Question

NO LINKS! PLEASE help me with this problem 3f​

Answer 1

`\textsf{a)} \quad a_1=-6, \quad d=4`

Explanation:

General formula of an arithmetic sequence

`\boxed{s_n=s_1+(n-1)d}`

where:

• sₙ is the nth term.
• s₁ is the first term.
• d is the common difference between terms.
• n is the position of the term.

Given terms:

• s₅ = 10
• s₈ = 22

Substitute the given terms into the formula to create two equations:

`\begin{aligned}\implies s_5=s_1+(5-1)d&=10\\s_1+4d&=10\end{aligned}`

`\begin{aligned}\implies s_8=s_1+(8-1)d&=22\\s_1+7d&=22\end{aligned}`

Subtract the first equation from the second equation to eliminate s₁:

`\begin{array}{rrclcl}& s_1 &+ &7d &= &22\\-& (s_1 &+ &4d& =& 10)\\\cline{2-6}&&&3d &=& 12\end{array}`

Solve for d:

`\implies 3d=12`

`\implies (3d)/(3)=(12)/(3)`

`\implies d=4`

Substitute the found value of d into one of the equations and solve for s₁:

`\implies s_1+4(4)=10`

`\implies s_1+16=10`

`\implies s_1=-6`

Therefore:

• s₁ = -6
• d = 4

Answer 2

Answer: Choice A

`s_1 = -6, \ d = 4`

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Work Shown:

a =
`s_1` = first term

d = common difference

Plug n = 5 into the arithmetic sequence formula to get...

`s_n = s_1 + d(n-1)\\\\s_5 = a + d(5-1)\\\\10 = a + 4d\\\\a+4d = 10\\\\a = 10-4d\\\\`

Do similar steps for n = 8

`s_n = s_1 + d(n-1)\\\\s_8 = a + d(8-1)\\\\22 = a + 7d\\\\a+7d = 22\\\\`

Now plug in a = 10-4d and solve for d.

`a+7d = 22\\\\10-4d+7d = 22\\\\10+3d = 22\\\\3d = 22-10\\\\3d = 12\\\\d = 12/3\\\\d = 4\\\\`

The common difference is d = 4. This narrows the answer choices between A and D.

Use d = 4 to find 'a'.

`a = 10-4d\\\\a = 10-4*4\\\\a = 10-16\\\\a = -6\\\\`

Alternatively you could use the other equations that had 'a' and 'd' in them.

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The first term is a = -6 and the common difference is d = 4

The arithmetic sequence is:

-6, -2, 2, 6, 10, 14, 18, 22, ...

The terms in bold represent the 5th and 8th terms respectively. They are highlighted to help confirm the answers.

It might help to make a table like this

`\begin{array}c \cline{1-2}n & s_n\\\cline{1-2}1 & -6\\\cline{1-2}2 & -2\\\cline{1-2}3 & 2\\\cline{1-2}4 & 6\\\cline{1-2}5 & \boldsymbol{10}\\\cline{1-2}6 & 14\\\cline{1-2}7 & 18\\\cline{1-2}8 & \boldsymbol{22}\\\cline{1-2}\end{array}`

or like this

`\begin{array}c \cline{1-9}n & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8\\\cline{1-9}s_n & -6 & -2 & 2 & 6 & \boldsymbol{10} & 14 & 18 & \boldsymbol{22}\\\cline{1-9}\end{array}`

Each time you need a new term, add on 4.