Diketahui Barisan aritmetika dengan u1+u10+u19=96 Suku ke 10 barisan tersebut adalah?

Berdasarkan [tex]U_n = a+(n-1)b[/tex], diperoleh
[tex]U_1 = a + (1-1)b = a + 0 \cdot b = a \\ U_2 = a + (2 - 1)b = a + 1 \cdot b = a + b \\ U_3 = a + (3 - 1)b = a + 2 \cdot b= a+2b [/tex]
Diperoleh
[tex] U_1 + U_{10} + U_{19} = 96 \\ a + a + 9b + a + 18b = 96 \\ 3a + 27b = 96 \\ 3(a + 9b) = 96 \\ a + 9b = \frac{96}{3} \\ a + (10 - 1)b = 32 \\ \therefore U_{10} = 32 [/tex]