# 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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