tentukan jumlah barisan aritmetika dai 4+9+14+19+.......+104

b=9-4=5
un=a+(n-1)b
104=4+(n-1)5
104-4=5n-5
100+5=5n
105=5n
n=21
sn =n/2(a+ un)
s25=21/2(4+104)
s25=21/2(108)
s25=1134

[tex]a = 4 \\ b = 9 - 4 = 5 \\ Un = 104 \\ a + (n - 1)b = 104 \\ 4 + 5(n - 1) = 104 \\ 5(n - 1) = 100 \\ n = 21 \\ S_{21} = \frac{21}{2}(2(4) + 20(5)) \\ = \frac{21}{2} (108) \\ = 1134[/tex]