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2 = 10 va a 5 = 22. Shu progressiyaning dastlabki sakkizta hadining yig’indisini toping. A) 162 B) 170 C) 115 D) 160 33. (98-11-26) Arifmetik progressiyaning uchinchi hadi va beshinchi hadi, mos ravishda 11 va 19 ga teng bo’lsa, S 10 ni toping. A) 210 B) 190 C) 230 D) 220 34. (98-11-75) (a n ) arifmetik progressiyada a 1 = 3, a 60 = 57 bo’lsa, progressiyaning dastlabki 60 ta hadi yig’indisi qanchaga teng bo’ladi? A) 1500 B) 3423 2 C) 1600 D) 1800 35. (00-4-22) Arifmetik progressiyaning beshinchi hadi 6 ga teng. Uning dastlabki to’qqizta hadi yig’indisi- ni toping. A) 36 B) 48 C) 54 D) 45 Yechish. 3-xossaga ko’ra a 1 + a 9 2 = a 5 = 6. 5-xossaga ko’ra S 9 = a 1 + a 9 2 · 9 = 6 · 9 = 54. Javob: 54 (C). 36. (99-4-28) Arifmetik progressiyaning o’n uchinchi hadi 5 ga teng. Uning dastlabki 25 ta hadlarining yig’indisini toping. A) 125 B) 100 C) 75 D) 225 37. (03-6-56) Arifmetik progressiyada a 10 = 56 bo’lsa, uning dastlabki 19 ta hadlari yig’indisini toping. A) 1024 B) 1032 C) 1056 D) 1064 38. (00-7-25) Arifmetik progressiyaning birinchi va to’qqizinchi hadlari yig’indisi 64 ga teng. Shu progressiyaning dastlabki 9 ta hadlari yig’indisi va beshinchi hadi ayirmasini toping. A) 256 B) 260 C) 270 D) 208 39. (03-4-19) 15 ta haddan iborat arifmetik progres- siyaning sakkizinchi hadi 18 ga teng. Shu pro- gressiyaning hadlari yig’indisini toping. A) 280 B) 270 C) 250 D) 300 40. (08-106-27, 08-123-27) Arifmetik progressiyning dastlabki 16 ta hadi yig’indisi 528 ga va a 16 = 63. Shu progressiyaning ayirmasini toping. A) 7 B) 4 C) 5 D) 6 41. (08-122-27) Arifmetik progressiyning oltinchi hadi 17 ga, dastlabki 16 ta hadi yig’indisi 392 ga teng. Shu progressiyaning 9-hadini toping. A) 24 B) 26 C) 13 D) 18 42. (97-2-36) (a n ) arifmetik progressiyaning dastlabki n ta hadi yig’indisi 120 ga teng. Agar a 3 +a n−2 = 30 bo’lsa, yig’indida nechta had qatnashgan? A) 6 B) 10 C) 8 D) 12 Yechish: 4-xossaga ko’ra a 1 + a n = a 3 + a n−2 = 30. Shartga ko’ra S n = a 1 + a n 2 · n = 30 2 · n = 15n = 120. Bu yerdan n = 8 ni olamiz. Javob: 8 (C). 43. (08-126-27) 25 va 4 sonlari orasiga shu sonlar bilan arifmetrik progressiya tashkil etadigan bir nechta son joylashtirilgan. Agar joylashtirilgan sonlarning yig’indisi 87 ga teng bo’lsa, nechta had joylashtirilgan. A) 6 B) 11 C) 12 D) 9 44. (98-8-27) Arifmetik progressiyaning uchinchi hadi 8 ga, to’rtinchi hadi 5 ga va dastlabki bir nechta hadlari yig’indisi 28 ga teng. Yig’indida nechta had qatnashgan? A) 10 B) 7 C) 11 D) 8 45. (99-6-54) Arifmetik progressiyaning dastlabki n ta hadining yig’indisi 91 ga teng. Agar a 3 = 9 va a 7 − a 2 = 20 ekanligi ma’lum bo’lsa, n ni toping. A) 7 B) 5 C) 3 D) 9 46. (03-12-63) 10; 15; 20; ... arifmetik progressiya- ning dastlabki nechta hadining yig’indisi 2475 ga teng bo’ladi? A) 40 B) 25 C) 30 D) 35 93 47. (97-7-27) 100 dan katta bo’lmagan 4 ga karrali barcha natural sonlarning yig’indisini toping. A) 1250 B) 1300 C) 1120 D) 1000 Yechish: Ma’limki, 4 ga karrali sonlar 4n, n = 1, 2, 3, . . . shaklda bo’lib, ular arifmetik progres- siya tashkil qiladi. Bu progressiya uchun a n = 4n dir. Bundan tashqari a 25 = 100. Demak, S 25 = 4 + 100 2 · 25 = 52 · 25 = 1300. Javob: 1300 (B). 48. (96-7-27) 100 dan katta bo’lmagan 3 ga karrali barcha natural sonlarning yig’indisini toping. A) 1683 B) 2010 C) 1500 D) 1080 49. (97-6-17) a n = 4n − 2 formula bilan berilgan ketma-ketlikning dastlabki 50 ta hadi yig’indisini toping. A) 4500 B) 5050 C) 3480 D) 5000 50. (97-11-17) Hadlari b n = 3n − 1 formula bilan berilgan ketma-ketlikning dastlabki 60 ta hadi- ning yig’indisini toping. A) 4860 B) 4980 C) 5140 D) 5430 51. (02-11-37) 9 ga bo’lganda, qoldig’i 4 ga teng bo’- ladigan barcha ikki xonali sonlarning yig’indisini toping. A) 527 B) 535 C) 536 D) 542 52. (03-9-26) 7 ga bo’lganda, qoldig’i 2 ga teng bo’la- digan barcha ikki xonali sonlarning yig’indisini toping. A) 640 B) 647 C) 650 D) 654 53. (03-1-70) Dastlabki mingta natural sonlarning o’rta arifmetigini toping. A) 500 B) 501 C) 501,5 D) 500,5 54. (00-2-11) 25 ta ketma-ket natural sonning yig’indisi 1000 ga teng. Bu sonlarning kichigi nechaga teng bo’ladi? A) 30 B) 28 C) 26 D) 27 55. (98-1-27) Arifmetik progressiyaning dastlabki 16 ta hadlari yig’indisi 840 ga va a 16 = 105 bo’lsa shu progressiyaning ayirmasini toping. A) 9 B) 7 C) 15 D) 5 56. (98-2-18) Arifmetik progressiyada S 20 − S 19 = −30 va d = −4 bo’lsa, a 25 ning qiymatini toping. A) −40 B) −50 C) −48 D) −56 Yechish: 6-qoidaga ko’ra, a 20 = S 20 − S 19 = −30. 2-qoidaga ko’ra, a 25 − a 20 = 5d. Bu ifodaga a 20 = −30 va d = −4 larni qo’yib, a 25 = −50 ni olamiz. Javob: −50 (B). 57. (00-3-44) Arifmetik progressiyaning dastlabki 13 ta hadi yig’indisi 104 ga teng. Yettinchi hadining kvadratini toping. A) 25 B) 36 C) 49 D) 64 58. (00-5-1) 1 dan 75 gacha bo’lgan toq sonlar yig’in- disi qanday raqam bilan tugaydi? A) 0 B) 2 C) 3 D) 4 59. (00-9-13) y; 3y + 5; 5y + 10; . . . arifmetik progres- siyaning dastlabki 8 ta hadi yig’indisi 396 ga teng. y ning qiymatini toping. A) 2 B) 3 C) 4 D) 5 60. (01-1-26) Agar soat 1 da bir marta, 2 da ikki marta va hokazo 12 da o’n ikki marta zang ursa, bir sutkada necha marta zang uradi? A) 72 B) 78 C) 108 D) 156 61. (01-5-28) Arifmetik progressiya uchun a 17 = 2 ga teng bo’lsa, S 21 − S 12 ni toping A) 18 B) 15 C) 16 D) 17 62. (02-1-55) Arifmetik progressiya birinchi o’nta ha- dining yig’indisi 140 ga teng bo’lsa, a 2 + a 9 ni aniqlang. A) 24 B) 26 C) 30 D) 28 63. (02-4-22) Agar arifmetik progressiya hadlari uchun a 1 + a 3 + · · · + a 19 = a 2 + a 4 + · · · + a 20 + 10 teng- lik o’rinli bo’lsa, arifmetik progressiyaning ayir- masini toping. A) 1 B) −1 C) 0 D) −2 Yechish: Ma’lumki, barcha n ≥ 2 lar uchun a n − a n−1 = d tenglik o’rinli. Masala shartida berilgan tenglikni −10 = (a 2 − a 1 ) + (a 4 − a 3 ) + · · · + (a 20 − a 19 ) shaklda yozib olamiz. Har bir qavs ichidagi ayirma d ga teng va qavslar soni 10 ta. Demak, −10 = 10d tenglik o’rinli. Bu yerdan d = −1 ni olamiz. Javob: −1 (B). 64. (02-4-18) Arifmetik progressiya hadlari uchun a 1 + a 3 + · · · + a 21 = a 2 + a 4 + · · · + a 20 + 15 tenglik o’rinli bo’lsa, a 11 ni toping. A) 11 B) 13 C) 15 D) 17 65. (03-5-27) Arifmetik progressiyaning oltinchi hadi 10 ga, dastlabki 16 ta hadining yig’indisi 200 ga teng. Bu progressiyaning 12-hadini toping. A) 16 B) 14 C) 18 D) 20 66. (03-8-50) Agar arifmetik progressiyada a 1 + a 2 + · · · + a 16 + a 17 = 136 bo’lsa, a 6 + a 12 ni hisoblang. A) 16 B) 10 C) 12 D) 14 8.2 Geometrik progressiya Birinchi hadi noldan farqli bo’lib, ikkinchi hadidan bosh- lab bir hadi o’zidan oldingi hadni shu ketma-ketlik uchun o’zgarmas va noldan farqli bo’lgan biror q songa ko’payti- rishdan hosil bo’lgan sonlar ketma-ketligi geometrik progressiya deyiladi. Masalan, 1) 1, 3, 9, . . . ; 2) 20, 10, 5, . . . ketma-ketliklar geometrik progressiya tashkil qi- ladi. Birinchi misolda q = 3, ikkinchisida q = 0, 5. 94 Geometrik progressiyani tashkil qiluvchi sonlar un- ing hadlari deyiladi va umumiy ko’rinishda b 1 , b 2 , b 3 , . . . , b n−1 , b n , . . . (1) yoziladi. Geometrik progressiyaning keyingi hadini hosil qilish uchun oldingi hadiga ko’paytiriladigan q son ge- ometrik progressiya maxraji deyiladi. Agar b 1 > 0 va q > 1 bo’lsa, progressiya o’suvchi deyiladi. Agar |q| < 1 bo’lsa, progressiya kamayuvchi, q < 0 bo’lsa, pro- gressiya ishorasi o’zgaruvchi deyiladi, q = 1 hol odatda qaralmaydi. Geometrik progressiyaning n − hadi b n quyidagi formula yordamida topiladi: b n = b 1 q (n−1) . Geometrik progressiya hadlarining xossalari. 1-xossa. Agar geometrik progressiyaning barcha hadlari musbat bo’lsa, u holda uning ikkinchi hadidan boshlab istalgan hadi o’ziga qo’shni bo’lgan ikki had- ning o’rta geometrik qiymatiga teng, ya’ni b n = p b n−1 b n+1 . 2-xossa. Chekli geometrik progressiyada boshidan va oxiridan teng uzoqlikda to’rgan hadlar ko’paytmasi chetki hadlar ko’paytmasiga teng, ya’ni b 1 b n = b 2 b n−1 = b 3 b n−2 = · · · = b k b n−k+1 . 3-xossa. Geometrik progressiyaning dastlabki n ta hadi yig’indisi S n = b 1 + b 2 + b 3 + · · · + b n−1 + b n bo’lsin. Geometrik progressiyaning dastlabki n ta hadi yig’indisi S n uchun quyidagi formulalar o’rinli: S n = b 1 − b n q 1 − q , S n = b n q − b 1 q − 1 S n = b 1 (q n − 1) q − 1 . 4-xossa. Cheksiz kamayuvchi geometrik progres- siya barcha hadlarining yig’indisi S uchun quyidagi for- mula o’rinli. S = b 1 + b 2 + · · · + b n + · · · = b 1 1 − q . 1. b n = b 1 · q n−1 , b n = q b n−1 . 2. b n : b m = q n−m , n > m. 3. b 2 n = b n−1 b n+1 , n ≥ 2. 4. b k b m = b p b q , k + m = p + q. 5. S n = b 1 (1 − q n ) 1 − q , S n = b n q − b 1 q − 1 , (q 6= 1). 6. S n − S n−1 = b n . 7. S = b 1 1 − q . 1. Agar geometrik progressiyada b 1 = 2, q = 3 bo’lsa, b 5 ni toping. A) 162 B) 158 C) 120 D) 254 Yechish: 1-xossaga ko’ra b 5 = b 1 q 4 . Endi b 1 va q larning qiymatlari qo’yib b 5 = 2·3 4 = 2·81 = 162 ni olamiz. Javob: 162 (A). 2. Agar geometrik progressiyada b 2 = 1, q = 2 bo’lsa, b 5 − b 4 ni hisoblang. A) 2 B) 5 C) 4 D) 8 3. Agar geometrik progressiyada b 3 = 10, q = 3 bo’lsa, b 5 ni toping. A) 2 B) 5 C) 4 D) 8 4. Geometrik progressiyada b 5 = 64, b 7 = 16 bo’lsa, geometrik progressiyaning maxrajini toping. A) 0, 2 B) 0, 5 C) 4 D) 2 5. Geometrik progressiyada b 3 b 5 = 64 bo’lsa, b 4 ni toping. A) 8 B) −8 C) ±8 D) 4 6. Agar ishorasi almashinuvchi geometrik progres- siyada b 3 = 4, b 7 = 9 bo’lsa, b 5 ni toping. A) 2 B) −6 C) 6 D) 5 7. (98-4-21) Nolga teng bo’lmagan x, y, z sonlar ko’r- satilgan tartibda ishorasi o’zgaruvchi geometrik progressiyani, x + y; y +z; z +x sonlar esa arifme- tik progressiyani tashkil etadi. Geometrik progres- siyaning maxrajini toping. A) −2 B) −1 C) −3 D) −4 Yechish: Geometrik progressiyaning maxraji q bo’lsin. U holda y = qx, z = q 2 x. Endi x + y, y+z, z+x sonlar arifmetik progressiya tashkil et- ganligi uchun 2(y+z) = x+y+z+x, ya’ni y+z = 2x bo’ladi. y, z larning o’rniga ularning ifodalar- ini qo’yib qx+q 2 x−2x = 0 tenglamani, bu yerdan esa x(q 2 + q − 2) = 0 ekanini topamiz. Masala shartiga ko’ra x 6= 0, shuning uchun q 2 +q−2 = 0 bo’ladi. Uning ildizlari q 1 = 1, q 2 = −2. Ge- ometrik progressiyaning ishorasi o’zgaruvchiligidan q = −2 ekani kelib chiqadi. Javob: −2 (A). 8. (08-120-27) O’suvchi geometrik progressiyaning birinchi hadi 2 ga, yettinchi va to’rtinchi hadlari- ning ayirmasi 1404 ga teng. Shu progressiyaning maxrajini toping. A) 2 B) 3 C) 2 √ 2 D) 4 9. (97-9-87) Geometrik progressiyaning dastlabki 6 ta hadi 2, b 2 , b 3 , b 4 , b 5 va 486 bo’lsa, b 2 +b 3 +b 4 +b 5 ni hisoblang. A) 200 B) 260 C) 230 D) 240 10. (98-7-38) Quyidagi ketma-ketliklardan qaysilari geometrik progressiyani tashkil etadi? 1) a n = 2x n 2) c n = ax n + 1 3) b n = ( 3 5 ) n A) 1;3 B) 2;3 C) hech biri D) 1;2;3 11. (00-2-21) Nechanchi hadidan boshlab −8; 4; −2; . . . geometrik progressiya hadlarining absolyut qiy- mati 0,001 dan kichik bo’ladi? A) 16 B) 12 C) 15 D) 14 12. (00-10-23) 64; 32; 16; . . . geometrik progressiya- ning to’qqizinchi hadi oltinchi hadidan nechtaga kam? A) 1,025 B) 1,5 C) 1,25 D) 1,75 95 Yechish: Berilgan geometrik progressiyada b 1 = 64, b 2 = 32. Bu yerdan q = 1/2 ni olamiz. U holda 1-xossadan b 6 = b 1 q 5 = 64 32 = 2, b 9 = b 1 q 8 = 64 256 = 1 4 . Ularning farqi b 6 −b 9 = 2−0, 25 = 1, 75. Javob: 1, 75 (D). 13. (02-4-23) Agar geometrik progressiya hadlari uchun b 1 b 3 · · · b 13 = b 2 b 4 · · · b 14 /128 tenglik o’rinli bo’lsa, progressiyaning maxrajini toping. A) 1 B) 2 C) 3 D) 4 14. (03-2-5) (b n ) geometrik progressiyada b 4 −b 2 = 24 va b 2 + b 3 = 6 bo’lsa, b 1 ning qiymatini toping. A) 0,4 B) 1 C) 1 1 5 D) 1 5 15. (03-6-57) Ikkinchi hadi 6 ga, birinchi uchta ha- dining yig’indisi 26 ga teng o’suvchi geometrik progressiyaning uchinchi va birinchi hadlari ayir- masini toping. A) 15 B) 16 C) 14 D) 13 16. (03-12-66) O’suvchi geometrik progressiyaning bi- rinchi hadi 3 ga, yettinchi va to’rtinchi hadlari- ning ayirmasi 168 ga teng. Shu progressiyaning maxrajini toping. A) 3 B) 3 2 C) √ 7 D) 2 3-4-xossalariga oid misollar 17. (00-9-40) x ning qanday qiymatlarida 0,(16); x va 0,(25) sonlar ishoralari almashinuvchi geometrik progressiyaning ketma-ket keluvchi hadlari bo’ladi? A) 0,(20) B) ± 0, (20) C) −0, (20) D) −0, (21) Yechish: Geometrik progressiyaning ishora alma- shinuvchiligidan x < 0 ekani, 3-xossasidan esa x 2 = 0, (16) · 0, (25) = 16 99 · 25 99 ekani kelib chiqadi. Shuning uchun x = − 4 · 5 99 = − 20 99 = −0, (20). Javob: −0, (20) (C). 18. (96-6-37) Geometrik progressiyada b 2 ·b 3 ·b 4 = 216 bo’lsa, uning uchinchi hadini toping. A) 12 B) 8 C) 4 D) 6 19. (97-8-36) b 3 ·b 4 ·b 5 = 64 ga teng bo’lgan geometrik progressiyaning to’rtinchi hadini toping. A) 10 B) 12 C) 4 D) 8 20. (00-3-46) Geometrik progressiyada uchinchi va yettinchi hadlarining ko’paytmasi 144 ga teng. Uning beshinchi hadini toping. A) 6 B) ± 12 C) −8 D) −12 21. (00-3-47) 1 3 va 1 48 sonlar orasiga shunday uchta musbat sonni joylashtiringki, natijada geometrik progressiya hosil bo’lsin. O’sha qo’yilgan uchta sonning yig’indisini toping. A) 0,5 B) 7 12 C) 0,375 D) 7 24 22. (00-7-26) Barcha hadlari musbat bo’lgan geometrik progressiyaning birinchi hadi 2 ga, beshinchi hadi 18 ga teng. Shu progressiyaning beshinchi va uchinchi hadlari ayirmasini toping. A) 10 B) 12 C) 8 D) 11 23. (02-8-9) 3 va 19683 sonlari o’rtasiga 7 ta shun- day musbat sonlar joylashtirilganki, hosil bo’lgan to’qqizta son geometrik progressiya tashkil etsin. 5-o’rinda turgan son nechaga teng? A) 243 B) 343 C) 286 D) 729 Dastlabki n ta hadi yig’indisiga oid misol- lar 24. Geometrik progrogressiyada b 1 = 3, q = 2. Shu progrogressiyaning daslabki oltita hadining yig’in- disini toping. A) 63 B) 189 C) 126 D) 184 Yechish: 5-xossaga ko’ra S 6 = b 1 (1 − q 6 ) 1 − q = 3(1 − 2 6 ) 1 − 2 = 3 · 63 = 189. Javob: 189 (B). 25. Geometrik progrogressiyada b 1 = 3, q = 2. Shu progrogressiyaning daslabki oltita hadining yig’in- disini toping. A) 63 B) 189 C) 126 D) 184 26. 81, 27, 9, . . . geometrik progressiyaning nechta ha- dining yig’indisi 121 ga teng bo’ladi. A) 3 B) 4 C) 5 D) 6 27. Geometrik progressiyada Download 1.09 Mb. Do'stlaringiz bilan baham: |
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