Question

1.tentukan hasil dari 1+2+3+...+48+49+50
2.3 x (-7) + (16):(-2)

Answer 1

1).
metode pasangan
1+2+3+...+48+49+50
=(1+50)+(2+49)+(3+48)+...+(25+26)
=51+51+51+...+51
=51×25
=1275

metode deret
a=1, b=1, Un=50
n=(Un-a+b)/1=(50-1+1)/1=50/1=50
Sn=n/2(a+Un)=50/2(1+50)=25×51=1275

2).
3 x (-7) + (16):(-2)
=(3×(-7))+(16:(-2))
=-21+(-8)
=-21-8
=-29