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[SPAM] Re: [obm-l] PA

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A PA =E9 igual a :

a0, a1, a2, .... a53

invertendo a ordem
a53, a52, a51....a0

somando os termos
a0, a1, a2, .... a53
a53, a52, a51....a0
=3D
a0+a53=3D a1+a52 =3D a2+a51....

Temos portanto 54 termos (a0+a53). Para chegar =E0 soma total, precisando
dividir por 2 por contamos duas vezes.

Somat=F3rio =3D 54 (a0+a53)/2 =3D 27 (a0+a53) =3D 1107

a0 + a53 =3D 1107/54

Acontece que
a53 =3D a0 + 53 n
n =E9 a raz=E3o da PA

a0 + a0 + 53 n =3D 1107/54

2 a0 + 53 n =3D 1107/54 (Equa=E7=E3o 1)

a23 =3D a0 + 23 n
a32 =3D a0 + 32 n
a32 - a23 =3D 9 n =3D 7
n =3D 7/9 (Equa=E7=E3o 2)

2ao + 53 (7/9) =3D 1107/54
2ao =3D 1107/54- 53*7/9

Multiplicando tudo por 6

12 ao =3D 1107 * 6/54 - 53*7*6/9
12 ao =3D 1107/9 - 2226/9
12 ao =3D -1119
ao =3D -1119/12

a23 =3D -1119/12 + 23 (7/9)
a32 =3D -1119/12 + 32 (7/9)

Seria isso?

2008/5/23 Thelio Gama <teliogama@xxxxxxxxx>:

> Boa tarde, professores,
>
> N=E3o consegui resolver esta PA:
>
> *A soma dos 54 termos de uma PA =E9 1107. Determine o valor dos termos  a=
23
> e   a32  sabendo que a diferen=E7a entre eles =E9 igual a 7.*
>
> Agrade=E7o a ajuda,
>
> Thelio
>
>

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A PA =E9 igual a :<br><br>a0, a1, a2, .... a53<br><br>invertendo a ordem<br=
>a53, a52, a51....a0<br><br>somando os termos<br>a0, a1, a2, .... a53<br>
a53, a52, a51....a0<br>
=3D<br>a0+a53=3D a1+a52 =3D a2+a51....<br><br>Temos portanto 54 termos (a0+=
a53). Para chegar =E0 soma total, precisando dividir por 2 por contamos dua=
s vezes.<br><br>Somat=F3rio =3D 54 (a0+a53)/2 =3D 27 (a0+a53) =3D 1107<br><=
br>a0 + a53 =3D 1107/54<br>
<br>Acontece que <br>a53 =3D a0 + 53 n <br>n =E9 a raz=E3o da PA<br><br>a0 =
+ a0 + 53 n =3D 1107/54<br><br>2 a0 + 53 n =3D 1107/54 (Equa=E7=E3o 1)<br><=
br>a23 =3D a0 + 23 n<br>a32 =3D a0 + 32 n<br>a32 - a23 =3D 9 n =3D 7<br>n =
=3D 7/9 (Equa=E7=E3o 2)<br>
<br>2ao + 53 (7/9) =3D 1107/54<br>2ao =3D 1107/54- 53*7/9<br><br>Multiplica=
ndo tudo por 6<br><br>12 ao =3D 1107 * 6/54 - 53*7*6/9<br>12 ao =3D 1107/9 =
- 2226/9<br>12 ao =3D -1119<br>ao =3D -1119/12<br><br>a23 =3D -1119/12 + 23=
 (7/9)<br>
a32 =3D -1119/12 + 32 (7/9)<br><br>Seria isso?<br><br><div class=3D"gmail_q=
uote">2008/5/23 Thelio Gama &lt;<a href=3D"mailto:teliogama@xxxxxxxxx";>teli=
ogama@xxxxxxxxx</a>&gt;:<br><blockquote class=3D"gmail_quote" style=3D"bord=
er-left: 1px solid rgb(204, 204, 204); margin: 0pt 0pt 0pt 0.8ex; padding-l=
eft: 1ex;">
<div>Boa tarde, professores,</div>
<div>&nbsp;</div>
<div>N=E3o consegui resolver esta PA:</div>
<div>&nbsp;</div>
<div><span style=3D"font-size: 10pt;"><font color=3D"#000099"><b>A soma dos=
 54 termos de uma PA =E9 1107. Determine o valor dos termos &nbsp;a<span><f=
ont size=3D"1">23</font></span>&nbsp; e &nbsp;&nbsp;<font face=3D"Times New=
 Roman">a<span><font size=3D"1">32</font></span></font>&nbsp; sabendo que a=
 diferen=E7a entre eles =E9 igual a 7.</b></font></span></div>


<div>&nbsp;</div>
<div>Agrade=E7o a ajuda,</div>
<div>&nbsp;</div><font color=3D"#888888">
<div>Thelio</div>
<div>&nbsp;</div>
</font></blockquote></div><br>

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