with $a = 0.36$ and $r = \frac{1}{100}$ (two digits repeat, so it moves two places). So the value is $\frac{0.36}{0.99} = \frac{4}{11}$ . If some digits do not repeat (like $0.4 \overline{7}$ ), they are a separate fraction and the GP of the repeating part is added after. A series of squares: the squares of the terms are a GP too, with $a^{2}$ and $r^{2}$ .

Spot itUse it when you see “sum to infinity”, “

\dots

” or a recurring decimal

0. \overline{ab}

, and

∣ r ∣ < 1

Steps

Find $a$ and $r$ ; check $∣ r ∣ < 1$ .
In a recurring decimal first separate the non-repeating part and write the first term of the repeating part with $r = \frac{1}{1 0 ^{k}}$ ( $k$ = the repeating digits).
Apply $\frac{a}{1 - r}$ ; bring to simplest form.
For reverse questions solve $S = \frac{a}{1 - r}$ for $a$ or $r$ .

Worked examples

Example 24 · Basic

Find the sum to infinity of the GP

18, 6, 2, \frac{2}{3}, \dots

Solution, step by step

1The common ratio is $r = \frac{6}{18} = \frac{1}{3}$ . It lies between −1 and 1, so the terms shrink fast enough for the sum to settle near a fixed number.
2Then $r^{n}$ goes towards 0 and $S_{n} = \frac{a ( 1 - r ^{n} )}{1 - r}$ becomes $S_{\infty} = \frac{a}{1 - r}$ .
3 $S_{\infty} = \frac{18}{1 - ( \frac{1}{3} )} = \frac{18}{\frac{2}{3}} = 27$ .

Answer $27$

Example 25 · Standard

An infinite GP has first term 12 and sum 16. Find its common ratio.

Solution, step by step

1In $S = \frac{a}{1 - r}$ put $S = 16$ and $a = 12$ : $16 = \frac{12}{1 - r}$ .
2Cross-multiply: $1 - r = \frac{12}{16} = \frac{3}{4}$ .
3 $r = 1 - \frac{3}{4} = \frac{1}{4}$ . (Check: $∣ r ∣ < 1$ , so the sum exists.)

Answer $\frac{1}{4}$

Example 26 · Standard

Convert

0. \overline{36}

(

0.363636 \dots

) into a fraction in simplest form.

Solution, step by step

1The 2 digits (36) repeat right from the start.
2The repeating part is a GP: $0.36 + 0.0036 + \dots$ . First term $a = \frac{9}{25}$ , ratio $r = \frac{1}{1 0 ^{2}} = \frac{1}{100}$ (it moves 2 place(s) each time).
3 $S_{\infty} = \frac{\frac{9}{25}}{1 - \frac{1}{100}} = \frac{4}{11}$ .
4In simplest form: $\frac{4}{11}$ .

Answer $\frac{4}{11}$

This is the same as simplifying

\frac{36}{99}

Example 27 · Advanced

Convert

0.4 \overline{7}

(

0.4777 \dots

) into a fraction in simplest form.

Solution, step by step

1The first 1 digits after the point (4) do not repeat, so keep them apart: $0.4 = \frac{2}{5}$ .
2The repeating part is a GP: $0.07 + 0.007 + \dots$ . First term $a = \frac{7}{100}$ , ratio $r = \frac{1}{1 0 ^{1}} = \frac{1}{10}$ (it moves 1 place(s) each time).
3 $S_{\infty} = \frac{\frac{7}{100}}{1 - \frac{1}{10}} = \frac{7}{90}$ ; total $\frac{2}{5} + \frac{7}{90} = \frac{43}{90}$ .
4In simplest form: $\frac{43}{90}$ .

Answer $\frac{43}{90}$

Practice

BasicTry it yourself first

Find the sum to infinity of the GP

24, 6, \frac{3}{2}, \frac{3}{8}, \dots

BasicTry it yourself first

Convert

0. \overline{27}

(

0.272727 \dots

) into a fraction in simplest form.

BasicTry it yourself first

An infinite GP has common ratio

\frac{2}{5}

and sum 15. Find its first term.

StandardTry it yourself first

Find the sum to infinity of the GP

10, - \frac{20}{3}, \frac{40}{9}, - \frac{80}{27}, \dots

StandardTry it yourself first

An infinite GP has first term 20 and sum 25. Find its common ratio.

StandardTry it yourself first

Convert

0.6 \overline{3}

(

0.6333 \dots

) into a fraction in simplest form.

StandardTry it yourself first

Find the sum to infinity of the series made of the squares of the terms of the GP

10, \frac{10}{3}, \frac{10}{9}, \dots

StandardTry it yourself first

Convert

0. \overline{135}

(

0.135135135 \dots

) into a fraction in simplest form.

AdvancedTry it yourself first

Find the sum to infinity of the series made of the squares of the terms of the GP

12, - 6, 3, \dots

AdvancedTry it yourself first

Convert

0.13 \overline{45}

(

0.13454545 \dots

) into a fraction in simplest form.

AdvancedTry it yourself first

An infinite GP has common ratio

- \frac{1}{3}

and sum

\frac{21}{2}

. Find its first term.

AdvancedTry it yourself first

Convert

0.08 \overline{3}

(

0.08333 \dots

) into a fraction in simplest form.

TrapTrap: (1) an infinite sum does not exist when

∣ r ∣ \geq 1

; (2) taking

r = \frac{1}{10}

when two digits repeat (it is

\frac{1}{100}

); (3) treating the sum of squares as

S_{\infty}^{2}

; (4) mixing the non-repeating digits into the repeating part.

Uses of GP

Bouncing balls, midpoint figures, spreading

Bouncing-ball distances, areas and perimeters of figures formed by joining midpoints, spreading, averages and doubling numbers.

The idea

These are the questions where a GP hides: you have to spot it — what is the first term, what is the ratio, how many terms (or infinitely many). In every story ask one question: “what multiple of the previous value is the next one?”

◭ Diagram

every rebound is

r

times the last height; the dot is the ball

Bouncing ball (height $h$ , rebound ratio $r$ ): it first falls $h$ ; after that every height $h r, h r^{2}, \dots$ is travelled up and also down, i.e. twice. Total distance = $h + 2 \times \frac{h r}{1 - r} = h \cdot \frac{1 + r}{1 - r}$ . If the distance up to the $n$ th touch of the ground is asked, take $h$ first, then $2 h r, 2 h r^{2}, \dots$ ( $n - 1$ such terms in all).

◭ Diagram

joining the midpoints of a triangle leaves a quarter of the area, with half the sides (half the perimeter)

◭ Diagram

joining the midpoints of a square gives half the area and

\frac{1}{2}

times the side

Figure	Ratio of areas	Ratio of perimeters
Triangle (midpoint triangle)	$\frac{1}{4}$	$\frac{1}{2}$
Square (midpoint square)	$\frac{1}{2}$	$\frac{1}{2}$

Spreading and growth: “each person tells $k$ new people” — round 0 has 1, then $k, k^{2}, \dots$ , in all $\frac{k ^{n + 1} - 1}{k - 1}$ . “Doubles every $T$ hours” — $P \cdot 2^{hours / T}$ . Averages: average × count = total; if the added amounts form a GP use the GP sum.

Spot itWhen something in the story keeps changing by one multiple (rebound, spread, doubling, halving, midpoints) it is a GP.

Steps

Identify the first term and the multiple by which it changes each time.
Decide: finitely many terms ( $S_{n}$ ) or infinitely many ( $S_{\infty}$ ).
Add the story’s own conditions (up and down twice for a ball; area versus perimeter for figures).
Apply the formula and write the answer with its unit.

Worked examples

Example 28 · Basic

A rumour starts with one person. Every person who learns it passes it to 3 new people in the next round. How many people know it after 6 rounds (the starter included)?

Solution, step by step

1Round 0 has 1 person, round 1 has 3, round 2 has 9, … the number is multiplied by 3 each round: a GP $1, 3, 9, \dots$ with 7 terms (rounds 0 to 6).
2 $S_{7} = \frac{1 \times ( 3 ^{7} - 1 )}{3 - 1} = \frac{2186}{2} = 1093$ .

Answer $1093$

Example 29 · Standard

A ball is dropped from a height of 120 m. After every bounce it rises to

\frac{3}{5}

of its previous height. What total distance does it travel before coming to rest (in m)?

Solution, step by step

1First fall: $h = 120$ . After each bounce the ball goes up to 72, $\frac{216}{5}$ , … and comes down the same, so each height is covered twice.
2The heights it rises form a GP: first term $h r = 72$ , ratio $r = \frac{3}{5}$ . Their sum is $\frac{h r}{1 - r} = \frac{72}{\frac{2}{5}} = 180$ .
3Total distance = first fall + 2 × (sum of the rises) = $120 + 2 \times 180 = 480$ . (As a formula, $h \cdot \frac{1 + r}{1 - r}$ .)

Answer $480$

Example 30 · Standard

The midpoints of the sides of a right triangle with sides 5, 12, 13 cm are joined to form a new triangle, and this is repeated for ever. Find the sum of the areas of all the triangles (the original included) in cm².

Solution, step by step

1Joining the midpoints cuts the triangle into four equal triangles, so the new triangle has $\frac{1}{4}$ of the area.
2The original area is $\frac{1}{2} \times 5 \times 12 = 30$ . The areas form a GP: $30, 7.5, 1.875, \dots$ with $r = \frac{1}{4}$ .
3 $S_{\infty} = \frac{30}{1 - \frac{1}{4}} = 30 \times \frac{4}{3} = 40$ cm².

Answer $40$

Example 31 · Advanced

The midpoints of the sides of a square of side 6 cm are joined to form a new square, and this is repeated for ever. Find the sum of the perimeters of all the squares (the original included) in cm.

Solution, step by step

1The new square’s side is $\frac{6}{2}$ (the hypotenuse of a right triangle with legs 6/2), so every side is $\frac{1}{2}$ times the old one.
2The perimeters shrink in the same ratio: $24, \frac{24}{2}, \dots$ , $r = \frac{1}{2}$ (less than 1). $S_{\infty} = \frac{24}{1 - \frac{1}{2}} = \frac{24 2}{2 - 1}$ .
3Rationalise the denominator (multiply top and bottom by $2 + 1$ ): $\frac{24 2 ( 2 + 1 )}{( 2 ) ^{2} - 1} = 242 (2 + 1) = 48 + 242$ cm.

Answer $48 + 242$

Example 32 · Advanced

The average of 6 numbers is 40. 3 is added to the first, 6 to the second, 12 to the third, and so on, each time 2 times the previous amount. Find the new average.

Solution, step by step

1The old total = 6 × 40 = 240.
2The amounts added $3, 6, 12, \dots$ form a GP (6 terms, $r = 2$ ) with sum $\frac{3 ( 2 ^{6} - 1 )}{2 - 1} = 189$ .
3The new total is 240 + 189 = 429; the new average is $\frac{429}{6} = \frac{143}{2}$ .

Answer $\frac{143}{2}$

Practice

BasicTry it yourself first

A culture starts with 500 bacteria. Their number doubles every 3 hours. How many bacteria are there after 18 hours?

BasicTry it yourself first

A video starts with one person. Every person who learns it passes it to 4 new people in the next round. How many people know it after 5 rounds (the starter included)?

BasicTry it yourself first

The midpoints of the sides of a square of side 10 cm are joined to form a new square, and this is repeated for ever. Find the sum of the areas of all the squares (the original included) in cm².

StandardTry it yourself first

A ball is dropped from a height of 90 m. After every bounce it rises to

\frac{1}{2}

of its previous height. What total distance does it travel before coming to rest (in m)?

StandardTry it yourself first

A ball is dropped from 80 m and bounces to

\frac{1}{2}

of its previous height each time. What total distance has it travelled when it touches the ground for the 4th time (in m)?

StandardTry it yourself first

The midpoints of the sides of a triangle with sides 9, 12, 15 cm are joined to form a new triangle, and this is repeated for ever. Find the sum of the perimeters of all the triangles (the original included) in cm.

StandardTry it yourself first

A culture has 250 bacteria and the number triples every 4 hours. How many hours does it take to reach 60,750?

StandardTry it yourself first

A rumour starts with one person. Every person who learns it passes it to 5 new people in the next round. How many people know it after 4 rounds (the starter included)?

AdvancedTry it yourself first

A ball is dropped from a height of 200 m. After every bounce it rises to

\frac{3}{7}

of its previous height. What total distance does it travel before coming to rest (in m)?

AdvancedTry it yourself first

A ball is dropped from 160 m and bounces to

\frac{3}{4}

of its previous height each time. What total distance has it travelled when it touches the ground for the 4th time (in m)?

AdvancedTry it yourself first

The midpoints of the sides of a right triangle with sides 8, 15, 17 cm are joined to form a new triangle, and this is repeated for ever. Find the sum of the areas of all the triangles (the original included) in cm².

AdvancedTry it yourself first

The midpoints of the sides of a square of side 9 cm are joined to form a new square, and this is repeated for ever. Find the sum of the perimeters of all the squares (the original included) in cm.

AdvancedTry it yourself first

The average of 5 numbers is 30. 4 is added to the first, 8 to the second, 16 to the third, and so on, each time 2 times the previous amount. Find the new average.

AdvancedTry it yourself first

A culture starts with 1200 bacteria. Their number doubles every 5 hours. How many bacteria are there after 30 hours?

TrapTrap: (1) counting the heights once for a ball (twice is needed) and the first fall twice; (2) using the area ratio for perimeters; (3) taking

n

terms for “until the

n

th touch” (it is

n - 1

); (4) leaving out the starter in a spread question.

AM, GM, HM and harmonic progression

Three means and how they relate

The three means of two numbers, $AM \geq GM \geq HM$ , harmonic progression and average speed.

The idea

Two positive numbers $a$ and $b$ have three “in-between values”. The AM (arithmetic mean) is the middle term of an AP between $a$ and $b$ ; the GM (geometric mean) of a GP; the HM (harmonic mean) of a progression whose reciprocals are in AP. A harmonic progression (HP) is recognised by exactly this: its terms’ reciprocals form an AP, so solve every HP question by taking reciprocals and working in the AP.

∑ Formula

AM = \frac{a + b}{2}, GM = ab, HM = \frac{2 ab}{a + b}

The three means of two numbers

$a, b$: two positive numbers

∑ Formula

AM \geq GM \geq HM

Their relation (equality only when

a = b

)

$AM, GM, HM$: the three means above

∑ Formula

GM^{2} = AM \times HM

Their relation (equality only when

a = b

)

$AM, GM, HM$: the three means above

Using AM ≥ GM: in an expression like $4 x + \frac{9}{x}$ ( $x > 0$ ) the product of the two terms is fixed, so AM ≥ GM gives the minimum directly: $\geq 2 4 \times 9 = 12$ . Average speed: if equal distances are covered at speeds $u$ and $v$ the average speed is the HM $\frac{2 uv}{u + v}$ — not the plain average, because more time is spent at the slow speed.

Spot itUse this for “mean”, “harmonic progression”, “minimum value of

a x + \frac{b}{x}

”, “average speed over equal distances”.

Steps

Identify which mean or progression is asked.
For an HP take reciprocals to get an AP.
Apply the AP formulas, then take the reciprocal of the answer.
Check: $AM \geq GM \geq HM$ must hold.

Worked examples

Example 33 · Basic

Find the harmonic mean (HM) of the positive numbers 4 and 16.

Solution, step by step

1HM is the reciprocal of the AM of the reciprocals: $\frac{2 ab}{a + b} = \frac{2 \times 4 \times 16}{4 + 16} = \frac{32}{5}$ .
2The three means are $AM = 10, GM = 8, HM = \frac{32}{5}$ : always $AM \geq GM \geq HM$ and $GM^{2} = AM \times HM$ (check: $10 \times \frac{32}{5} = 64 = 64$ ).

Answer $\frac{32}{5}$

Example 34 · Standard

Find the 12th term of the harmonic progression (HP)

\frac{1}{3}, \frac{1}{7}, \frac{1}{11}, \dots

Solution, step by step

1By definition the reciprocals of the terms of an HP form an AP. Write the reciprocals: $3, 7, 11, \dots$ .
2This AP has $a = 3, d = 4$ , so its 12th term is $3 + 11 \times 4 = 47$ .
3The 12th term of the HP is its reciprocal: $\frac{1}{47}$ .

Answer $\frac{1}{47}$

Example 35 · Standard

Find the minimum value of

4 x + \frac{9}{x}

for

x > 0

Solution, step by step

1Take the two positive terms $4 x$ and $\frac{9}{x}$ and use AM ≥ GM: $\frac{4 x + \frac{9}{x}}{2} \geq 4 x \cdot \frac{9}{x}$ .
2On the right $x$ cancels: $4 \times 9 = 36 = 6$ , so $4 x + \frac{9}{x} \geq 2 \times 6 = 12$ .
3Equality holds when the two terms are equal: $4 x = \frac{9}{x}$ , $x^{2} = \frac{9}{4}$ . So the minimum $12$ is really reached.

Answer $12$

Example 36 · Advanced

Two positive numbers have AM 10 and GM 8. Find the larger number.

Solution, step by step

1Let the numbers be $x, y$ . From $\frac{x + y}{2} = 10$ : $x + y = 20$ ; from $x y = 8$ : $x y = 64$ .
2Using $(x - y)^{2} = (x + y)^{2} - 4 x y$ : $(x - y)^{2} = 400 - 256 = 144$ , so $x - y = 12$ .
3 $x = \frac{20 + 12}{2} = 16$ and $y = 4$ (check: 4×16=64=8²).

Answer $16$

Example 37 · Advanced

A car covers two equal distances at 30 and 45 km/h. Find the average speed for the whole journey (km/h).

Solution, step by step

1Call each distance $d$ : the times are $\frac{d}{30}$ and $\frac{d}{45}$ ; the total distance is $2 d$ .
2Average speed = $\frac{2 d}{\frac{d}{30} + \frac{d}{45}} = \frac{2}{\frac{1}{30} + \frac{1}{45}} = \frac{2 \times 30 \times 45}{30 + 45}$ .
3 $\frac{2700}{75} = 36$ km/h. (This is the HM of 30 and 45, not the plain average 37.5.)

Answer $36$

Practice

BasicTry it yourself first

Find the arithmetic mean (AM) of the positive numbers 9 and 25.

BasicTry it yourself first

Find the geometric mean (GM) of the positive numbers 8 and 18.

BasicTry it yourself first

Find the 15th term of the harmonic progression (HP)

\frac{1}{2}, \frac{1}{7}, \frac{1}{12}, \dots

StandardTry it yourself first

Find the harmonic mean (HM) of the positive numbers 2 and 32.

StandardTry it yourself first

The 4th term of an HP is

\frac{1}{11}

and the 9th term is

\frac{1}{26}

. Find its 20th term.

StandardTry it yourself first

Find the minimum value of

25 x + \frac{4}{x}

for

x > 0

StandardTry it yourself first

A car covers two equal distances at 40 and 60 km/h. Find the average speed for the whole journey (km/h).

StandardTry it yourself first

Find the 20th term of the harmonic progression (HP)

\frac{1}{1}, \frac{1}{4}, \frac{1}{7}, \dots

AdvancedTry it yourself first

Two positive numbers have AM 17 and GM 15. Find the larger number.

AdvancedTry it yourself first

Two positive numbers have AM 25 and GM 7. Find the larger number.

AdvancedTry it yourself first

The 3rd term of an HP is

\frac{1}{13}

and the 8th term is

\frac{1}{33}

. Find its 15th term.

AdvancedTry it yourself first

Find the minimum value of

2 x + \frac{18}{x}

for

x > 0

AdvancedTry it yourself first

A car covers three equal distances at 20, 30 and 60 km/h. Find the average speed for the whole journey (km/h).

AdvancedTry it yourself first

A car covers three equal distances at 24, 36 and 72 km/h. Find the average speed for the whole journey (km/h).

TrapTrap: (1) applying AP formulas directly to HP terms (take reciprocals first); (2) using

\frac{u + v}{2}

for average speed; (3) taking the AM ≥ GM minimum without checking that the two terms can be equal; (4) forgetting the final reciprocal in the answer.

Summary, mixed practice, revision

All methods in one place

One table of all methods, exam-style mixed questions and revision cards.

Summary of all methods

Which method when

One page to pick the method as soon as you read a question.

Method	Use when	Key formula / move
AP: term	changes by equal steps	$T_{n} = a + (n - 1) d$
AP: count	first and last known	$n = \frac{l - a}{d} + 1$
AP: sum	sum of the first $n$ terms	$\frac{n}{2} (a + l) = \frac{n}{2} [2 a + (n - 1) d]$
Middle term	sum of consecutive numbers	middle = sum ÷ count; $a - d, a, a + d$
$S_{n}$ given	$S_{n} = p n^{2} + q n$	$T_{n} = S_{n} - S_{n - 1}$ ; $a = p + q$ , $d = 2 p$
Inserting means	between two numbers	$d = \frac{B - A}{k + 1}$ ; $r^{k + 1} = \frac{B}{A}$
Condition puzzles	$S_{p} = S_{q}$ ; $a_{i}$ के योग	$S_{p + q} = 0$ ; $a_{1} + a_{N}$ pairs
Counting multiples	“or”, “and”, “neither”	LCM; $n (A) + n (B) - n (A \cap B)$
GP: term	multiplied by the same each time	$T_{n} = a r^{n - 1}$
GP: sum	first $n$ terms	$\frac{a ( r ^{n} - 1 )}{r - 1}$ / $\frac{a ( 1 - r ^{n} )}{1 - r}$
Infinite GP	$∣ r ∣ < 1$	$S_{\infty} = \frac{a}{1 - r}$
Recurring decimal	$0. \overline{ab}$	$a = 0. ab$ , $r = \frac{1}{100}$
Bouncing ball	rebound ratio $r$	$h \cdot \frac{1 + r}{1 - r}$
Midpoint figures	triangle / square	area $\frac{1}{4}$ / $\frac{1}{2}$
AM, GM, HM	means of two numbers	$\frac{a + b}{2}$ , $ab$ , $\frac{2 ab}{a + b}$
HP	reciprocals in AP	reciprocals → AP → flip back

Here: $a$ = first term; $d$ = common difference; $r$ = common ratio; $n$ = number of terms; $l$ = last term; $S_{n}$ = sum of the first $n$ terms; $p, q$ = the coefficients in $S_{n} = p n^{2} + q n$ ; $A, B$ = the numbers between which means are inserted; $k$ = the number of means; $h$ = the ball’s height; $a, b$ (in AM-GM) = two positive numbers.

TrickAfter reading ask: “does it grow by adding or by multiplying?” — AP or GP. “Out to infinity?” —

\frac{a}{1 - r}

. “Are the reciprocals in a plain progression?” — HP.

Mixed practice — set 1

Basic to standard

AP and GP questions mixed; spot the method first.

These questions are not in order — just like the exam. First decide: AP, GP or something else?

BasicTry it yourself first

Find the 31st term of the progression

- 9, - 4, 1, 6, \dots

BasicTry it yourself first

Find the sum of the first 26 terms of

18, 22, 26, 30, \dots

BasicTry it yourself first

How many terms are there in

15, 21, 27, \dots, 267

BasicTry it yourself first

Find the 9th term of the geometric progression (GP)

7, 14, 28, 56, \dots

BasicTry it yourself first

Find the sum of the first 9 terms of the GP

3, 6, 12, 24, \dots

BasicTry it yourself first

Find the sum to infinity of the GP

30, 5, \frac{5}{6}, \frac{5}{36}, \dots

BasicTry it yourself first

The first row of a hall has 30 seats and every next row has 4 more seats than the one before. How many seats are there in 22 rows altogether?

BasicTry it yourself first

The sum of 7 consecutive even numbers is 336. Find the largest of them.

StandardTry it yourself first

Which term of the progression

- 13, - 6, 1, 8, \dots

is 190?

StandardTry it yourself first

The 5th term of an AP is 25 and the 12th term is 53. Find its 30th term.

StandardTry it yourself first

The 2nd term of a GP is 6 and the 5th term is 48 (all terms positive). Find its 9th term.

StandardTry it yourself first

How many multiples of 11 are there from 120 to 980 (both included)?

StandardTry it yourself first

Convert

0. \overline{45}

(

0.454545 \dots

) into a fraction in simplest form.

StandardTry it yourself first

Find the average of all multiples of 8 between 100 and 500 (both included).

StandardTry it yourself first

A video starts with one person. Every person who learns it passes it to 3 new people in the next round. How many people know it after 5 rounds (the starter included)?

StandardTry it yourself first

A ball is dropped from a height of 100 m. After every bounce it rises to

\frac{1}{4}

of its previous height. What total distance does it travel before coming to rest (in m)?

Mixed practice — set 2

Standard to advanced

Mixed questions from the hard end of the exam.

StandardTry it yourself first

Find the sum:

7 + 13 + 19 + \dots + 253

StandardTry it yourself first

An infinite GP has first term 18 and sum 27. Find its common ratio.

StandardTry it yourself first

The sum of the first

n

terms of an AP is

S_{n} = 6 n^{2} - 4 n

. Find its 14th term.

StandardTry it yourself first

A culture starts with 800 bacteria. Their number doubles every 6 hours. How many bacteria are there after 36 hours?

AdvancedTry it yourself first

In an AP the sum of the first 11 terms equals the sum of the first 24 terms. If its first term is 51, find the 35th term.

AdvancedTry it yourself first

For the terms of an AP,

a_{4} + a_{10} + a_{15} + a_{22} + a_{27} + a_{33} = 330

. Find the sum of its first 36 terms.

AdvancedTry it yourself first

The ratio of the sums of the first n terms of two APs is

(9 n + 2) : (4 n + 7)

. What is the ratio of their 9th terms?

AdvancedTry it yourself first

How many terms of

54, 48, 42, 36, \dots

are needed for the sum to be zero?

AdvancedTry it yourself first

How many numbers from 100 to 1000 (both included) are divisible by neither 8 nor 12?

AdvancedTry it yourself first

Find the sum of all numbers from 1 to 200 that are divisible by 5 or 8.

AdvancedTry it yourself first

A total of 7473 digits were used to print the page numbers of a book. How many pages does it have?

AdvancedTry it yourself first

Four numbers in AP add up to 56, and the product of the first and the last is 115. Find the largest.

AdvancedTry it yourself first

Three positive numbers are in GP. Their product is 1000 and their sum is 39. Find the largest.

AdvancedTry it yourself first

Insert 3 geometric means between 6 and 3750 (all positive). Find the 2nd mean.

AdvancedTry it yourself first

Convert

0.27 \overline{6}

(

0.27666 \dots

) into a fraction in simplest form.

AdvancedTry it yourself first

Find the sum to infinity of the series made of the squares of the terms of the GP

9, 6, 4, \dots

AdvancedTry it yourself first

The midpoints of the sides of a right triangle with sides 20, 21, 29 cm are joined to form a new triangle, and this is repeated for ever. Find the sum of the areas of all the triangles (the original included) in cm².

AdvancedTry it yourself first

The average of 7 numbers is 24. 5 is added to the first, 10 to the second, 20 to the third, and so on, each time 2 times the previous amount. Find the new average.

AdvancedTry it yourself first

Two positive numbers have AM 15 and GM 12. Find the larger number.

AdvancedTry it yourself first

The 5th term of an HP is

\frac{1}{11}

and the 12th term is

\frac{1}{25}

. Find its 30th term.

AdvancedTry it yourself first

A car covers three equal distances at 15, 30 and 60 km/h. Find the average speed for the whole journey (km/h).

Revision — cards and traps

Flip and recall

Read the front, think, then flip.

Trap 1Trap 1: writing $T_{n} = a + n d$ . It is

n - 1

Trap 2Trap 2: forgetting the $+ 1$ in a count. Posts are one more than gaps.

Trap 3Trap 3: using the product for “divisible by both”. Use the LCM.

Trap 4Trap 4: an infinite sum with $∣ r ∣ \geq 1$ . Then the sum does not exist.

Trap 5Trap 5: counting ball heights once. Both up and down.

Trap 6Trap 6: using AP formulas straight on an HP. Reciprocals first.

Trap 7Trap 7: a plain average for average speed. Use the HM for equal distances.