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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a) i) Tunapewa kwamba neno la kwanza la AP (Arithmetic Progression) ni a = 2. Jumla ya maneno 8 ya kwanza ya AP ni S_8 = 156. Tunahitaji kupata tofauti ya kawaida, d. Step 1: Tumia fomula ya jumla ya maneno n ya AP, ambayo ni S_n = (n)/(2)(2a + (n-1)d). Badilisha maadili uliyopewa: S_8 = 156, n = 8, na a = 2. 156 = (8)/(2)(2(2) + (8-1)d) Step 2: Rahisisha na utatue kwa d. 156 = 4(4 + 7d) 156 = 16 + 28d 156 - 16 = 28d 140 = 28d d = (140)/(28) d = 5 Tofauti ya kawaida ni 5. a) ii) Tunapewa kwamba jumla ya maneno n ya AP ni S_n = 116. Kutoka sehemu (i), tunajua a = 2 na d = 5. Tunahitaji kupata n. Step 1: Tumia tena fomula ya jumla ya maneno n ya AP: S_n = (n)/(2)(2a + (n-1)d). Badilisha maadili yanayojulikana: S_n = 116, a = 2, na d = 5. 116 = (n)/(2)(2(2) + (n-1)5) Step 2: Rahisisha mlinganyo. 116 = (n)/(2)(4 + 5n - 5) 116 = (n)/(2)(5n - 1) Panua pande zote mbili kwa 2: 232 = n(5n - 1) 232 = 5n^2 - n Step 3: Panga upya kuwa mlinganyo wa quadratic. 5n^2 - n - 232 = 0 Step 4: Tatua mlinganyo wa quadratic kwa n kwa kutumia fomula ya quadratic n = -b ± sqrt(b^2 - 4ac)2a. Hapa, a = 5, b = -1, c = -232. n = -(-1) ± sqrt((-1)^2 - 4(5)(-232))2(5) n = 1 ± sqrt(1 + 4640)10 n = 1 ± sqrt(4641)10 Kwa kuwa n lazima iwe nambari kamili chanya kwa idadi ya maneno katika AP, na sqrt(4641) si nambari kamili, hakuna thamani kamili ya n inayokidhi masharti haya. n ≈ (1 ± 68.12)/(10) n ≈ (1 + 68.12)/(10) ≈ 6.912 au n ≈ (1 - 68.12)/(10) ≈ -6.712 Kwa kuwa n lazima iwe nambari kamili chanya, hakuna suluhisho kamili la n kwa jumla ya 116. Hakuna thamani kamili chanya ya n kwa jumla ya 116. Tuma swali linalofuata 📸