Yig’indisi 35 ga teng bo’lgan uchta son o’suvchi geometrik progressiyaning dastlabki uchta hadlaridir

1-misol
Yig’indisi 35 ga teng bo’lgan uchta son o’suvchi geometrik progressiyaning dastlabki uchta hadlaridir . agar shu sonlardan mos ravishda 2, 2 va 7sonlarini ayirilsa, hosil bo’lgan sonlar arifmetik progressiyaning ketma - ket hadlari bo’ladi . Arifmetik progressiyaning dastlabki 10 ta hadining yig’indisini toping.
A)245 B)275 C)255 D) 265 E)235
2-misol 4;9;14;…..arifmetik progressiyaning sakkizinchi hadi to’rtinchi hadidan nechtaga ortiq
A)16 B)18 C)20 D)22 E)24
3-misol
Ikkinchi hadi 6 ga teng birinchi uchta hadining yig’indisi 26 ga teng o’suvchi geometrik progressiyaning uchinchi va birinchi hadlari ayirmasini toping
A)15 B)16 C)14 D)13 E)12
4-misol Arifmetik progressiyaning birinchi hadi 6 ga, oxirgi hadi esa 39 ga teng. Agar progressiyaning ayirmasi butun sonlardan iborat bo’lib u 2dan katta va 6 dan kichik bo’lsa oxirgi hadidan oldingi hadlar sonini toping
A)12 B)11 C)10 D)9 E)13
5-misol
Agar geometrik progressiyada + + =17 bo’lsa, ni toping
A)4 B)3 C)2 D)1 E)6
6-misol
Kamayuvchi geometrik progressiya tashkil etuvchi uchta sondan uchinchisi 18 ga teng . Bu son o’rniga 10soni olinsa , uchta son arifmetik progressiya tashkil etadi. Birinchi sonni toping.
A)50 B)60 C)40 D)27 E)36
7-misol
2, , va sonlari o’suvchi geometrik progressiyaning dastlabki uchta hadidan iborat.Agar bu progressiyaning ikkinchi hadiga 4 qo’shilsa , hosil bo’lgan sonlar arifmetik progressiyaning dastlabki uchta hadini tashkil etadi. Berilgan progressiyaning maxrajini toping.
A)3 B)2 C)2,5 D)3,5 E)1,5
8-misol To’rtta banderolni jo’natish haqi uchun jami 120 so’mlik 4ta Har xil pochta markasi kerakk bo’ladi.Agar markalaring baholari arifmetik progressiyani tashkil etib, eng qimmat marka eng arzonidan 3marta qimmat tursa eng qimmatining bahosi necha so’m?
A)50 B)45 C)62 D)54 E)48
9-misol
Geometrik progressiyaningmaxraji 3 ga , dastlabki 4 ta hadlari yig’indisi 80 ga teng . Uning to’rtinchi hadini toping.
A)24 B)32 C)54 D)27 E)57
10-misol 150 dan katta bo’lmagan7 ga karrali barcha natural sonlarning yig’indisini toping
A)1450 B)1617 C)1803 D)1517 E)1950
11-misol
Geometrik progressiyaning dastlabki uchta hadi yig’indisi -26 ga , dastlabki to’rttasiniki esa -80 ga teng .Agar shu progressiyaning birinchi hadi -2 ga teng bo’lsa, uning maxraji qanchaga teng bo’ladi?
A)3 B)-3 C)-2 D)2 E)4
12-misol Arifmetik progressiyaning hadlari 19 ta. Uning o’rta hadi 21 ga teng. Shu progressiyaning hadlari yig’indisini toping

1. 
   398 B)399 C)400 D)384 E)392
   

13-misol
Ishorasi almashinuvchi geometrik progressiyaning birinchi hadi 2 ga , uchinchi hadi 8 ga teng . Shu progressiyaning dastlabki 6 ta hadi yig’indisini toping.
A)20 B)-20 C)-42 D)42 E)-64
14-misol 1;8;22;43….sonlar ketma-ketligi shunday xususiyatga egaki, ikkita qo’shni hadlarinig ayirmasi 7;14;21;… arifmetik progressiyani tashil etadi. Berilgan ketma- ketlikning nechanchi hadi 35351 ga teng? A)97 B)99 C)101 D)103 E) 107

15-misol
5 ta haddan iborat geometrik progressiyaning hadlari yig’indisi birinchi hadini hisobga olmaganda 30 ga oxirgisini hisobga olmaganda 15 ga teng. Shu progressiyaning uchinchi hadini toping

A)6 B)8 C)4 D)10 E)12
16-misol 9 ga bo’lganda qoldig’I 4 ga teng bo’ladigan barcha ikki xonali sonlarning yig’indisi toping
A)527 B)535 C)536 D)542 E)545

17. 
    misol
    

Oltita haddan iborat geometrik progressiyaning dastlabki uchta hadining yig’indisi 168 ga, keyingi uchtasiniki esa 21 ga teng. Shu progressiyaning maxrajini toping
A)0.25 B)0.2 C)0.5 D)2 E)3
18-misol 1 dan 75 gacha bo’lgan natural sonlardan kvadratini 3 ga bo’lganda 1 qoldiq chiqadigan sonlar yig’indisini toping A)1875 B)925 C)1900 D)2850 E)2125
19-misol
Yig’indisi 15 ga teng bo’lgan 3 ta son arifmetik progressiyaning dastlabki uchta hadidir. Agar shu sonlarga mos ravishda 1:3va 9 sonlari q’shilsa , hosil bo’lgan sonlar o’suvchi geometric progressiyaning ketma- ket hadlari bo’ladi. Geometrik progressiyaning dastlabki 6 ta hadi yig’indisini toping.
A)252 B)256 C)248 D)254 E)250
20-misol 1*4+2*8+3*12+….+30*120 yig’indida har bir qo’shiluvchining ikkinchi ko’paytuvchisi bittadan kamaytirilsa bu yig’indi qanchaga kamayadi
A)60 B)120 C)210 D)375 E)465
21-misol
Geometrik progressiyaning dastlabki 6 ta hadi yig’indisi 1820 ga , maxraji esa 3 ga teng.Shu progressiyaning birinchi va beshinchi hadlari yig’indisini toping.
A)164 B) 246 C)328 D)410 E)492
22-misol Natural sonlar qatori har bir sonning kvadrati bilan tugaydigan quyidagi qismlarga ajratilgan :1, (2,3,4), (5,6,7,8,9), (10,11,12,13,14,15,16),… 10-qismdagi sonlar yig’indisini toping
A)1758 B)1800 C)1626 D)1729 E)1913
23-misol
Cheksiz kamayuvchi geometrik progressiyaning hadlari yig’indisi 8 ga , dastlabki 4 tasiniki esa ga teng . Agar uning barcha hadlari musbat bo’lsa progressiyaning birinchi hadini toping.
A)2 B)1,5 C)1 D)3 E)3,5
24-misol