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U3 => a + 2b = 14 _ eliminasikan
4b = 12
b = 12/4
b = 3
subtitusikan
a + 2b = 14
a + 2(3) = 14
a + 6 = 14
a = 14 - 6
a = 8
jumlah suku
Sn = n/2(2a+(n-1)b)
S18 = 18/2 ( 2 x 8 + ( 18-1) 3)
= 9 ( 16 + 57)
= 9 ( 73)
= 657
jawabnnya 657
u7=26
jumlah 18 suku pertama?
u3=a+2b=14...........(i)
u7=a+6b=26............(ii)
dari persamaan (i)dan(ii),diperoleh
a+6b=26
a+2b=14
---------------------- -
4b=12 b=3..................(iii)
subtitusi persamaan (iii) ke persamaan (i), diperoleh:
a+2b=14
a+2.3=14
a+6=14
a=14-6=8
jumlah 18 suku pertama(n=18)=
Sn=N/2(2a+(n-1)b)
Sn=18/2(2.8+(18-1)3
Sn=9(16+17).3
Sn=9.99
Sn=9801