Arifmetik progressiya: matematika darsi

Bola birinchi kuni $5$ ta so'z yodlab, keyingi har kuni oldingisidan $3$ ta ko'p yodlasa, kunlik sonlar $5,\ 8,\ 11,\ 14,\ \ldots$ bo'ladi. Yigirmanchi kunni sanab o'tirmasdan $a_{20}=5+19\cdot 3=62$ deb topish mumkin, $20$ kunlik jami esa $670$ ta so'z. Bu darsda arifmetik progressiyaning har bir formulasi, shartlari va imtihonda uchraydigan usullari to'liq beriladi.

Ta'rif

Ikkinchi hadidan boshlab har bir hadi oldingi hadga bir xil $d$ sonni qo'shishdan hosil bo'ladigan ketma-ketlik arifmetik progressiya deyiladi: $a_{n+1}=a_n+d$ . $d$ son progressiyaning ayirmasi: $d=a_{n+1}-a_n$ .

Ayirma	Progressiya	Misol
Ayirma $d>0$	Progressiyao'suvchi	Misol $2,\ 5,\ 8,\ 11,\ \ldots$ ( $d=3$ )
Ayirma $d<0$	Progressiyakamayuvchi	Misol $10,\ 6,\ 2,\ -2,\ \ldots$ ( $d=-4$ )
Ayirma $d=0$	Progressiyao'zgarmas	Misol $7,\ 7,\ 7,\ \ldots$ ( $d=0$ )

Ayirma keyingi haddan oldingi hadni ayirib topiladi, teskarisi emas: $10,\ 6,\ 2$ da $d=6-10=-4$ . Ayirma va hadlar kasr yoki manfiy bo'lishi mumkin.

n-hadi formulasi

$a_2=a_1+d$ , $a_3=a_1+2d$ , $a_4=a_1+3d$ . Birinchi haddan $n$ -hadgacha $n-1$ ta qadam bor, shuning uchun:

a_n=a_1+(n-1)d

n\in\mathbb{N}

Birinchi had va ayirma orqali istalgan had

a_n=a_k+(n-k)d

Istalgan ma'lum haddan boshlab:

a_{10}=a_6+4d

a_3=a_6-3d

d=\frac{a_m-a_k}{m-k}

m\ne k

Ikki had ma'lum bo'lsa,

a_8

a_4

dan

d=\dfrac{a_8-a_4}{4}

Formulani ochsak, $a_n=dn+(a_1-d)$ , ya'ni $a_n$ ifoda $n$ ga nisbatan chiziqli. Aksincha, $a_n=kn+b$ ko'rinishdagi har qanday ketma-ketlik ayirmasi $k$ bo'lgan arifmetik progressiya. Masalan, $a_n=4n-1$ da $a_1=3$ , $d=4$ .

Hadlar sonini topish: $a_n=a_1+(n-1)d$ dan $d\ne 0$ da:

n=\frac{a_n-a_1}{d}+1

Natija natural son bo'lsagina, berilgan son progressiyaning hadi

$5,\ 8,\ 11,\ \ldots$ progressiyada $101$ soni $33$ -had, chunki $\dfrac{96}{3}+1=33$ . $100$ esa had emas, chunki $\dfrac{95}{3}$ butun son emas.

Dastlabki n ta had yig'indisi

Yig'indini ikki marta, oldin to'g'ri, keyin teskari tartibda yozib qo'shamiz. Ustma-ust turgan har bir juftlik bir xil $a_1+a_n$ ni beradi, juftliklar $n$ ta, demak $2S_n=(a_1+a_n)n$ :

S_n=\frac{a_1+a_n}{2}\cdot n

Birinchi va oxirgi had ma'lum bo'lsa

S_n=\frac{2a_1+(n-1)d}{2}\cdot n

Birinchi had va ayirma ma'lum bo'lsa

Formulalar $d$ ning har qanday qiymatida o'rinli, $d=0$ da ham: $S_n=na_1$ . Tez-tez uchraydigan yig'indilar:

Yig'indi	Natija	Misol
Yig'indi $1+2+3+\ldots+n$	Natija $\dfrac{n(n+1)}{2}$	Misol $1+2+\ldots+100=5050$
Yig'indi $1+3+5+\ldots+(2n-1)$	Natija $n^2$	Misol $1+3+\ldots+19=10^2=100$
Yig'indi $2+4+6+\ldots+2n$	Natija $n(n+1)$	Misol $2+4+\ldots+20=10\cdot 11=110$

Karrali sonlar ham progressiya tuzadi. Uch xonali $7$ ga karrali sonlar $105,\ 112,\ \ldots,\ 994$ :

n=\frac{994-105}{7}+1=128

S_n=\frac{105+994}{2}\cdot 128=70336

Avval hadlar soni, keyin yig'indi

O'rta arifmetik xossasi

Ikkinchi hadidan boshlab har bir had qo'shni hadlarning o'rta arifmetigiga teng: $a_n=\dfrac{a_{n-1}+a_{n+1}}{2}$ , $n\ge 2$ . Umuman, $a_n=\dfrac{a_{n-k}+a_{n+k}}{2}$ , $1\le k<n$ .

Sababi: $a_{n-k}=a_n-kd$ , $a_{n+k}=a_n+kd$ va yig'indida $kd$ qisqaradi. Xossa teskarisiga ham to'g'ri, shuning uchun u progressiyaning belgisi: uchta $x,\ y,\ z$ son shu tartibda arifmetik progressiya tashkil etishi uchun $2y=x+z$ bo'lishi zarur va yetarli. Bu yerda hech qanday ishora sharti yo'q.

2y=x+z

x,\ y,\ z

arifmetik progressiya bo'lishining zarur va yetarli sharti. Hadlar ifoda bilan berilganda shu tenglama tuziladi

Indekslar yig'indisi xossasi

Ikki hadni $a_1$ va $d$ orqali yozib qo'shamiz:

a_m+a_n=2a_1+(m+n-2)d

Ikki had yig'indisi faqat indekslar yig'indisiga bog'liq

Shuning uchun $m+n=k+l$ bo'lsa:

a_m+a_n=a_k+a_l

Masalan,

2+11=5+8

, shuning uchun

a_2+a_{11}=a_5+a_8

a_{k-p}+a_{k+p}=2a_k

a_4+a_8=2a_6

O'rtadagi indeks:

\dfrac{4+8}{2}=6

. To'rtta had bo'lsa, ikkitadan juftlanadi

Xususan, chetlardan bir xil uzoqlikdagi hadlar yig'indisi teng:

a_1+a_n=a_2+a_{n-1}

=a_3+a_{n-2}=\ldots

Shundan $a_1$ va $d$ ni bilmasdan ham yig'indi topiladi:

S_{2k-1}=(2k-1)\cdot a_k

Toq sondagi hadlar yig'indisi: hadlar soni karra o'rtadagi had

Masalan, $a_8=14$ bo'lsa, $S_{15}=15\cdot 14=210$ . $a_1$ va $d$ alohida noma'lum bo'lsa ham javob aniq. Juft sondagi hadlarda o'rtada ikki had turadi: $S_{2k}=k(a_k+a_{k+1})$ .

Ikki yig'indi ayirmasi oraliqdagi hadlar yig'indisidir:

S_m-S_k=a_{k+1}+\ldots+a_m

$S_{21}-S_8$ da $a_9$ dan $a_{21}$ gacha $13$ ta had bor, o'rtasi $a_{15}$ , demak $S_{21}-S_8=13a_{15}$ .

Yig'indi formulasi berilganda

a_1=S_1

a_n=S_n-S_{n-1}

n\ge 2

Dastlabki

n

ta had yig'indisidan

n-1

ta had yig'indisini ayiramiz

$S_n$ ning formulasini $n$ bo'yicha ochsak, $S_n=\dfrac{d}{2}n^2+\left(a_1-\dfrac{d}{2}\right)n$ . Demak, arifmetik progressiya uchun $S_n=An^2+Bn$ ozod hadsiz bo'ladi, bunda $d=2A$ , $a_1=A+B$ .

Ozod had bo'lsa, ketma-ketlik birinchi hadidan boshlab progressiya emas. $S_n=n^2+1$ da $a_1=2$ , $a_2=3$ , $a_3=5$ : ayirmalar har xil.

Masalan, $S_n=3n^2+2n$ bo'lsa:

a_n=6n-1

d=6

a_1=5

d=2\cdot 3

a_1=3+2

. Tekshiruv:

S_2=16=5+11

Toq va juft nomerli hadlar

Har bir juft nomerli had bilan undan oldingi toq nomerli had ayirmasi $d$ ga teng: $a_{2k}-a_{2k-1}=d$ . Bundan ikki holat kelib chiqadi:

Hadlar soni	Toq nomerlilar va juft nomerlilar	Natija
Hadlar sonijuft, $2m$	Toq nomerlilar va juft nomerlilarikkalasida $m$ tadan had	Natija $S_{\text{juft}}-S_{\text{toq}}=md$
Hadlar sonitoq, $2m+1$	Toq nomerlilar va juft nomerlilartoq nomerlilar $m+1$ ta, juft nomerlilar $m$ ta	Natija $S_{\text{toq}}-S_{\text{juft}}=a_{m+1}$

Toq nomerli hadlarning o'zi ham progressiya: $a_1,\ a_3,\ a_5,\ \ldots$ ning ayirmasi $2d$ . Xuddi shunday, har $p$ -hadni olsak, ayirmasi $pd$ bo'lgan progressiya chiqadi. $2m+1$ ta hadda $S_{\text{toq}}=(m+1)a_{m+1}$ , $S_{\text{juft}}=ma_{m+1}$ : masalan, $1,2,3,4,5$ da $9=3\cdot 3$ , $6=2\cdot 3$ .

Shu fikr ishorasi almashib keladigan yig'indida ham ishlaydi:

a_1-a_2+\ldots+a_{2m+1}=a_{m+1}

Har bir juftlik $-d$ ni, $m$ ta juftlik $-md$ ni beradi va $a_{2m+1}-md=a_{m+1}$ .

a_1-a_2+\ldots+a_{19}=a_{10}=25

Masalan,

a_1=7

d=2

Hadlarni qulay belgilash

Hadlar soni	Belgilash	Foydasi
Hadlar soni $3$ ta	Belgilash $a-d,\ a,\ a+d$	Foydasiyig'indisi $3a$
Hadlar soni $5$ ta	Belgilash $a-2d,\ a-d,\ a,\ a+d,\ a+2d$	Foydasiyig'indisi $5a$ , kvadratlari yig'indisi $5a^2+10d^2$
Hadlar soni $4$ ta	Belgilash $a-3t,\ a-t,\ a+t,\ a+3t$	Foydasiyig'indisi $4a$ , ayirma $d=2t$

Beshta had yig'indisi $15$ , kvadratlari yig'indisi $85$ bo'lsa: $5a=15$ , $a=3$ ; $45+10d^2=85$ , $d^2=4$ . O'suvchi progressiyada $d=2$ va sonlar $-1,\ 1,\ 3,\ 5,\ 7$ .

Yig'indilar bloklari

Ketma-ket $n$ tadan olingan hadlar yig'indilari ham arifmetik progressiya tuzadi: $S_n,\ S_{2n}-S_n,\ S_{3n}-S_{2n}$ ning ayirmasi $n^2d$ , chunki har bir had $n$ qadam, ya'ni $nd$ ga oshadi va bunday hadlar $n$ ta.

Masalan, $S_5=15$ , $S_{10}=80$ bo'lsa, bloklar:

15,\ 65,\ 115

S_{15}=15+65+115=195

65-15=50=5^2\cdot 2

. Shu progressiya:

a_1=-1

d=2

Ayirmalari progressiya bo'lgan ketma-ketlik

$2,\ 5,\ 11,\ 20,\ \ldots$ o'zi progressiya emas, lekin qo'shni hadlari ayirmalari $3,\ 6,\ 9,\ \ldots$ progressiya. Bunday holda birinchi hadga $n-1$ ta ayirmaning yig'indisi qo'shiladi:

x_n=2+\frac{3+3(n-1)}{2}(n-1)

=2+\frac{3n(n-1)}{2}

Kvadratlar ayirmasidan tuzilgan yig'indilar ham progressiyaga keladi, 5-misolga qarang.

Ishlangan misollar

1-misol

Arifmetik progressiyada $a_3=11$ , $a_9=29$ . $a_{20}$ va $S_{20}$ ni toping.

$d=\frac{29-11}{9-3}=3$
Uchinchi haddan to'qqizinchi hadgacha $6$ ta qadam bor.
$a_1=a_3-2d=11-6=5$
Uchinchi haddan ikki qadam orqaga qaytamiz.
$a_{20}=5+19\cdot 3=62$
Yigirmanchi hadgacha $19$ ta qadam.
$S_{20}=\frac{5+62}{2}\cdot 20=670$
Chetki hadlarning o'rta arifmetigi karra hadlar soni.

Javob: $a_{20}=62$ , $S_{20}=670$

2-misol

Arifmetik progressiyada $a_4+a_6+a_{12}+a_{14}=52$ . $S_{17}$ ni toping.

$4+14=6+12=2\cdot 9$
Indekslarni juftlaymiz: o'rtasi $9$ .
$a_4+a_{14}=a_6+a_{12}=2a_9$
Indekslar yig'indisi xossasi.
$4a_9=52,\quad a_9=13$
$a_1$ va $d$ ni alohida topish shart emas.
$S_{17}=17\cdot 13=221$
$17$ ta hadning o'rtasi $a_9$ .

Javob: $221$

3-misol

Arifmetik progressiyaning dastlabki $24$ ta hadida toq nomerli hadlar yig'indisi $120$ , juft nomerlilari yig'indisi $156$ . $a_1$ ni toping.

$12d=156-120=36,\quad d=3$
Ikkala guruhda $12$ tadan had, har bir juft nomerli had oldingi toq nomerli haddan $d$ ga katta.
$S_{\text{toq}}=12a_1+(0+2+\ldots+22)d$
Toq nomerli hadlar $a_1$ , $a_1+2d$ , $\ldots$ , $a_1+22d$ , ularni qo'shamiz.
$0+2+\ldots+22=132$
Juft sonlar yig'indisi: $11\cdot 12$ .
$12a_1+132\cdot 3=120,\quad a_1=-23$
$12a_1=120-396=-276$ .
$a_{24}=-23+23\cdot 3=46$
Tekshirish uchun oxirgi hadni topamiz.
$S_{24}=\frac{-23+46}{2}\cdot 24=276$
Tekshiruv: $120+156=276$ .

Javob: $-23$

4-misol

O'suvchi arifmetik progressiyada $a_2+a_6=20$ , $a_2\cdot a_6=64$ . $S_{10}$ ni toping.

$t^2-20t+64=0$
Yig'indi va ko'paytma berilgan: Viyet teoremasiga ko'ra $a_2$ , $a_6$ shu tenglamaning ildizlari.
$t=4\ \text{yoki}\ t=16$
Tekshiruv: $4+16=20$ , $4\cdot 16=64$ .
$a_2=4,\quad a_6=16$
Progressiya o'suvchi, kichik had oldin keladi.
$4d=12,\quad d=3,\quad a_1=1$
Ikkinchi haddan oltinchi hadgacha $4$ qadam.
$S_{10}=\frac{2\cdot 1+9\cdot 3}{2}\cdot 10=145$
Ikkinchi yig'indi formulasi.

Javob: $145$

5-misol

Hisoblang:

60^2-57^2+56^2-53^2+\ldots

+4^2-1^2

$(4k)^2-(4k-3)^2=3(8k-3)$
Har bir juftlikda katta son $4$ ga karrali, kvadratlar ayirmasini ochamiz.
$4k=60,\quad k=15$
$k=1,\ 2,\ \ldots,\ 15$ , ya'ni $15$ ta juftlik.
$3(8\cdot 1-3)=15$
Juftliklar qiymati ayirmasi $24$ bo'lgan progressiya. Bu uning birinchi hadi, $k=1$ .
$3(8\cdot 15-3)=351$
Oxirgi hadi, $k=15$ .
$\frac{15+351}{2}\cdot 15=2745$
Progressiya yig'indisi.

Javob: $2745$

6-misol

Arifmetik progressiyada $a_1=2$ , $a_4=5+x$ , $a_{10}=8+4x$ . $a_{20}$ ni toping.

$3d=(5+x)-2=3+x$
Birinchi haddan to'rtinchi hadgacha $3$ qadam.
$9d=(8+4x)-2=6+4x$
Birinchi haddan o'ninchi hadgacha $9$ qadam.
$3(3+x)=6+4x,\quad x=3$
Birinchi tenglikni $3$ ga ko'paytirib, $9d$ larni tenglashtiramiz.
$d=\frac{3+3}{3}=2$
$3d=3+x$ ga $x=3$ ni qo'yamiz.
$a_{20}=2+19\cdot 2=40$
Tekshiruv: $a_4=8$ , $a_{10}=20$ .

Javob: $40$

7-misol

Arifmetik progressiyaning yettinchi hadi $23$ . O'n birinchi hadini uchinchi hadiga bo'lganda bo'linma $3$ , qoldiq $2$ chiqadi. $a_1$ va $d$ ni toping.

$a_{11}=3a_3+2$
Bo'linuvchi = bo'luvchi karra bo'linma qo'shuv qoldiq.
$a_1+10d=3a_1+6d+2$
Hadlarni $a_1$ va $d$ orqali yozamiz.
$a_1=2d-1$
Soddalashtiramiz: $2a_1=4d-2$ .
$a_1+6d=23$
Ikkinchi shart: yettinchi had $23$ .
$8d-1=23,\quad d=3$
$a_1=2d-1$ ni qo'yamiz.
$a_1=5,\quad a_3=11,\quad a_{11}=35$
$a_1=2d-1$ va $d=3$ dan.
$35=3\cdot 11+2$
Tekshiruv: qoldiq $2$ bo'luvchi $11$ dan kichik.

Javob: $a_1=5$ , $d=3$

Ko'p uchraydigan xatolar

Noto'g'ri	To'g'ri
$a_{10}=a_1+10d$	$a_{10}=a_1+9d$ : qadamlar soni $n-1$
$a_3+a_5=a_8$	$a_3+a_5=2a_4$ . $a_3+a_5=a_8$ faqat $a_1=d$ bo'lganda bajariladi
$a_4$ , $a_8$ dan $d=a_8-a_4$	$d=\dfrac{a_8-a_4}{4}$
$5,\ 8,\ \ldots,\ 35$ da hadlar soni $\dfrac{35-5}{3}=10$	$\dfrac{35-5}{3}+1=11$
$S_{20}-S_9$ ichida $a_9$ ham bor deyish	$a_{10}$ dan $a_{20}$ gacha, ya'ni $11$ ta had: $S_{20}-S_9=11a_{15}$
$a_n=S_n-S_{n-1}$ ni $n=1$ da ham qo'llash	$a_1=S_1$ , formula $n\ge 2$ uchun
$10,\ 6,\ 2$ da $d=4$	$d=6-10=-4$

Eslab qoling

$a_n=a_1+(n-1)d$ , $a_n=a_k+(n-k)d$ , $d=\dfrac{a_m-a_k}{m-k}$ .
$S_n=\dfrac{a_1+a_n}{2}\cdot n$ .
$S_n=\dfrac{2a_1+(n-1)d}{2}\cdot n$ .
$a_n=\dfrac{a_{n-1}+a_{n+1}}{2}$ ; uch son uchun $2y=x+z$ .
$m+n=k+l$ bo'lsa, $a_m+a_n=a_k+a_l$ ; $S_{2k-1}=(2k-1)a_k$ .
$a_1=S_1$ ; $n\ge 2$ da $a_n=S_n-S_{n-1}$ .
$S_n=An^2+Bn$ da $d=2A$ .
Uchta had: $a-d,\ a,\ a+d$ . Hadlar soni: $\dfrac{a_n-a_1}{d}+1$ , $d\ne 0$ .