HOTS Zone : Barisan dan Deret Aritmetika [2]
Table of Contents

Tipe
No.
Masing-masing bilangan dari sekumpulan- 44.505
- 42.705
- 37.855
- 35.555
ALTERNATIF PENYELESAIAN
\(\begin{aligned}
10\times21+10\times22+\cdots+10\times30+11\times21+\cdots+11\times30+\cdots+20\times30&=10\times(21+22+\cdots+30)+11\times(21+22+\cdots+30)+\cdots+20\times(21+22+\cdots+30)\\
&=(10+11+\cdots+20)\times(21+22+\cdots+30)\\
&=\left(\dfrac{11(10+20)}2\right)\times\left(\dfrac{10(21+30)}2\right)\\[4pt]
&=165\times255\\
&=\boxed{\boxed{42075}}
\end{aligned}\)
Jadi, jumlah semua hasil kalinya adalah 42.705.
JAWAB: B
JAWAB: B
No.
Hasil dari operasi $2025+\left(3+\dfrac1{2020}\right)+\left(3+\dfrac3{2020}\right)+\left(3+\dfrac5{2020}\right)+\cdots+\left(3+\dfrac{4039}{2020}\right)=$ ....
- 22.222
- 20.202
- 10.105
- 10.100
- 10.010
ALTERNATIF PENYELESAIAN
Banyak suku dari $\left(3+\dfrac1{2020}\right)$ sampai $\left(3+\dfrac{4039}{2020}\right)$
$n=\dfrac{4039-1}{2}+1=2020$
\(\begin{aligned} 2025+\left(3+\dfrac1{2020}\right)+\left(3+\dfrac3{2020}\right)+\left(3+\dfrac5{2020}\right)+\cdots+\left(3+\dfrac{4039}{2020}\right)&=2025+2020\times3+\dfrac{1+3+5+\cdots+4039}{2020}\\[4pt] &=2025+6060+\dfrac{\dfrac{2020}2(1+4039)}{2020}\\[4pt] &=8085+2020\\ &=\boxed{\boxed{10105}} \end{aligned}\)
$n=\dfrac{4039-1}{2}+1=2020$
\(\begin{aligned} 2025+\left(3+\dfrac1{2020}\right)+\left(3+\dfrac3{2020}\right)+\left(3+\dfrac5{2020}\right)+\cdots+\left(3+\dfrac{4039}{2020}\right)&=2025+2020\times3+\dfrac{1+3+5+\cdots+4039}{2020}\\[4pt] &=2025+6060+\dfrac{\dfrac{2020}2(1+4039)}{2020}\\[4pt] &=8085+2020\\ &=\boxed{\boxed{10105}} \end{aligned}\)
Jadi, hasil dari operasi $2025+\left(3+\dfrac1{2020}\right)+\left(3+\dfrac3{2020}\right)+\left(3+\dfrac5{2020}\right)+\cdots+\left(3+\dfrac{4039}{2020}\right)=10{.}105$.
JAWAB: C
JAWAB: C
Post a Comment