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SIMAK UI 2010

Solusi SIMAK UI 2010 Kode 509 [IPA] – Nomor 4


Jumlah\ p\ suku\ pertama\ dari\ suatu\ barisan\ aritmetika\ ialah\ q\ dan\\jumlah\ q\ suku\ pertama\ ialah\ p,\ maka\ jumlah\ (p+q)\ suku\ pertama\\dari\ barisan\ tersebut\ adalah\ ....
(A)\quad p+q
(B)\quad \frac{p+q}{2}
(C)\quad p+q+1
(D)\quad -(p+q)
(E)\quad -(p+q+1)
Jawaban: D
Dengan menggunakan rumus jumlah n suku pertama deret aritmetika

S_n=\frac{n}{2}\Big(2a+(n-1)b\Big)
diperoleh:

S_p=q
\frac{p}{2}\big(2a+(p-1)b\big)=q
p(2a+pb-b)=2q
2a+pb-b=\frac{2q}{p}\quad ..............\quad (1)

S_q=p
\frac{q}{2}\big(2a+(q-1)b\big)=p
q(2a+qb-b)=2p
2a+qb-b=\frac{2p}{q}\quad ..............\quad (2)
Selisih (1) dan (2) menghasilkan:

(p-q)b=\frac{2q}{p}-\frac{2p}{q}=\frac{2(q^2-p^2 )}{pq}=\frac{-2(p^2-q^2)}{pq}=\frac{-2(p-q)(p+q)}{pq}
b=\frac{-2(p+q)}{pq}\quad ..............\quad (3)
Substitusi (3) ke (1) menghasilkan:

2a=\frac{2q}{p}-(p-1)b=\frac{2q}{p}-(p-1)\big(\frac{-2(p+q)}{pq}\big)
2a=\frac{2q}{p}+\frac{2(p-1)(p+q)}{pq}
a=\frac{q}{p}+\frac{(p-1)(p+q)}{pq}=\frac{q^2+(p-1)(p+q)}{pq}\quad ..............\quad (4)
sehingga:

\begin{array} {lcl} S_{p+q}&=&\frac{p+q}{2}\Big(2\Big(\frac{q^2+(p-1)(p+q)}{pq}\Big)\Big)+(p+q-1)\Big(\frac{-2(p+q)}{pq}\Big)\\\\&=&(p+q)\Big(\frac{q^2+(p-1)(p+q)}{pq}-\frac{(p+q-1)(p+q)}{pq}\Big)\\\\&=&(p+q)\bigg(\frac{q^2+(p+q)\big(p-1-(p+q-1)\big)}{pq}\bigg)\\\\&=&(p+q)\Big(\frac{q^2+(p+q)(-q)}{pq}\Big)\\\\&=&(p+q)\Big(\frac{q^2-pq-q^2}{pq}\Big)\\\\&=&(p+q)(-1)\\\\&=&-(p+q)\end{array}

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Guru Matematika SMK

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