This mathematics problem involves applying core mathematical principles and formulas. Below you will find a complete step-by-step solution with detailed explanations for each step, helping you understand not just the answer but the method behind it.
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Haya, twende na swali la 17. 17. Nambari ya kwanza ya mfuatano wa kihesabu (A.P) unaoongezeka inatolewa na (2x - 1) na tofauti ya kawaida ni x. Zao la nambari ya 1^st na 3^rd ni 21. a) Tafuta thamani ya x. Step 1: Andika nambari ya kwanza (a) na tofauti ya kawaida (d) ya A.P. a = 2x - 1 d = x Step 2: Tafuta nambari ya 1^st (T_1) na nambari ya 3^rd (T_3) kwa kutumia fomula T_n = a + (n-1)d. T_1 = a = 2x - 1 T_3 = a + 2d = (2x - 1) + 2(x) = 2x - 1 + 2x = 4x - 1 Step 3: Tumia masharti yaliyotolewa kwamba zao la T_1 na T_3 ni 21. T_1 × T_3 = 21 (2x - 1)(4x - 1) = 21 Step 4: Panua na tatua mlinganyo wa quadratic. 8x^2 - 2x - 4x + 1 = 21 8x^2 - 6x + 1 = 21 8x^2 - 6x - 20 = 0 Gawanya kwa 2: 4x^2 - 3x - 10 = 0 Step 5: Tatua mlinganyo wa quadratic kwa kutumia factorization. 4x^2 - 8x + 5x - 10 = 0 4x(x - 2) + 5(x - 2) = 0 (4x + 5)(x - 2) = 0 Hivyo, 4x + 5 = 0 au x - 2 = 0. x = -(5)/(4) au x = 2 Step 6: Angalia masharti ya "mfuatano wa kihesabu unaoongezeka". Hii inamaanisha tofauti ya kawaida (d) lazima iwe chanya. Ikiwa x = -(5)/(4), basi d = -(5)/(4), ambayo inafanya mfuatano kupungua. Hivyo, x = -(5)/(4) si jibu sahihi. Ikiwa x = 2, basi d = 2, ambayo inafanya mfuatano kuongezeka. Hivyo, x = 2 ndio jibu sahihi. Jibu la mwisho ni: x = 2 b) Ikiwa nambari ya mwisho ya A.P ni 105, kokotoa jumla ya nambari hizo. Step 1: Kutoka sehemu (a), tunajua x = 2. Tumia hii kupata a na d. a = 2x - 1 = 2(2) - 1 = 4 - 1 = 3 d = x = 2 Step 2: Nambari ya mwisho (L au T_n) ni 105. Tumia fomula T_n = a + (n-1)d kutafuta idadi ya nambari (n). 105 = 3 + (n-1)2 105 = 3 + 2n - 2 105 = 2n + 1 104 = 2n n = 52 Step 3: Kokotoa jumla ya nambari (S_n) kwa kutumia fomula S_n = (n)/(2)(a + L). S_52 = (52)/(2)(3 + 105) S_52 = 26(108) S_52 = 2808 Jibu la mwisho ni: S_52 = 2808 c) Kutokana na kwamba nambari ya 1^st, 4^th na 13^th za A.P zinaunda nambari tatu za kwanza mfululizo za mfuatano wa kijiometri (G.P), Bainisha: (i) Nambari ya 12^th ya G.P Step 1: Tafuta nambari ya 1^st, 4^th na 13^th za A.P kwa kutumia a=3 na d=2. T_1 = a = 3 T_4 = a + 3d = 3 + 3(2) = 3 + 6 = 9 T_13 = a + 12d = 3 + 12(2) = 3 + 24 = 27 Hivyo, G.P ni 3, 9, 27, Step 2: Tambua nambari ya kwanza ya G.P (A) na uwiano wa kawaida (R). A = 3 R = (9)/(3) = 3 Step 3: Kokotoa nambari ya 12^th ya G.P kwa kutumia fomula T_n = A R^n-1. T_12 = A R^12-1 = 3 × 3^11 T_12 = 3^1 × 3^11 = 3^12 T_12 = 531441 Jibu la mwisho ni: T_12 = 531441 (ii) Jumla ya nambari 12 za kwanza za G.P Step 1: Tumia fomula ya jumla ya nambari n za kwanza za G.P, S_n = (A(R^n - 1))/(R - 1) (kwa sababu R > 1). S_12 = 3(3^12 - 1)3 - 1 Step 2: Badilisha thamani ya 3^12 na kurahisisha. S_12 = (3(531441 - 1))/(2) S_12 = (3(531440))/(2) S_12 = 3 × 265720 S_12 = 797160 Jibu la mwisho ni: S_12 = 797160 That's 2 down. 3 left today — send the next one.