Arithmetic Sequences (Sum)

Question 1 of 5

Find the sum of the sequence

$6+11+16+…+236$

$S_n=$ (5687)

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Sum of an Arithmetic Sequence

$S_{\color{#9a00c7}{n}}=\frac{\color{#9a00c7}{n}}{2}[\color{#e65021}{a}+\color{#00880A}{l}]$

Common Difference Formula

$d=U_2-U_1=U_3-U_2$

General Rule of an Arithmetic Sequence

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

First, solve for the value of $d$ .

$\color{#00880A}{d}$	$=$	$U_2-U_1$
	$=$	$11-6$	Substitute the first and second term
	$=$	$5$

Next, substitute the known values to the general rule and solve for $n$

$\text(Nth term)$ $[U_n]$	$=$	$236$
$\text(First term)$ $[a]$	$=$	$6$
$\text(Common Difference)$ $[d]$	$=$	$5$

$U_{\color{#9a00c7}{n}}$	$=$	$\color{#e65021}{a}+[(\color{#9a00c7}{n}-1)\color{#00880A}{d}]$
$236$	$=$	$\color{#e65021}{6}+[(\color{#9a00c7}{n}-1)\color{#00880A}{5}]$	Substitute known values
$236$	$=$	$6+5n-5$	Distribute
$236$ $-1$	$=$	$1+5n$ $-1$	Subtract $1$ from both sides
$235$ $divide5$	$=$	$5n$ $divide5$	Divide both sides by $5$
$47$	$=$	$n$
$n$	$=$	$47$

Finally, substitute the known values to the sum formula

$\text(Number of terms) [n]$	$=$	$47$
$\text(First Term) [a]$	$=$	$6$
$\text(Last Term) [l]$	$=$	$236$

$S_{\color{#9a00c7}{n}}$	$=$	$\frac{\color{#9a00c7}{n}}{2}[\color{#e65021}{a}+\color{#00880A}{l}]$

$S_{\color{#9a00c7}{47}}$	$=$	$\frac{\color{#9a00c7}{47}}{2}[\color{#e65021}{6}+\color{#00880A}{236}]$	Substitute known values

	$=$	$\frac{47}{2}[242]$	Evaluate

	$=$	$\frac{11374}{2}$

	$=$	$5687$

$S_n=5687$

Question 2 of 5

Find the sum of the first

$20$ terms given that:

$U_1=8$

$U_18=76$

$S_20=$ (920)

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Sum of an Arithmetic Sequence

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

General Rule of an Arithmetic Sequence

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

First, substitute the known values to the general rule and solve for $d$

$\text(18th term)$ $[U_18]$	$=$	$76$
$\text(First term)$ $[a]$	$=$	$8$

$U_{\color{#9a00c7}{n}}$	$=$	$\color{#e65021}{a}+[(\color{#9a00c7}{n}-1)\color{#00880A}{d}]$
$U_{\color{#9a00c7}{18}}$	$=$	$\color{#e65021}{8}+[(\color{#9a00c7}{18}-1)\color{#00880A}{d}]$	Substitute known values
$76$ $-8$	$=$	$8+17d$ $-8$	Subtract $8$ from both sides
$68$ $divide17$	$=$	$17d$ $divide17$	Divide both sides by $17$
$4$	$=$	$d$
$d$	$=$	$4$

Finally, substitute the known values to the sum formula

$\text(Number of terms) [n]$	$=$	$20$
$\text(First Term) [a]$	$=$	$8$
$\text(Common Difference) [d]$	$=$	$4$

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

$S_{\color{#9a00c7}{20}}$	$=$	$\frac{\color{#9a00c7}{20}}{2}[2\cdot\color{#e65021}{8}+(\color{#9a00c7}{20}-1)\color{#00880A}{4}]$	Substitute known values

	$=$	$10[16+(19\cdot4)]$	Evaluate
	$=$	$10(16+76)$
	$=$	$10*92$
	$=$	$920$

$S_20=920$

Question 3 of 5

Find the sequence given that:

$S_10=145$

$S_20=590$

1. $1+4+7+10...$
2. $1+5+9+13...$
3. $1+11+21+31...$
4. $1+3+5+7...$

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Sum of an Arithmetic Sequence

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

General Rule of an Arithmetic Sequence

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

First, use the sum formula to both sums and transform them into general rule form

$S_10$

$S_{\color{#9a00c7}{n}}$	$=$	$\frac{\color{#9a00c7}{n}}{2}[2\color{#e65021}{a}+(\color{#9a00c7}{n}-1)\color{#00880A}{d}]$
$S_{\color{#9a00c7}{10}}$	$=$	$\frac{\color{#9a00c7}{10}}{2}[2\color{#e65021}{a}+(\color{#9a00c7}{10}-1)\color{#00880A}{d}]$	Substitute known values
$145$ $divide5$	$=$	$5(2a+9d)$ $divide5$	Divide both side by $5$
$29$	$=$	$2a+9d$

$S_20$

$S_{\color{#9a00c7}{20}}$	$=$	$\frac{\color{#9a00c7}{20}}{2}[2\color{#e65021}{a}+(\color{#9a00c7}{20}-1)\color{#00880A}{d}]$	Substitute known values
$590$ $divide10$	$=$	$10(2a+19d)$ $divide10$	Divide both side by $10$
$59$	$=$	$2a+19d$

Next, solve for the value of $d$ by subtracting the $S_10$ ’s general rule form from the $S_20$ ’s general rule form

$59$ $-$ $29$	$=$	$($ $2a+19d$ $)-($ $2a+9d$ $)$
$30$ $divide10$	$=$	$10d$ $divide10$	Divide both sides by $10$
$3$	$=$	$d$
$d$	$=$	$3$

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

$29$	$=$	$2a+9$ $d$
$29$	$=$	$2a+9($ $3$ $)$	Substitute $d=5$
$29$ $-27$	$=$	$2a+27$ $-27$	Subtract $27$ from both sides
$2$ $divide2$	$=$	$2a$ $divide2$	Divide both sides by $2$
$1$	$=$	$a$
$a$	$=$	$1$

Finally, start with $a=1$ and keep adding $d=3$ to its value to get the sequence

$U_1$	$=$	$1$
$U_2$	$=$	$1+$ $3$	$=$	$4$
$U_3$	$=$	$4+$ $3$	$=$	$7$
$U_4$	$=$	$7+$ $3$	$=$	$10$

$1+4+7+10…$

Question 4 of 5

Given that

$S_n=301$ , find the value of n

$2+5+8…$

$n=$ (14)

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Sum of an Arithmetic Sequence

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

Common Difference Formula

$d=U_2-U_1=U_3-U_2$

Quadratic Formula

$n=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

First, solve for the value of $d$ .

$\color{#00880A}{d}$	$=$	$U_2-U_1$
	$=$	$5-2$	Substitute the first and second term
	$=$	$3$

Next, substitute the known values to the sum formula

$\text(Sum of terms) [S_n]$	$=$	$301$
$\text(First term) [a]$	$=$	$2$
$\text(Common difference) [d]$	$=$	$3$

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

$301$	$=$	$\frac{\color{#9a00c7}{n}}{2}[2\cdot\color{#e65021}{2}+(\color{#9a00c7}{n}-1)\color{#00880A}{3}]$	Substitute known values

$301$ $times2$	$=$	$n/2 [4+3n-3]$ $times2$	Multiply both sides by $2$

$602$	$=$	$n[1+3n]$
$602$	$=$	$3n^2+n$	Distribute
$602$ $-602$	$=$	$3n^2+n$ $-602$	Subtract $602$ from both sides
$0$	$=$	$3n^2+n-602$
$3n^2+n-602$	$=$	$0$

Finally, use the quadratic formula to solve for

$n$

$a$	$=$	$3$
$b$	$=$	$1$
$c$	$=$	$-602$

$n$	$=$	$\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

	$=$	$\frac{-1\pm \sqrt{(-1)^2-4\cdot3\cdot(-603)}}{2(3)}$	Substitute known values

	$=$	$\frac{-1\pm \sqrt{1+7224}}{6}$	Evaluate

	$=$	$\frac{-1\pm \sqrt{7225}}{6}$

	$=$	$\frac{-1\pm 85}{6}$

Solve for the two values of

$n$ .

$n$	$=$	$\frac{-1+85}{6}$

	$=$	$\frac{84}{6}$

	$=$	$14$

$n$	$=$	$\frac{-1-85}{6}$

	$=$	$\frac{-86}{6}$

	$=$	$-14.\overline{33}$

Since the negative value has a repeating decimal, it is considered irrational.

Hence, the value of

$n$ is

$14$ .

$n=14$

Question 5 of 5

Given that

$S_n=39$ , find the value of the last term

$1 1/2+1 3/4+2+2 1/4…$

1. $3 1/4$
2. $5 1/2$
3. $4 1/2$
4. $2 1/4$

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Sum of an Arithmetic Sequence

$S_{\color{#9a00c7}{n}}=\frac{\color{#9a00c7}{n}}{2}[\color{#e65021}{a}+\color{#00880A}{l}]$

Common Difference Formula

$d=U_2-U_1=U_3-U_2$

General Rule of an Arithmetic Sequence

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

First, solve for the value of $d$ .

$\color{#00880A}{d}$	$=$	$U_2-U_1$
	$=$	$1 3/4-1 1/2$	Substitute the first and second term

	$=$	$1/4$

Next, substitute the known values to the sum formula

$\text(Sum of terms) [S_n]$	$=$	$39$

$\text(First term) [a]$	$=$	$1 1/2$

$\text(Common difference) [d]$	$=$	$1/4$

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

$39$	$=$	$\frac{\color{#9a00c7}{n}}{2}\left[2\cdot\color{#e65021}{1\frac{1}{2}}+(\color{#9a00c7}{n}-1)\color{#00880A}{\frac{1}{4}}\right]$	Substitute known values

$39$ $times2$	$=$	$n/2 [3+n/4-1/4]$ $times2$	Multiply both sides by $2$

$78$	$=$	$n[11/4+n/4]$

$78$	$=$	$(11n)/4+(n^2)/4$	Distribute

$78$ $times4$	$=$	$((11n)/4+(n^2)/4)$ $times4$	Multiply both sides by $4$

$312$ $-312$	$=$	$11n+n^2$ $-312$	Subtract $312$ from both sides
$0$	$=$	$n^2+11n-312$
$n^2+11n-312$	$=$	$0$

Since the equation is in standard form

$(ax^2+bx+c=0)$ we can factorise using the cross method.

$n^2$ $+11$

$n$ $-312$

$=0$

To factorise, we need to find two numbers that add to

$11$ and multiply to

$-312$

$24$ and

$-13$ fit both conditions

$24 + (-13)$	$=$	$11$
$(24) xx (-13)$	$=$	$-312$

Read across to get the factors.

$(n+24)(n-13)$

Since we need a positive value for

$n$ , we need to use $n=13$ .

Finally, substitute the known values to the general rule

$\text(Number of terms)$ $[n]$	$=$	$13$

$\text(First term)$ $[a]$	$=$	$1 1/2$

$\text(Common Difference)$ $[d]$	$=$	$1/4$

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

$U_{\color{#9a00c7}{13}}$	$=$	$\color{#e65021}{1\frac{1}{2}}+[(\color{#9a00c7}{13}-1)\color{#00880A}{\frac{1}{4}}]$	Substitute known values

	$=$	$1 1/2+12/4$	Evaluate

	$=$	$4 1/2$

$\text(Last term)=4 1/2$

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Quiz summary

Information

1. Question

Hint

Great Work!

Sum of an Arithmetic Sequence

Common Difference Formula

General Rule of an Arithmetic Sequence

2. Question

Hint

Correct!

Sum of an Arithmetic Sequence

General Rule of an Arithmetic Sequence

3. Question

Hint

Keep Going!

Sum of an Arithmetic Sequence

General Rule of an Arithmetic Sequence

4. Question

Hint

Fantastic!

Sum of an Arithmetic Sequence

Common Difference Formula

Quadratic Formula

5. Question

Hint

Excellent!

Sum of an Arithmetic Sequence

Common Difference Formula

General Rule of an Arithmetic Sequence

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