AP - simple

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

Result

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

Solution:

a10=1323=129a_{10}=132-3 = 129
a9=1293=126a_9=129-3 = 126
a8=1263=123a_8=126-3 = 123
a7=1233=120a_7=123-3 = 120
a6=1203=117a_6=120-3 = 117
a5=1173=114a_5=117-3 = 114
a4=1143=111a_4=114-3 = 111
a3=1113=108a_3=111-3 = 108
a2=1083=105a_2=108-3 = 105
a1=1053=102a_1=105-3 = 102



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