alimov-9-gdz (Алгебра - 9 класс - Алимов), страница 9
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an + 1 – an не зависит от n, то это – арифметическая прогрессия.4) an + 1 = 2(2 – n);an + 1 – an = 2(2 – n) – 2(3 – n) = 4 − 2/ n/ − 6 + 2/ n/ = −2 ,т.к. an + 1 – an не зависит от n, то это – арифметическая прогрессия.374.1) an = a1 + (n – 1)d, n = 15, поэтомуа15 = a1 + 14d = 2 + 14 ⋅ 3 = 2 + 42 = 44.Ответ: а15 = 44.2) an = a1 + (n – 1)d, n = 20, тогда a20 = a1 + 19d;а20 = 3 + 19 ⋅ 4 = 3 + 76 = 79.Ответ: a20 = 79.3) an = a1 + (n – 1)d, n = 18, тогда а18 = a1 + 17d;а18 = – 3 + 17 ⋅ ( – 2) = – 37.Ответ: a18 = – 37.4) an = a1 + (n – 1)d, n = 11, тогда а11 = a1 + 10d;а11 = – 2 + 10 ⋅ ( – 4) = – 42.Ответ: a11 = – 42.375.1) а1 = 1; а2 = 6;d = 6 – 1 = 5;an = a1 + (n – 1)d = 1 + (n – 1) ⋅ 5;an = 5n – 4;1222) а1 = 25; а2 = 21;d = 21 – 25 = – 4;an = a1 + (n – 1)d=25 + (n – 1) ⋅ ( – 4);an = – 4n + 29;3) а1 = – 4; а2 = – 6;d = – 6 – ( – 4) = – 2;an = a1 + (n – 1)d == – 4 + (n – 1) ⋅ ( – 2);an = – 2n – 2;4) а1 = 1; а2 = – 4;d = – 4 – 1 = – 5;an = a1 + (n – 1)d == 1 + (n – 1) ⋅ ( – 5);an = – 5n + 6.376.а1 = 44; d = 38 – 44 = – 6;an = a1 + (n – 1)d.
Тогда – 22 = 44 + (n – 1) ⋅ ( – 6);0 = 66 – 6n + 6; 6n = 50 + 22;6n = 72; n = 12.377.a1 = – 18; a2 = – 15; d = – 15 – ( – 18) = 3;an = a1 + (n – 1)d.Тогда 12 = – 18 + (n – 1) ⋅ 3;30 = 3n – 3; 3n = 33; n = 11.Ответ: 12 является членом аn.378.a1 = 1; a2 = – 5; d = – 5 – 1 = – 6;Тогда – 59 = 1 + (n – 1) ⋅ ( – 6);– 60 = – 6n + 6;6n = 66;an = a1 + (n – 1)d.Значит – 46 = 1 + (n – 1) ⋅ ( – 6);0 = 47 – 6n + 6;6n = 53;n = 11;n=8а11 = – 59является членом an.значит, – 46 не являетсячленом an.5– не натуральное,6379.1) an = а1 + (n – 1)d;а16 = а1 + 15 ⋅ d, т.к. a1 = 7, a16 = 67, то67 = 7 + 15d; 15d = 60.
Отсюда d = 4.2) a9 = а1 + 8d, т.к. a1 = – 4, a9 = 0, то10 = – 4 + 8d; 8d = 4. Тогда d = .2380.1) а9 = 12.Т.к. а9 = а1 + 8 ⋅ d, то12 = а1 + 8 ⋅ 1,5;а1 = 12 – 12;а1 = 0.2) а7 = – 4.Т.к. а7 = а1 + 6 ⋅ d, то– 4 = а1 + 6 ⋅ 1,5;а1 = – 4 – 9;а1 = – 13.123381.1) d = – 3; а11 = 20.Т.к.
а11 = а1 + 10d, то20 = а1 + 10 ⋅ ( – 3);а1 = 20 + 30 = 50;а1 = 50;382.1) если а3 = 13; а6 = 22.Т.к. а6 = а3 + 8d, то22 = 13 + 3 ⋅ d.Тогда 3d = 9и d = 3;а3 = а1 + 2d;13 = а1 + 2 ⋅ 3;а1 = 13 – 6.Получим а1 = 7.Значит аn = а1 + (n – 1)d;аn = 7 + (n – 1) ⋅ 3.Итак, аn = 3n + 4.2) а21 = – 10; a22 = – 5,5;d = а22 – а21 = – 5,5 – ( – 10) = 4,5.Т.к. a21 = а1 + 20 ⋅ d, то– 10 = а1 + 20 ⋅ 4,5;а1 = – 10 – 90 = – 100.2) если а2 = – 7; а7 = 18.Т.к.
а7 = а2 + 5d, то18 = – 7 + 5d.Значит 5d = 25и d = 5;а2 = а1 + d;а1 = – 7 – 5.Получим а1 = – 12.Значит аn = а1 + (n – 1)d;аn = – 12 + (n – 1) ⋅ 5.Итак, аn = 5n – 17.383.а1 = 15; a2 = 13. Тогда d = 13 – 15 = – 2.Т.к. an = a1 + (n – 1)d, то an = 15 + (n – 1) ( – 2);an = – 2n + 17. Т.к. an < 0, то – 2n + 17 < 0; – 2n < – 17.Тогда n > 8,5, т.е. при n ≥ 9 an<0.384.12Т.к. an = а1 + (n – 1)d, то аn = – 10 + (n – 1)⋅ ;an =1111n – 10 . Если an < 2, то n – 10 < 2;2222n – 21 < 4, n<25. Т.е. при n ≤ 25; an<2.385.1) если а8 = 126, а10 = 146;а9 =а 8 + а 10, тогда2126 + 146 272а9 == 136;=22d = a9 – a8,d = 136 – 126 = 10;1242) если а8 = – 64, а10 = – 50;а9 =а 8 + а 10, тогда2−64 − 50 −114а9 == – 57;=22d = a9 – a8;d = – 57 – ( – 64) = – 57 + 64 = 7;3) если а8 = – 7, а10 = 3;а9 =а 8 + а 102−7 + 3 −4=== – 2;22d = a9 – a8= – 2 – ( – 7) = 5;4) если а8 = 0,5, а10 = – 2,5;а9 =а 8 + а 102=0,5 − 2,5= – 1;2d = a9 – a8 = – 1 – 0,5 = – 1,5.386.Запишем данные условия: а5 = а1 + 4d.Тогда а5 = 4,9 + 4 ⋅ 9,8 = 44,1 (м).387.Т.к.
an = a1 + (n – 1)d,то 105 = 15 + (n – 1) ⋅ 10;90 = 10n – 10;10n = 100, отсюда n = 10.Ответ: 10 дней.388.an + ak = а1 + (n – 1)d + a1 + (k – 1)d = 2a1 + (n + k – 2)d,но an – ℓ + ak + ℓ = а1 + (n – ℓ – 1)d + a1 + (k + ℓ – 1)d = 2a1 + (n + k – 2)d,тогда an + ak = an – 1 + ak + 1 , доказано,поэтому а10 + а5 = а10 – 3 + а5 – 3 = а7 + а8 = 30.Ответ: а10 + а5 = 30.389.а n + k + a n − k а n + a n 2a n= аn (из предыдущего номера),==222тогда а20 =а 10 + a 302=120= 60.2390.1) а1 = 1, an = 20, n = 50;Sn =а1 + a n1 + 20⋅ n ; S 50 =⋅ 50 = (1 + 20) ⋅25 = 21 ⋅ 25 = 525;222) а1 = 1, an = 200, n = 100;S100 =1 + 200⋅ 100 = 201⋅50 = 10050;23) а1 = – 1, an = – 40, n = 20;S 20 =−1 − 40⋅ 20 = – 41⋅10 = – 410;24) а1 = 2, an = 100, n = 50;S 50 =2 + 100⋅ 50 = 102⋅25 = 2550.2125391.an = 98; a1 = 2; d = 1. Т.к.
an = a1 + (n – 1)d, то98 = 2 + (n – 1) ⋅ 1;96 = n – 1; n = 97;S 97 =2 + 98⋅ 97 = 50 ⋅ 97 = 4850.2392.а1 = 1; d = 2; an = 133.Т.к. an = a1 + (n – 1)d , то133 = 1 + (n – 1) ⋅ 2;132 = 2n – 2; n = 67;S67 =1 + 133⋅ 67 = 67 ⋅ 67 = 4489.2393.1) а1 = – 5; d = 0,5;2) а1 =Sn =2a1 + (n − 1)d⋅n ;2Sn =S12 =2 ⋅ (−5) + 11 ⋅ 0,5⋅ 12 =2S12= ( – 10 + 5,5) ⋅ 6 = – 27;1; d = – 3;22a1 + (n − 1)d⋅n ;212 ⋅ + 11 ⋅ (−3)2=⋅ 12 =2= (1 – 33) ⋅ 6 = – 192.394.1) а1 = 9; d = а2 – а1 = 13 – 9 = 4;S11 =2 ⋅ 9 + 10 ⋅ 42а1 + 10d(18 + 40) ⋅11⋅ 11 =⋅ 11 == 29 ⋅ 11 = 319;2222) а1 = – 16; d = а2 – а1 = – 13 – ( – 16) = 6 S12 ==2а 1 + 11d⋅ 12 =22 ⋅ (−16) + 11 ⋅ 6⋅ 12 = (−32 + 66) ⋅ 6 = 6 ⋅ 34 = 204.2395.1) а1 = 3; d = 3; an = 273.Т.к.
an = a1 + (n – 1)d, то 273 = 3 + (n – 1) ⋅ 3;270 = 3n – 3; 3n = 273.Тогда n = 91.S 91 =126a 1 + a 912⋅ 91 =3 + 273⋅ 91 = 138 ⋅ 91 = 12558.22) а1 = 90; d = 80 – 90 = – 10; an = – 60.Т.к. an = a1 + (n – 1)d, то– 60 = 90 – 10n + 10;10n = 100 + 60 = 160.Т.е. n = 16;S16 =a 1 + a 162⋅ 16 = (90 – 60) ⋅ 8 = 30 ⋅ 8 = 240.127396.a) а1 = 10; d = 1; an = 99.Т.к. an = a1 + (n – 1)d, то99 = 10 + n – 1.
Тогда n = 90;S 90 =a 1 + a 902⋅ 90 =10 + 99⋅ 90 = 109 ⋅ 45 = 4905.2б) а1 = 100; d = 1; an = 999.Т.к. an = a1 + (n – 1)d, то999 = 100 + n – 1. Т.е. n = 900;S 900 =a 1 + a 9002⋅ 900 =100 + 999⋅ 900 = 1099 ⋅ 450 = 494550.2397.1) а1 = 3⋅1 + 5 = 8; а50 = 3⋅50 + 5 = 155;S50 =а 1 + а 502⋅ 50 =8 + 155⋅ 50 = 163 ⋅ 25 = 4075;22)а1 = 7 + 2 = 9; а50 = 7 + 2⋅50 = 107;S50 =а 1 + а 502⋅ 50 =9 + 107⋅ 50 = 116 ⋅ 25 = 2900.2398.а1 = 7, а = аn + 1 – an = – 3, a9 = 7 – 3 ⋅ 8 = – 17.7 − 17Тогда S9 =⋅ 9 = – 45.2399.а1 = 3; d = 1.Т.к.
S n =2a 1 + ( n − 1)d6 + ( n − 1)⋅n ;⋅ n , то 75 =22150 = 6n + n2 – n;n2 + 5n – 150 = 0. Решим:n1 = 10, n2 = – 15 – не натуральное.Ответ: 10.400.1) а1 = 10; n = 14; S14 = 1050.Т.к. S14 =2а 1 + 13d⋅ 14 , то212562) а1 = 2 ; n = 10; S10 = 90 .Т.к. S10 =2а 1 + 9 ⋅ d⋅ 10 , то21271050 =20 + 13d⋅ 14 .2Отсюда 1050 = 7(20 + 13d);910 = 91d и d = 10.25= 4 + 9d ⋅ 5 .6 351Отсюда 90 – 23 = 45d;63145d = 67 и d = 1,5.290Тогда a14 = a1 + 13d;Тогда a10 = a1 + 9d;a14 = 10 + 130 = 140;a10 = 2401.1) а7 = 21; S7 = 205.2) а11 = 92; S11 = 22.Т.к. S 7 =205 =a1 + a 72⋅ 7 , тоa 1 + 21⋅7 ;2410 = 7а1 + 147;7а1 = 263.4Тогда а1 = 37 .7Т.к.
а7 = а1 + 6d, то421 = 37 + 6d;746d = – 16 ;758d=–.21Итак d = – 2511+ 13 = 15 .326a 1 + a 11⋅ 11 , то2a + 92⋅ 11 ;22 = 12Т.к. S11 =44 = (а1 + 92) ⋅ 11;а1 + 92 = 4.Тогда а1 = – 88.Т.к. a11 = a1 + 10d, то92 = – 88 + 100d;180 = 10d.Итак d = 18.16.21402.an = 12; d = 1; a1 = 1. Т.к. an = a1 + (n – 1)d, то 12 = 1 + n – 1.Тогда n = 12. S12 =a 1 + a 121 + 12⋅ 12 ; S12 =⋅ 12 = 13 ⋅ 6 = 78 (брёвен).22403.а3 + а9 = а1 + 2 + a11 – 2 = a1 + а11 = 8 (из предыдущих задач).a + a 11S11 = 1⋅11 .2128Тогда S11 =8⋅11 = 44.2404.2a 1 + 4d2a + 9d⋅ 5 , т.к.
S10 = 1⋅ 10 ,222/ (a 1 + 2d )то 65 =⋅ 5 , то 230 = (2а1 + 9d) ⋅ 5.2/Тогда 13 = a1 + 2d. Тогда 2а1 + 9d = 46 : 2 ,Т.к. S 5 =а1 + 2d = 13 5d = 20получим ; ;2a1 + 9d = 46 a1 + 2d = 13d = 4.a1 = 5405.2a 1 + 11d⋅ 12 ; S12 = 6 ⋅ (2a1 + 11d). Тогда22a + 7 d2a + 3dS8 − S 4 = 1⋅8 − 1⋅ 4 = 4 ⋅ ( 2a 1 + 7d ) − 2 ⋅ ( 2a 1 + 3d ) =22S12 == 8a1 + 28d – 4a1 – 6d = 4a1 + 22d;3(S8 – S4) = 3⋅(4a1 + 22d) = 3⋅2(2a1 + 11d) = 6⋅(2a1 + 11d),получили: S12 = 3(S8 – S4).407.1) b1 = 12, q = 2;b2 = b1⋅q = 12⋅2 = 24;b3 = 24⋅2 = 48;b4 = 48⋅2 = 96;b5 = 192;2) b1 = – 3, q = – 4;b2 = b1⋅q = – 3⋅( – 4) = 12;b3 = 12⋅( – 4) = – 48;b4 = – 48⋅( – 4) = 192;b5 = 192⋅( – 4) = – 768.408.1) bn = 3 ⋅ 2n.Пусть bn + 1 = 3 ⋅ 2n + 1.Тогдат.к.b n +1bn=3 ⋅ 2 n +13⋅ 2n=3/ ⋅ 2 n ⋅ 23/ ⋅ 2/ n=2,bn +1не зависит от n то bn – геометрическая прогрессия.bn2) bn = 5n + 3.Пусть bn + 1 = 5n + 4.Тогдаb n +1 5 n + 4 5/ n ⋅ 5 4b=== 5 т.к.
n +1 не зависит от n, тоn3n3+/bnbn55/ ⋅ 5/129bn – геометрическая прогрессия.13) bn = n −23.1Пусть bn + 1 = n −13bn +1b=n1 31 3n −1n −2=;n−1n−21 1 ⋅ 3 31 1 ⋅ 3 3=11 3−1=b1т.к. n +1 не зависит от n,3bnто bn – геометрическая прогрессия.4) bn =15 n −1.Пусть bn + 1 =1bn +1bт.к.nn= 515 n −115n;1n1= 5= ,15⋅5n5bn +1не зависит от n, то bn – геометрическая прогрессия.bn409.1) Т.к. bn = b1 ⋅ qn – 1 , тоb4 = b1 ⋅ q3 , b4 = 3 ⋅ 103 = 3000.2) Т.к.
bn = b1 ⋅ qn – 1 , то11 6 4.b7 = b1 ⋅ q6 = 4 ⋅ = =62163) Т.к. bn = b1 ⋅ qn – 1 , тоb5 = b1 ⋅ q4 = 1 ⋅ ( – 2)4 = 16.4) Т.к. bn – b1 ⋅ q5, то1 5−3b6 = b1 ⋅ q5 = – 3 ⋅ − = 3− 243=410.1) b1 = 4; q = 3; Т.к. bn = b1 ⋅ qn – 1,то bn = 4 ⋅ 3 n – 1;1301.812) b1 = 3; q =1 n −11; Т.к. bn = b1 ⋅ qn – 1 , то bn = 4 ⋅ ;333) b1 = 4; q = –1 n −11; Т.к.
bn = b1 ⋅ qn – 1 , то bn = 4 ⋅ − ;4 44) b1 = 3; q = –4 n −14; Т.к. bn = b1 ⋅ qn – 1 , то bn = 3 ⋅ − .3 3411.1) b1 = 6; b2 = 12, … , bn = 192;q=b2 12= 2.=6b1Т.к. bn = b1 ⋅ qn – 1 , то192 = 6 ⋅ 2n – 1 , но 32 = 25, значит,32 = 2n – 1 , 25 = 2n – 1;5 = n – 1;n = 6;2) b1 = 4; b2 = 12, … , bn = 324;q=12= 3.4Т.к.
