Sonli ketma-ketlik haqida tushuncha. Ta’rif

Arifmetik progressiyaning n- hadi formulasi.
Agar APning birinchi hadi a₁ va ayirmasi d ma’lum bo’lsa, uning keyingi hadlarini ta’rif bo’yicha toppish mumkin.
a₂=a₁+d
a₃=a₂+d=a₁+2d
a₄=a₃+d=a₁+3d
a₅=a₄+d=a₁+4d
…
…
a_n=a₁+(n-1)d formula kelib chiqadi.


a_n=a₁+(n-1)d APning n-hadi formulasi d-di.

Bu formulada to’rtta kattalik a₁,a_n,d va n lar ishtirok etayapdi, ulardan ixtiyoriy uchtasi ma’lum bo’lsa, to’rtinchisini toppish mumkin

a₁=a_n-(n-1)d;
d= ;
n= ;
Masalan:1) agar APda a₁=-13, d=2 bo’lsa, a₃₁=?
Yechish: a₃₁=a₁+(31-1)d=-13+30∙2=47. Javob: a₃₁=47
2) agar APda a₄₅=68, d=1,5 bo’lsa, a₁=?
Yechish: a₁=a₄₅-(45-1)d=68-44∙1,5=2. Javob: 2
3) agar APda a₁=-3, d=2 bo’lsa, uning nechanchi hadi 197 ga teng bo’ladi?
Yechish: a_n=197, n= = =101.
Javob: 101
Teorema: APning dastlabki n ta hadining yig’indisi
formula bilan hisoblanadi.
Masalan: 1) APda a₁=3, d=2 bo’lsa, S₅₀=?
Yechish: a₅₀=3+49∙2=101
S₅₀= =2600. Javob: 2600.
Eslatma: Agar APning birinchi hadi a₁ va ayirmasi d=0 bo’lsa, dastlabki n ta hadining yig’indisi S_n=a₁∙n formula bilan topiladi.
Masalan: 1) 3,3,3,3,… ketma-ketlikning dastlabki 20 ta hadining yig’indisini toping?
Yechish: a₁=3, d=0, n=20, S₂₀=3∙20=60. Javob:60
2) Dastlabki nechta natural sonlarning yig’indisi 253 ga teng bo’ladi?
Yechish: a₁=1, d=1, S_n=253, a_n=n, n=?
253=
n²+n-506=0
n₁=22
n₂=-23 chet il. Javob: 22
Eslatma: formuladagi a_n ni o’rniga a_n=a₁+(n-1)d ni qo’ysak,

to’rtinchisini toppish mumkin.

Masalan:1) GPda b₁=3, q=-2, bo’lsa, b₇=?
Yechish: b₇=b₁q⁶=3∙(-2)⁶=192.
Javob: 192
2) GPda b₇=320, q=2 bo’lsa, b₁=?
Yechish: b₇=b₁∙q⁶, b₁=
Javob: 5
3) Gpda b₅=162, b₁=2 bo’lsa q ni toping?
Yechish: b₅=b₁q⁴, q= =±3. javob: ±3
4) 486 soni 2,6,18,…ketma-ketlikning nechanchi hadi bo’ladi?
Yechish: b_n=486, b₁=2, q=3, n=?
b_n=b₁q^n-1
486=2∙3^n-1
243=3^n-1
3⁵=3^n-1
n=6, Javob: 6