Mavzu: arifmetik progressiya

a₁ + d,

a₂ + d = a₁ + 2d,

a₃ + d = a₁ + 3d va h.k.

Umuman,

a_n = a₁ + (n -1)d, (1)

chunki arifmetik progressiyaning n- hadi uning birinchi hadiga d sonini (n - 1) marta qo‘shish natijasida hosil qilinadi.

(1) formula arifmetik progressiyaning n-hadi formulasi deyiladi.

2 - m a s a 1 a. Agar a₁ = -6 va d = 4 bo‘lsa, arifmetik progressiyaning yuzinchi hadini toping.

(1) formula bo‘yicha: a₁₀₀ = -6 + (100-1)-4= 390.

3-masaIa. 99 soni 3, 5, 7, 9, ... arifmetik progressiyaning hadi. Shu hadning nomerini toping.

Aytaylik, n - izlangan nomer bo‘lsin. a₁ = 3 va d = 2 bo‘lgani uchun a_n = a₁ + (n - 1)d formulaga ko‘ra: 99 = 3 + (n - 1) • 2. Shuning uchun 99 = 3 + In - 2; 98 = 2n, n = 49.

J a vob: n = 49.

4 - masa1a. Arifmetik progressiyada a₈ = 130 va a₁₂ = 166. n-hadining formulasini toping.

(1) formuladan foydalaniib, topamiz:

a₈ = a₁+ 7d, a₁₂ = a₁ + 11d.

a₈ va a₁₂ larning berilgan qiymatlarini qo‘yib, a_x va d ga nisbatan tenglamalar

sistemasini hosil qilamiz:Demak, a₁ = 130 - Id = 130 - 63 = 67. Progressiya n-hadi Ikkinchi tenglamadan birinchi tenglamani ayirib, hosil qilamiz:

4d = 36, d = 9

formulasini yozamiz:an = 67 + 9(n - 1) = 67 + 9n - 9 = 58 + 9n.

Javob: a = 9n + 58. a
O‘tilgan mavzuni mustahkamlash:

1. Savollar: 1) Sonli ketma-ketlik deb nimaga aytiladi?

2) Arifmetik progressiya deb nimaga aytiladi?

3) Arifmetik progressiyaning ayirmasi deb nimaga aytiladi?

4) Arifmetik progressiyaning n hadi formulasini ayting.

2. Misollar.

395. Agar:

l) a₁ = 2 va d = 5; bo‘lsa, arifmetik progressiyaning dastlabki beshta hadini yozing.

a₂= a₁+d=2+5=7; a₃= a₂+d=7+5=12; a₄= a₃+d=12+5=17;

a₅= a₄+d=17+5=22;

Javob:_2,_7,_12,_17,_22.'>Javob: 2, 7, 12, 17, 22.
2) a₁ = -3 va d = 2 bo‘lsa, arifmetik progressiyaning dastlabki beshta hadini yozing.

a₂= a₁+d=-3+2=-1; a₃= a₂+d=-1+2=1; a₄= a₃+d=1+2=3; a₅= a₄+d=3+2=5;

Javob: -3, -1, 1, 3, 5.
396. n- hadining formulasi bilan berilgan quyidagi ketma-ketlik arifmetik progressiya bo‘lishini isbotlang:

1) a_n = 3 - 4n;

a₁ =3-4=-1; a₂ =3-4*2=3-8=-5;

a₃ =3-4*3=3-12=-9 a₄ =3-4*4=3-16=-13.

d=a₂-a₁=a₃-a₂=a₄-a₃=-4;
397. Arifmetik progressiyada:

1) agar a₁ = 2, d = 3 bo‘lsa, a₁₅ ni toping;

a₁₅=a₁+14d=2+14*3=2+42=44;

Javob: a₁₅=44;

2) agar a₁ = 3, d = 4 bo‘lsa, a₂₀ ni toping;

a₂₀=a₁+19d=3+19*4=3+76=79;

Javob: a₂₀=79;
Darsni yakunlash:

Darsda faol qatnashgan o‘quvchilarni baholash va rag‘batlantirish.

Uyga vazifa: 396, 397(3-4).