Arithmetic Sequence Problems

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Question 1 of 3

Find the sequence given that:

U_{5} + U_{13} = 126

$U_5+U_13=126$

U_{9} + U_{17} = 182

$U_9+U_17=182$

1. $7 + 11 + 15 + 19 ...$ $7+11+15+19...$
2. $7 + 15 + 23 + 31 ...$ $7+15+23+31...$
3. $7 + 12 + 17 + 22 ...$ $7+12+17+22...$
4. $7 + 14 + 21 + 28 ...$ $7+14+21+28...$

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General Rule of an Arithmetic Sequence

U_{n} = a + [(n - 1) d]

$U_{\color{#9a00c7}{n}}=\color{#e65021}{a}+[(\color{#9a00c7}{n}-1)\color{#00880A}{d}]$

First, transform the

5 t h

$5th$ and

13 t h

$13th$ terms into general rule form

5th Term

$U_{\color{#9a00c7}{n}}$	$=$ $=$	$\color{#e65021}{a}+[(\color{#9a00c7}{n}-1)\color{#00880A}{d}]$
$U_{\color{#9a00c7}{5}}$	$=$ $=$	$\color{#e65021}{a}+[(\color{#9a00c7}{5}-1)\color{#00880A}{d}]$	Substitute known values
$U_{5}$ $U_5$	$=$ $=$	$a + 4 d$ $a+4d$	Distribute

13th Term

$U_{\color{#9a00c7}{n}}$	$=$ $=$	$\color{#e65021}{a}+[(\color{#9a00c7}{n}-1)\color{#00880A}{d}]$
$U_{\color{#9a00c7}{13}}$	$=$ $=$	$\color{#e65021}{a}+[(\color{#9a00c7}{13}-1)\color{#00880A}{d}]$	Substitute known values
$U_{13}$ $U_13$	$=$ $=$	$a + 12 d$ $a+12d$	Distribute

Next, add the first pair general forms

$U_{5} + U_{13}$ $U_5+U_13$	$=$ $=$	$(a + 4 d) + (a + 12 d)$ $(a+4d)+(a+12d)$
$126$ $126$ $\div 2$ $divide2$	$=$ $=$	$2 a + 16 d$ $2a+16d$ $\div 2$ $divide2$	Divide both sides by $2$ $2$
$63$ $63$	$=$ $=$	$a + 8 d$ $a+8d$

Next, transform the

9 t h

$9th$ and

17 t h

$17th$ terms into general rule form

9th Term

$U_{\color{#9a00c7}{n}}$	$=$ $=$	$\color{#e65021}{a}+[(\color{#9a00c7}{n}-1)\color{#00880A}{d}]$
$U_{\color{#9a00c7}{9}}$	$=$ $=$	$\color{#e65021}{a}+[(\color{#9a00c7}{9}-1)\color{#00880A}{d}]$	Substitute known values
$U_{9}$ $U_9$	$=$ $=$	$a + 8 d$ $a+8d$	Distribute

17th Term

$U_{\color{#9a00c7}{n}}$	$=$ $=$	$\color{#e65021}{a}+[(\color{#9a00c7}{n}-1)\color{#00880A}{d}]$
$U_{\color{#9a00c7}{17}}$	$=$ $=$	$\color{#e65021}{a}+[(\color{#9a00c7}{17}-1)\color{#00880A}{d}]$	Substitute known values
$U_{17}$ $U_17$	$=$ $=$	$a + 16 d$ $a+16d$	Distribute

Next, add the second pair general forms

$U_{9} + U_{17}$ $U_9+U_17$	$=$ $=$	$(a + 4 d) + (a + 12 d)$ $(a+4d)+(a+12d)$
$182$ $182$ $\div 2$ $divide2$	$=$ $=$	$2 a + 24 d$ $2a+24d$ $\div 2$ $divide2$	Divide both sides by $2$ $2$
$91$ $91$	$=$ $=$	$a + 12 d$ $a+12d$

Next, solve for the value of $d$ $d$ by subtracting the first combined general rule form from the second combined general rule form

$91$ $91$ $-$ $-$ $63$ $63$	$=$ $=$	$($ $($ $a + 12 d$ $a+12d$ $) - ($ $)-($ $a + 8 d$ $a+8d$ $)$ $)$
$28$ $28$ $\div 4$ $divide4$	$=$ $=$	$4 d$ $4d$ $\div 4$ $divide4$	Divide both sides by $4$ $4$
$7$ $7$	$=$ $=$	$d$ $d$
$d$ $d$	$=$ $=$	$7$ $7$

Next, substitute $d$ $d$ to one of the combined general rule forms to solve for $a$ $a$

$63$ $63$	$=$ $=$	$a + 8$ $a+8$ $d$ $d$
$63$ $63$	$=$ $=$	$a + 8 ($ $a+8($ $7$ $7$ $)$ $)$	Substitute $d = 7$ $d=7$
$63$ $63$ $- 56$ $-56$	$=$ $=$	$a + 56$ $a+56$ $- 56$ $-56$	Subtract $56$ $56$ from both sides
$7$ $7$	$=$ $=$	$a$ $a$
$a$ $a$	$=$ $=$	$7$ $7$

Finally, start with $a = 7$ $a=7$ and keep adding $d = 7$ $d=7$ to its value to get the sequence

$U_{1}$ $U_1$	$=$ $=$	$7$ $7$
$U_{2}$ $U_2$	$=$ $=$	$7 +$ $7+$ $7$ $7$	$=$ $=$	$14$ $14$
$U_{3}$ $U_3$	$=$ $=$	$14 +$ $14+$ $7$ $7$	$=$ $=$	$21$ $21$
$U_{4}$ $U_4$	$=$ $=$	$21 +$ $21+$ $7$ $7$	$=$ $=$	$28$ $28$

7 + 14 + 21 + 28 \dots

$7+14+21+28…$

7 + 14 + 21 + 28 \dots

$7+14+21+28…$

Question 2 of 3

The short end of a fence has a height of

140

$140$ cm. After about

71

$71$ pieces of timber, the height of the fence is

210

$210$ cm. Find:

(i)

$(i)$ The difference in height between each fence

(i i)

$(ii)$ The total height of the fences in metres

$(i)$ $(i)$ $d =$ $d=$ (1) $cm$ $\text(cm)$

$(i i)$ $(ii)$ $S_{n} =$ $S_n=$ (124.25) $m$ $\text(m)$

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Sum of an Arithmetic Sequence

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

$S_{\color{#9a00c7}{n}}=\frac{\color{#9a00c7}{n}}{2}[2\color{#e65021}{a}+(\color{#9a00c7}{n}-1)\color{#00880A}{d}]$

General Rule of an Arithmetic Sequence

U_{n} = a + [(n - 1) d]

$U_{\color{#9a00c7}{n}}=\color{#e65021}{a}+[(\color{#9a00c7}{n}-1)\color{#00880A}{d}]$

(i)

$(i)$ Finding the difference in height between each fence (

d

$d$ )

Substitute the known values to the general rule

$Number of terms$ $\text(Number of terms)$ $[n]$ $[n]$	$=$ $=$	$71$ $71$
$First term$ $\text(First term)$ $[a]$ $[a]$	$=$ $=$	$140$ $140$

$U_{\color{#9a00c7}{n}}$	$=$ $=$	$\color{#e65021}{a}+[(\color{#9a00c7}{n}-1)\color{#00880A}{d}]$
$U_{\color{#9a00c7}{71}}$	$=$ $=$	$\color{#e65021}{140}+[(\color{#9a00c7}{71}-1)\color{#00880A}{d}]$	Substitute known values
$210$ $210$ $- 140$ $-140$	$=$ $=$	$140 + 70 d$ $140+70d$ $- 140$ $-140$	Subtract $140$ $140$ from both sides
$70$ $70$ $\div 70$ $divide70$	$=$ $=$	$70 d$ $70d$ $\div 70$ $divide70$	Divide both sides by $70$ $70$
$1$ $1$	$=$ $=$	$d$ $d$
$d$ $d$	$=$ $=$	$1 cm$ $1 \text(cm)$

(i i)

$(ii)$ Finding the total height of the fences in metres (

S_{n}

$S_n$ )

Substitute the known values to the sum formula

$Number of terms [n]$ $\text(Number of terms) [n]$	$=$ $=$	$71$ $71$
$First Term [a]$ $\text(First Term) [a]$	$=$ $=$	$140$ $140$
$Common Difference [d]$ $\text(Common Difference) [d]$	$=$ $=$	$1$ $1$

$S_{\color{#9a00c7}{n}}$	$=$ $=$	$\frac{\color{#9a00c7}{n}}{2}[2\color{#e65021}{a}+(\color{#9a00c7}{n}-1)\color{#00880A}{d}]$

$S_{\color{#9a00c7}{71}}$	$=$ $=$	$\frac{\color{#9a00c7}{71}}{2}[2\cdot\color{#e65021}{140}+(\color{#9a00c7}{71}-1)\color{#00880A}{1}]$	Substitute known values

	$=$ $=$	$35\frac{1}{2}(280+70)$	Evaluate

	$=$ $=$	$35\frac{1}{2}(350)$

	$=$ $=$	$12425 cm$ $12 425 \text(cm)$

Finally, convert the centimetres into metres

1 metre

$1 \text(metre)$

=

$=$

100 centimetres

$100 \text(centimetres)$

12425 \times \frac{1}{100}

$12425xx1/(100)$

=

$=$

124.25 cm

$124.25 \text(cm)$

(i) d = 1 cm

$(i) d=1 \text(cm)$

(i i) S_{n} = 124.25 m

$(ii) S_n=124.25 \text(m)$

Question 3 of 3

A stack of cans has

1

$1$ can at the top and then each row has

2

$2$ more cans than the previous row. Find:

(i)

$(i)$ The number of cans in the

21 s t

$21st$ row

(i i)

$(ii)$ The row with

61

$61$ cans

(i i i)

$(iii)$ The number of rows if there are a total of

1296

$1296$ cans

$(i)$ $(i)$ $U_{21} =$ $U_21=$ (41)

$(i i)$ $(ii)$ $n =$ $n=$ (31)

$(i i i)$ $(iii)$ $n =$ $n=$ (36)

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Sum of an Arithmetic Sequence

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

$S_{\color{#9a00c7}{n}}=\frac{\color{#9a00c7}{n}}{2}[2\color{#e65021}{a}+(\color{#9a00c7}{n}-1)\color{#00880A}{d}]$

General Rule of an Arithmetic Sequence

U_{n} = a + [(n - 1) d]

$U_{\color{#9a00c7}{n}}=\color{#e65021}{a}+[(\color{#9a00c7}{n}-1)\color{#00880A}{d}]$

(i)

$(i)$ Finding the number of cans in the

21 s t

$21st$ row (

U_{21}

$U_21$ )

Substitute the known values to the general rule

$Number of terms$ $\text(Number of terms)$ $[n]$ $[n]$	$=$ $=$	$21$ $21$
$First term$ $\text(First term)$ $[a]$ $[a]$	$=$ $=$	$1$ $1$
$Common Difference$ $\text(Common Difference)$ $[d]$ $[d]$	$=$ $=$	$2$ $2$

$U_{\color{#9a00c7}{n}}$	$=$ $=$	$\color{#e65021}{a}+[(\color{#9a00c7}{n}-1)\color{#00880A}{d}]$
$U_{\color{#9a00c7}{21}}$	$=$ $=$	$\color{#e65021}{1}+[(\color{#9a00c7}{21}-1)\color{#00880A}{2}]$	Substitute known values
	$=$ $=$	$1 + (20) 2$ $1+(20)2$	Evaluate
	$=$ $=$	$1 + 40$ $1+40$
	$=$ $=$	$41$ $41$

(i i)

$(ii)$ Finding the row with

61

$61$ cans (

n

$n$ )

Substitute the known values to the general rule and solve for

n

$n$

$Nth term$ $\text(Nth term)$ $[U_{n}]$ $[U_n]$	$=$ $=$	$61$ $61$
$First term$ $\text(First term)$ $[a]$ $[a]$	$=$ $=$	$1$ $1$
$Common Difference$ $\text(Common Difference)$ $[d]$ $[d]$	$=$ $=$	$2$ $2$

$U_{\color{#9a00c7}{n}}$	$=$ $=$	$\color{#e65021}{a}+[(\color{#9a00c7}{n}-1)\color{#00880A}{d}]$
$61$	$=$ $=$	$\color{#e65021}{1}+[(\color{#9a00c7}{n}-1)\color{#00880A}{2}]$	Substitute known values
$61$ $61$	$=$ $=$	$1 + (2 n - 2)$ $1+(2n-2)$	Distribute
$61$ $61$ $+ 1$ $+1$	$=$ $=$	$- 1 + 2 n$ $-1+2n$ $+ 1$ $+1$	Add $1$ $1$ to both sides
$62$ $62$ $\div 2$ $divide2$	$=$ $=$	$2 n$ $2n$ $\div 2$ $divide2$	Divide both sides by $2$ $2$
$31$ $31$	$=$ $=$	$n$ $n$
$n$ $n$	$=$ $=$	$31$ $31$

(i i i)

$(iii)$ Finding the number of rows if

S_{n} = 1296

$S_n=1296$ (

n

$n$ )

Substitute the known values to the sum formula and solve for

n

$n$

$Sum of terms$ $\text(Sum of terms)$ $[S_{n}]$ $[S_n]$	$=$ $=$	$1296$ $1296$
$First Term [a]$ $\text(First Term) [a]$	$=$ $=$	$1$ $1$
$Common Difference [d]$ $\text(Common Difference) [d]$	$=$ $=$	$2$ $2$

$S_{\color{#9a00c7}{n}}$	$=$ $=$	$\frac{\color{#9a00c7}{n}}{2}[2\color{#e65021}{a}+(\color{#9a00c7}{n}-1)\color{#00880A}{d}]$

$1296$	$=$ $=$	$\frac{\color{#9a00c7}{n}}{2}[2\cdot\color{#e65021}{1}+(\color{#9a00c7}{n}-1)\color{#00880A}{2}]$	Substitute known values

$1296$ $1296$	$=$ $=$	$\frac{n}{2}(2+2n-2)$	Evaluate

$1296$ $1296$ $\times 2$ $times2$	$=$ $=$	$[\frac{n}{2} (2 n)]$ $[n/2(2n)]$ $\times 2$ $times2$	Multiply both sides by $2$ $2$

$2592$ $2592$	$=$ $=$	$n (2 n)$ $n(2n)$	$\frac{2}{2} = 1$ $2/2=1$

$2592$ $2592$ $\div 2$ $divide2$	$=$ $=$	$2 n^{2}$ $2n^2$ $\div 2$ $divide2$	Divide both sides by $2$ $2$
$\sqrt{1296}$ $sqrt1296$	$=$ $=$	$\sqrt{n^{2}}$ $sqrt(n^2)$	Find the square root of both sides
$36$ $36$	$=$	$n$
$n$	$=$	$36$

$(i) U_21=41$

$(ii) n=31$

$(iii) n=36$

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1. Question

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Great Work!

General Rule of an Arithmetic Sequence

2. Question

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Nice Job!

Sum of an Arithmetic Sequence

General Rule of an Arithmetic Sequence

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Well Done!

Sum of an Arithmetic Sequence

General Rule of an Arithmetic Sequence

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