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, hebu tutatue maswali hayo. Swali la 32: Kigezo cha 5 cha A.P. ni 19 na kigezo cha 12 ni 40. Tafuta kigezo cha 20. Fomula ya kigezo cha n cha A.P. ni a_n = a + (n-1)d. Step 1: Andika milinganyo kutokana na taarifa uliyopewa. Kigezo cha 5: a_5 = a + (5-1)d = a + 4d = 19 (Mlinganyo 1) Kigezo cha 12: a_12 = a + (12-1)d = a + 11d = 40 (Mlinganyo 2) Step 2: Tatua kwa tofauti ya kawaida (d) na kigezo cha kwanza (a). Toa Mlinganyo 1 kutoka Mlinganyo 2: (a + 11d) - (a + 4d) = 40 - 19 7d = 21 d = (21)/(7) d = 3 Badilisha d=3 katika Mlinganyo 1: a + 4(3) = 19 a + 12 = 19 a = 19 - 12 a = 7 Step 3: Tafuta kigezo cha 20 (a_20). a_20 = a + (20-1)d a_20 = 7 + (19)3 a_20 = 7 + 57 a_20 = 64 Kigezo cha 20 ni 64. --- Swali la 33: Kigezo cha 3 cha G.P. ni 12 na kigezo cha 6 ni 96. Tafuta kigezo cha 8. Fomula ya kigezo cha n cha G.P. ni a_n = ar^n-1. Step 1: Andika milinganyo kutokana na taarifa uliyopewa. Kigezo cha 3: a_3 = ar^3-1 = ar^2 = 12 (Mlinganyo 1) Kigezo cha 6: a_6 = ar^6-1 = ar^5 = 96 (Mlinganyo 2) Step 2: Tatua kwa uwiano wa kawaida (r) na kigezo cha kwanza (a). Gawanya Mlinganyo 2 kwa Mlinganyo 1: (ar^5)/(ar^2) = (96)/(12) r^3 = 8 r = [3]8 r = 2 Badilisha r=2 katika Mlinganyo 1: a(2)^2 = 12 4a = 12 a = (12)/(4) a = 3 Step 3: Tafuta kigezo cha 8 (a_8). a_8 = ar^8-1 a_8 = 3 × (2)^7 a_8 = 3 × 128 a_8 = 384 Kigezo cha 8 ni 384. --- Swali la 34: Jumla ya vigezo vya kwanza vya A.P. ni 200. Ikiwa kigezo cha kwanza ni 5 na kigezo cha mwisho ni 35, tafuta idadi ya vigezo na tofauti ya kawaida. Fomula ya jumla ya vigezo n vya kwanza vya A.P. ni S_n = (n)/(2)(a + l). Fomula ya kigezo cha n cha A.P. ni a_n = a + (n-1)d. Step 1: Tafuta idadi ya vigezo (n). Tunapewa S_n = 200, a = 5, na l = 35. 200 = (n)/(2)(5 + 35) 200 = (n)/(2)(40) 200 = 20n n = (200)/(20) n = 10 Step 2: Tafuta tofauti ya kawaida (d). Tunajua a_n = l = 35, a = 5, na n = 10. 35 = 5 + (10-1)d 35 = 5 + 9d 35 - 5 = 9d 30 = 9d d = (30)/(9) d = (10)/(3) Idadi ya vigezo ni 10 na tofauti ya kawaida ni (10)/(3). --- Swali la 35: Kigezo cha 4 cha G.P. ni 24 na kigezo cha 7 ni 192. Tafuta jumla ya vigezo 8 vya kwanza. Fomula ya kigezo cha n cha G.P. ni a_n = ar^n-1. Fomula ya jumla ya vigezo n vya kwanza vya G.P. ni S_n = (a(r^n - 1))/(r - 1) (kwa r ≠ 1). Step 1: Andika milinganyo kutokana na taarifa uliyopewa. Kigezo cha 4: a_4 = ar^4-1 = ar^3 = 24 (Mlinganyo 1) Kigezo cha 7: a_7 = ar^7-1 = ar^6 = 192 (Mlinganyo 2) Step 2: Tatua kwa uwiano wa kawaida (r) na kigezo cha kwanza (a). Gawanya Mlinganyo 2 kwa Mlinganyo 1: (ar^6)/(ar^3) = (192)/(24) r^3 = 8 r = [3]8 r = 2 Badilisha r=2 katika Mlinganyo 1: a(2)^3 = 24 8a = 24 a = (24)/(8) a = 3 Step 3: Tafuta jumla ya vigezo 8 vya kwanza (S_8). Tumia fomula S_n = (a(r^n - 1))/(r - 1) na n=8, a=3, na r=2. S_8 = (3(2^8 - 1))/(2 - 1) S_8 = (3(256 - 1))/(1) S_8 = 3(255) S_8 = 765 Jumla ya vigezo 8 vya kwanza ni 765. 3 done, 2 left today. You're making progress.