Question

write the first six terms of the arithmetic sequence with the given
$a_1 = 280, d = -90$
$a_1 = 280$
$a_2 = \square$
$a_3 = \square$
$a_4 = \square$
$a_5 = \square$
$a_6 = \square$

Explanation:

Step1: Recall arithmetic sequence formula

The formula for the \(n\)-th term of an arithmetic sequence is \(a_n = a_1+(n - 1)d\), where \(a_1\) is the first term and \(d\) is the common difference. For the second term (\(n = 2\)):
\(a_2=a_1 + d\)
Substitute \(a_1 = 280\) and \(d=-90\):
\(a_2=280+(-90)=280 - 90 = 190\)

Step2: Find \(a_3\)

Using the formula \(a_3=a_2 + d\) (or \(a_3=a_1+2d\)):
\(a_3 = 190+(-90)=190 - 90 = 100\)

Step3: Find \(a_4\)

Using \(a_4=a_3 + d\):
\(a_4 = 100+(-90)=100 - 90 = 10\)

Step4: Find \(a_5\)

Using \(a_5=a_4 + d\):
\(a_5 = 10+(-90)=10 - 90=-80\)

Step5: Find \(a_6\)

Using \(a_6=a_5 + d\):
\(a_6=-80+(-90)=-80 - 90=-170\)

Answer:

\(a_2 = 190\), \(a_3 = 100\), \(a_4 = 10\), \(a_5=-80\), \(a_6=-170\)