# AP - simple

Find the first ten members of the sequence if a11 = 132, d = 3.

Correct result:

a10 =  129
a9 =  126
a8 =  123
a7 =  120
a6 =  117
a5 =  114
a4 =  111
a3 =  108
a2 =  105
a1 =  102

#### Solution:

$a_{10}=132-3=129$
$a_9=129-3=126$
$a_8=126-3=123$
$a_7=123-3=120$
$a_6=120-3=117$
$a_5=117-3=114$
$a_4=114-3=111$
$a_3=111-3=108$
$a_2=108-3=105$
$a_1=105-3=102$

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