Sonli ketma-ketlik haqida tushuncha. Ta’rif

Geometric progressiyaning xossalari. 1-xossa: GPning ixtiyoriy hadi o’zidan teng uzoqlashgan hadlarning o’rta geometrigiga teng. Bu xossa barcha hadlari musbat bo’lga GP uchun o’rinli Masalan: 1) GPda b6=14, b14=56 bo’lsa, b10=? Yechish: b10= . Javob: 28 2) GPda b10=20, b18=80, bo’lsa b16 ni toping? Yechish: b14= b16= . Javob: 40 2-xossa: Agar bn,bm,bk va bt lar biror GPning hadlari bo’lib, n+m=k+t bo’lsa, u holda bn ∙ bm = bk ∙ bt bo’ladi. Masalan: 1) Agar GPda b5∙b6∙b7=216 bo’lsa, b6=? Yechish: b6∙b6=b5∙b7 ; b5∙b6∙b7=b6∙(b5∙b7)=b6∙b6∙b6=216 =216, b6=6. Javob: 6 2) Agar PGda b1+b9=5, =17 bo’lsa, b4∙b6=? Yechish: =(b1+b9)^2-2b1b9=17 5^2-2b1b9=17; b1b9=b4b6 25-2 b4b6=17 b4b6=4. Javob: 4. Geometrik progressiyaning dastlabki n ta hadini hisoblash formulasi. b1, b2, b3,..bn-1, bn, bn+1,.. GPning dastlabki n ta hadining yig’indisi deb, b1+ b2+ b3+…+bn-1+ bn. yig’indiga aytiladi. Bu yig’indi Sn deb belgilanadi. Sn= b1+ b2+ b3+…+bn-1+ bn S1=b1; S2=b1+b2; S3=b1+b2+b3; S2-S1=b2; S3-S2=b3 Sn-Sn-1=bn; Sn+1-Sn=bn+1; 5-misol: 5+15+45+…1215 yig’indida qatnashuvchi sonlar GPni tashkil qilsa, shu yig’indini toping. Yechish: b1=5, q=3, bn=1215 Sn= J: 1820 Cheksiz kamayuvchi geometrik progressiya Ta’rif: Cheksiz kamayuvchi geometrik progressiya deb maxrajining moduli birdan kichik bo’lgan GPga aytiladi Demak, ChkGPda q , 0< <1, q 0 bo’ladi‌‌. Misollar: 1, ChkGP q= 1, - ChkGP q=- Chiksiz kamayuvchi bo’lgan GP larning barcha hadlari yi’g’indisini topish mumkin. Teorema: birinchi hadi b1 va maxraji q bo’lgan ChkGP ning barcha hadlari yig’indisi ushbu formula bilan topiladi S= 1-misol: yig’indini hisoblang. Yechish: Ular ChkGPni tashkil qiladi. b1=1, q= S= J: 1,5 2-misol: yig’indini hisoblang. Yechish: S= J: 3-misol: Hisoblang J: 4-misol: ChkGP ning hadlari yig’indisi 8 ga, dastlabki 4tasiniki esa gat teng. Agar uning barcha hadlari musbat bo’lsa, progressiyaning birinchi hadini toping S4= ; q^4= ; q= ; b1=4; J: b1=4