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B. Find the value of x; then write the first 5 terms of
the arithmetic sequence, given a₁ and d.
1) a₁ = x, d=3x, a₁ = 25
2) a₁ = x, d = ½ x, a₁₁ = 30
3) a
3
1=x, d=8x, a7=10
4) a₁ =x, d=x, a₁ = 36
5) a₁ =x, d = −5x, ag = -11.5
12
​

Sagot :

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Answer:

Problem 1:

Given:

[tex](a_1 = x ), ( d = 3x ), ( a_1 = 25 )[/tex]

[tex]x = 25[/tex]

The common difference ( d ) is:

[tex]d = 3x = 3 \times 25 = 75 [/tex]

First five terms of the sequence:

[tex]a_1 = 25[/tex]

[tex]a_2 = a_1 + d = 25 + 75 = 100[/tex]

[tex]a_3 = a_2 + d = 100 + 75 = 175[/tex]

[tex]a_4 = a_3 + d = 175 + 75 = 250[/tex]

[tex]a_5 = a_4 + d = 250 + 75 = 325[/tex]

[tex] \text{First five terms:} 25, 100, 175, 250, 325 [/tex]

Problem 2:

Given:

[tex] ( a_1 = x ), ( d = \frac{1}{2}x ), ( a_{11} = 30 )[/tex]

[tex] \text{Since} ( a_1 = x ),[/tex]

The ( n )-th term of an arithmetic sequence is given by:

[tex]a_n = a_1 + (n-1)d[/tex]

Substitute ( n = 11 ):

[tex]a_{11} = x + (11-1) \left( \frac{1}{2} x \right) = 30[/tex]

[tex]30 = x + 10 \left( \frac{1}{2} x \right)[/tex]

[tex]30 = x + 5x[/tex]

[tex]30 = 6x[/tex]

[tex]x = 5[/tex]

The common difference ( d ) is:

[tex]d = \frac{1}{2}x = \frac{1}{2} \times 5 = 2.5[/tex]

First five terms of the sequence:

[tex]a_1 = 5 [/tex]

[tex]a_2 = a_1 + d = 5 + 2.5 = 7.5[/tex]

[tex]a_3 = a_2 + d = 7.5 + 2.5 = 10[/tex]

[tex]a_4 = a_3 + d = 10 + 2.5 = 12.5[/tex]

[tex]a_5 = a_4 + d = 12.5 + 2.5 = 15[/tex]

First five terms: [ 5, 7.5, 10, 12.5, 15 ]

Problem 3:

Given:

[tex]( a_1 = x ), ( d = 8x ), ( a_7 = 10 )[/tex]

[tex]{ \text{We need to find ( x ) such that }( a_7 = 10 ).}[/tex]

The ( n )-th term of an arithmetic sequence is given by:

[tex]a_n = a_1 + (n-1)d[/tex]

Substitute ( n = 7 ):

[tex]a_7 = x + (7-1) \left( 8x \right) = 10[/tex]

[tex]10 = x + 6 \left( 8x \right)[/tex]

[tex]10 = x + 48x[/tex]

[tex]10 = 49x[/tex]

[tex]x = \frac{10}{49} [/tex]

The common difference ( d ) is:

[tex]d = 8x = 8 \left( \frac{10}{49} \right) = \frac{80}{49}[/tex]

First five terms of the sequence:

[tex]a_1 = \frac{10}{49}[/tex]

[tex]a_2 = a_1 + d = \frac{10}{49} + \frac{80}{49} = \frac{90}{49}[/tex]

[tex]a_3 = a_2 + d = \frac{90}{49} + \frac{80}{49} = \frac{170}{49}[/tex]

[tex]a_4 = a_3 + d = \frac{170}{49} + \frac{80}{49} = \frac{250}{49} [/tex]

[tex]a_5 = a_4 + d = \frac{250}{49} + \frac{80}{49} = \frac{330}{49}[/tex]

First five terms:

[tex]\frac{10}{49}, \frac{90}{49}, \frac{170}{49}, \frac{250}{49}, \frac{330}{49}[/tex]

Problem 4:

Given:

[tex]( a_1 = x ), ( d = x ), ( a_1 = 36 )[/tex]

The common difference ( d ) is:

[tex]d = x = 36[/tex]

First five terms of the sequence:

[tex]a_1 = 36[/tex]

[tex]a_2 = a_1 + d = 36 + 36 = 72

[/tex]

[tex]a_3 = a_2 + d = 72 + 36 = 108[/tex]

[tex]a_4 = a_3 + d = 108 + 36 = 144[/tex]

[tex]a_5 = a_4 + d = 144 + 36 = 180[/tex]

First five terms:

[tex] [ 36, 72, 108, 144, 180 ][/tex]

Problem 5:

Given:

[tex]{( a_1 = x ), ( d = -5x ), ( a_{12} = -11.5 )}[/tex]

[tex] {\text{We need to find ( x ) such that} ( a_{12} = -11.5 ).}[/tex]

The ( n )-th term of an arithmetic sequence is given by:

[tex]a_n = a_1 + (n-1)d [/tex]

Substitute ( n = 12 ):

[tex]{a_{12} = x + (12-1) \left( -5x \right) = -11.5}[/tex]

[tex]-11.5 = x - 55x[/tex]

[tex]-11.5 = -54x[/tex]

[tex]x = \frac{11.5}{54} \approx \frac{23}{108}[/tex]

The common difference ( d ) is:

[tex]d = -5x = -5 \left( \frac{23}{108} \right) = \frac{-115}{108} [/tex]

First five terms of the sequence:

[tex]a_1 = \frac{23}{108}[/tex]

[tex]{a_2 = a_1 + d = \frac{23}{108} + \frac{-115}{108} = \frac{23 - 115}{108} = \frac{-92}{108} = \frac{-46}{54}}[/tex]

[tex]{a_3 = a_2 + d = \frac{-46}{54} + \frac{-115}{108} = \frac{-92 - 115}{108} = \frac{-207}{108} = -1.9167}[/tex]

[tex]{a_4 = a_3 + d = -1.9167 + \frac{-115}{108} = -1.9167 - 1.0648 = -2.9815}[/tex]

[tex]{a_5 = a_4 + d = -2.9815 + \frac{-115}{108} = -2.9815 - 1.0648 = -4.0463}[/tex]

First five terms:

[tex] \frac{23}{108}, \frac{-46}{54}, -1.9167, -2.9815, -4.0463[/tex]

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