What would the second and third terms of an arithmetic sequence be if the first term is 6 and the fourth is 33?

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What would the second and third terms of an arithmetic sequence be if the first term is 6 and the fourth is 33?

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Answer 1 · 2018-05-16T13:07:06

The second term = 15, and the third term = 24

Explanation:

An arithmetic sequence can be written on the form
$s_{n} = a n + b$ $s_n=an+b$
We know that $s_{1} = 6$ $s_1=6$ and $s_{4} = 33$ $s_4=33$
Therefore: $s_{4} - s_{1}$ $s_4-s_1$ =(4a+b)-(a+b) $= 33 - 6 = 27$ $=33-6=27$
$3 a = 27$ $3a=27$
$a = 9$ $a=9$
As $6 = 9 + b$ $6=9+b$ ,
$b = - 3$ $b=-3$
Therefore $s_{n} = 9 n - 3$ $s_n=9n-3$
We, therefore, get:
$s_{1} = 1 \cdot 9 - 3 = 6$ $s_1=1*9-3=6$
$s_{2} = 2 \cdot 9 - 3 = 15$ $s_2=2*9-3=15$
$s_{3} = 3 \cdot 9 - 3 = 24$ $s_3=3*9-3=24$
$s_{4} = 4 \cdot 9 - 3 = 33$ $s_4=4*9-3=33$

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What would the second and third terms of an arithmetic sequence be if the first term is 6 and the fourth is 33?

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