Question

interpole dez meios aritméticos entre 8 e 63

Answer 1

a2= 8 + ( 2 -1) .5  =  13
a3 = 8 + ( 3 -1) .5 =  18
a4 = 8 + ( 4 -1) .5  =  23
a5 = 8 + ( 5 -1) .5  =  28
a6 = 8 + ( 6 -1) .5  =  33
a7 = 8 + ( 7 -1) .5  =  38
a8 = 8 + ( 8 -1) .5  =  43
a9 = 8 + ( 9 -1) .5  =  48
a10 = 8 + ( 10 -1) .5 =  53
an = 8 + ( 11 -1) .5  =  58


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PA = ( 8, 13, 18, 23, 28, 33, 38, 43, 48, 53, 58, 63) 

Answer 2

$calcular~~raz\tilde{a}o \\ \\ a_n=63 \\ a_1=8 \\ n=10+2=12~~(10~~que~~interpola~~mais~~2~\mapsto a_1~~e~~a_n) \\ r=? \\ \\ a_n=a_1+(n-1)r \\ \\ 63=8+(12-1)r \\ 63=8+11r \\ 63-8=11r \\ 55=11r \\ r=55\div11 \\ r=5$


$a_1=8 \\ a_2=a_1+r=8+5=13 \\ a_3=a_1+2r=8+10=18 \\ a_4=a_1+3r=8+15=23 \\ a_5=a_1+4r=8+20=28 \\ a_6=a_1+5r=8+25=33 \\ a_7=a_1+6r=8+30=38 \\ a_8=a_1+7r=8+35=43 \\ a_9=a_1+8r=8+40=48 \\ a_{10}=a_1+9r=8+45=53 \\ a_{11}=a_1+10r=8+50=58 \\ a_{12}=63 \\ \\ P.A.(8,\underline{13},\underline{18},\underline{23},\underline{28},\underline{33},\underline{38},\underline{43},\underline{48},\underline{53},\underline{58},63)$