jika Barisan aritmatika memiliki U2=7 dan U6=19,Maka hitunglah beda(b),suku pertama (a),suku ke empat puluh satu(U41),dan jumlah suku ke-50! Tolong Bantu Jawab Yaa Semuanya,Thank You🙏

<p>Diketahui:</p><p>U₂= 7</p><p>U₆=19</p><p>Ditanya:</p><p>Beda (b)=....</p><p>Suku pertama (a)=.....</p><p>suku ke empat puluh satu(U41)=....</p><p>jumlah suku ke-50 (s₅₀)=.....</p><p>&nbsp;</p><p>Rumus suku ke-n=a+(n–1)b masukkan suku yang sudah diketahui.</p><p>U₂= a+ (2–1)b</p><p>7 = a+b &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; (i)</p><p>&nbsp;</p><p>U₆= a+ (6–1)b</p><p>19= a+ 5b &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;(ii)</p><p>&nbsp;</p><p>Eliminasi persamaan (i) dan (ii)</p><p>a + b = 7</p><p><u>a + 5b=19 –</u></p><p>&nbsp; &nbsp;–4b = –12</p><p>&nbsp; &nbsp; &nbsp; &nbsp; b = –12 : –4 &nbsp; &nbsp;</p><p>&nbsp; &nbsp; &nbsp; &nbsp; b = 3</p><p>&nbsp;</p><p>Substitusi nilai b=3 pada persamaan (i)</p><p>a + 3 = 7</p><p>&nbsp; &nbsp; &nbsp; &nbsp;a = 7 – 3</p><p>&nbsp; &nbsp; &nbsp; &nbsp;a = 4</p><p>&nbsp;</p><p>U_n=a + (n–1)b</p><p>U₄₁= 4 + (41–1)3</p><p>&nbsp; &nbsp; &nbsp; = 4 + 120</p><p>&nbsp; &nbsp; &nbsp; =124</p><p>&nbsp;</p><p>S_n= ½n (2a + (n–1)b)</p><p>S₅₀=½.50 (2.4 + (50–1)3)</p><p>&nbsp; &nbsp; &nbsp;= 25 (8 + 147)</p><p>&nbsp; &nbsp; &nbsp;= 25 (155)</p><p>&nbsp; &nbsp; &nbsp;= 3.875</p><p>&nbsp;</p><p>Jadi, Beda(b) = &nbsp;3</p><p>Suku pertama (a)= 4</p><p>suku ke empat puluh satu(U₄₁)= 124</p><p>jumlah suku ke-50 (S₅₀)= 3.875</p>