MathTest.uz
Algebra

Arifmetik progressiya yig'indisi

ortacha 165 daqiqa arifmetik progressiya yig‘indisiarithmetic seriesS_npartial sumGauss pairingo‘rtacha hadsigmaketma-ketlikjamlanmachiziqli o‘sishword problemsseries

Nima uchun muhim?

Ko‘p real vaziyatlarda bizga bitta had emas, ma’lum vaqtgacha yig‘ilgan umumiy natija kerak bo‘ladi: jami ishlab chiqarish, jami o‘rindiq, umumiy mashq soni, bosqichma-bosqich jamg‘arma. Arifmetik progressiya yig‘indisi doimiy o‘zgarishning kumulyativ natijasini tez hisoblashni o‘rgatadi va sigma yozuvi, ketma-ketliklar hamda keyingi geometrik progressiya yig‘indisi uchun tayanch yaratadi.

O'quv maqsadlari

  • a_n va S_n belgilarini farqlash
  • Dastlabki n had yig‘indisini partial sum sifatida tushuntirish
  • S_n=n(a_1+a_n)/2 formulasini qo‘llash
  • S_n=n[2a_1+(n-1)d]/2 formulasini qo‘llash
  • Juftlash usuli bilan yig‘indi formulasini asoslash
  • Birinchi va oxirgi had o‘rtachasini n ga ko‘paytirish g‘oyasini tushuntirish
  • n toq bo‘lganda markaziy had orqali yig‘indini topish
  • n juft bo‘lganda simmetrik juftlardan foydalanish
  • a_1,d,n ma’lum bo‘lsa S_n ni topish
  • a_1,a_n,n ma’lum bo‘lsa S_n ni topish
  • S_n va S_{n-1} orqali a_n ni tiklash
  • Partial-sum formulasidan ketma-ketlik hadlarini tiklash
  • Berilgan S_n dan noma’lum n ni topish
  • Berilgan S_n dan a_1 yoki d ni topish
  • Chekli AP ning ma’lum indeks oralig‘idagi yig‘indisini hisoblash
  • a_p dan a_q gacha yig‘indini partial-sum farqi orqali topish
  • Simmetrik had juftlari yig‘indisidan tez foydalanish
  • Sigma notationni AP yig‘indisiga o‘tkazish
  • Chiziqli an formula uchun yig‘indini kvadratik n-formaga keltirish
  • Yig‘indi o‘sishining hadlar bilan bog‘lanishini talqin qilish
  • Real hayotdagi doimiy o‘suvchi miqdorlarning jami qiymatini model qilish
  • Yig‘indi va individual had formulalarini aralashtirmaslik
Arifmetik progressiyada a_n bitta hadni bildiradi, S_n esa birinchi n ta hadning yig‘indisi: S_n=a_1+a_2+...+a_n. Agar hadlar doimiy d qadam bilan o‘zgarsa, birinchi va oxirgi hadni, ikkinchi va oxiridan ikkinchi hadni juftlasak har bir juftlik bir xil a_1+a_n yig‘indini beradi. Shu symmetry Gauss juftlash usulining asosi va S_n=n(a_1+a_n)/2 formulasiga olib keladi. Oxirgi had alohida berilmagan bo‘lsa Topic49 dagi a_n=a_1+(n-1)d formulasini qo‘yib, S_n=n[2a_1+(n-1)d]/2 ko‘rinishini olamiz. Demak AP yig‘indisi aslida “hadlar soni × hadlarning o‘rtacha qiymati”dir. Partial sums yana bitta kuchli g‘oya beradi: S_n-S_{n-1}=a_n. Shuning uchun yig‘indi formulasidan original ketma-ketlikni tiklash ham mumkin. Bu mavzuda formulani yodlashdan ko‘ra qaysi ma’lumotlar berilganini tanib, eng qisqa equivalent formulani tanlash muhim.

Ta'riflar

Arifmetik qator · Arithmetic series

Arifmetik progressiya hadlarini qo‘shishdan hosil bo‘lgan chekli yoki kontekstga qarab qisman yig‘indi.

Progressiyaning hadlari ro‘yxat bo‘lsa, ularning yig‘indisi qator yoki series sifatida qaraladi.

Misol: $3+7+11+15$

Bu emas: $3,7,11,15$ — bu hadlar ro‘yxati, yig‘indi emas.

💡 Bu mavzuda asosan chekli partial sums ishlatiladi.

Partial sum · Qisman yig‘indi

Birinchi n had yig‘indisi S_n=a_1+a_2+...+a_n.

n-gacha jamlangan umumiy qiymat.

Misol: $S_4=2+5+8+11=26$

Bu emas: $a_4=11$ — bitta had.

💡 S_n va a_n ni ajrating.

S_n belgisi · Sum notation

S_n — ketma-ketlikning birinchi n hadining yig‘indisi.

Past indeks n nechta had qo‘shilganini bildiradi.

Misol: $S_{10}$ — birinchi 10 had yig‘indisi.

Bu emas: $a_{10}$ — 10-hadning o‘zi.

💡 Eng ko‘p uchraydigan notation xatosi shu.

Birinchi had · First term

Yig‘indiga kiradigan birinchi AP hadi a_1.

Juftlash formulasidagi chap endpoint.

Misol: $a_1=4$

Bu emas: S_1 bilan bir xil qiymat bo‘lishi mumkin, lekin belgi ma’nosi boshqa.

💡 S_1=a_1.

Oxirgi had · Last term

Dastlabki n had yig‘indisidagi oxirgi qiymat a_n.

Juftlash formulasidagi o‘ng endpoint.

Misol: $a_{20}=61$

Bu emas: n hadlar soni, a_n qiymat.

💡 a_n=a_1+(n-1)d.

Hadlar soni · Number of terms

Yig‘indiga kiradigan hadlar umumiy soni n.

Birinchi haddan oxirgi hadgacha nechta son qo‘shilayotganini bildiradi.

Misol: $S_{25}$ da n=25.

Bu emas: Oxirgi indeks 25 bo‘lsa, 0-based model bilan aralashtirmang.

💡 Standart Topic50 indekslash 1 dan.

Umumiy ayirma · Common difference

AP ning qo‘shni hadlari ayirmasi d.

Yig‘indi formulasida oxirgi hadni tiklash uchun ishlatiladi.

Misol: $4,7,10,...$ uchun d=3.

Bu emas: Foiz o‘sishi d emas.

💡 Topic49 prerequisite.

Juftlash usuli · Pairing method

Yig‘indining boshidan va oxiridan simmetrik hadlarni juftlab, teng juft yig‘indilardan foydalanish usuli.

a_1+a_n, a_2+a_{n-1}, ... bir xil qiymat beradi.

Misol: $1+2+...+100$ ni 1+100,2+99,... qilib juftlash.

Bu emas: Tasodifiy hadlarni juftlash teng sum bermasligi mumkin.

💡 Gauss pairing deb ham ataladi.

Simmetrik juftlar · Symmetric pairs

Indekslar yig‘indisi n+1 bo‘lgan a_k va a_{n+1-k} hadlar.

Boshidan k-chi va oxiridan k-chi had.

Misol: $a_3+a_{n-2}=a_1+a_n$

Bu emas: Indekslar yig‘indisi n+1 bo‘lmasa bu ayni symmetry emas.

💡 Har juft sum a_1+a_n.

O‘rtacha had qiymati · Average term value

Chekli AP hadlarining arifmetik o‘rtachasi (a_1+a_n)/2.

AP simmetriyasi sabab barcha hadlar average’i endpointlar average’iga teng.

Misol: $2,5,8,11$ average=6.5=(2+11)/2.

Bu emas: Geometrik o‘rta emas.

💡 S_n=n·average.

Markaziy had · Middle term

n toq bo‘lganda indeks (n+1)/2 dagi yagona o‘rta had.

Simmetrik juftlar markaz atrofida joylashadi.

Misol: n=9 da a_5 markaziy had.

Bu emas: n juft bo‘lsa yagona markaziy had yo‘q.

💡 n toq holatda average=a_{(n+1)/2}.

Sigma notation · Sigma yozuvi

∑ belgisi yordamida indeks bo‘yicha hadlarni jamlash yozuvi.

Uzun yig‘indini ixcham yozadi.

Misol: $\sum_{k=1}^{n} a_k=S_n$

Bu emas: $\sum$ ko‘paytirish belgisi emas.

💡 k — dummy index.

Jamlangan qiymat · Cumulative total

Bosqichlargacha yig‘ilib kelgan umumiy natija.

Har yangi bosqichda oldingi totalga navbatdagi had qo‘shiladi.

Misol: $S_n=S_{n-1}+a_n$

Bu emas: $a_n$ning o‘zi cumulative total emas.

💡 Real-life interpretation uchun muhim.

Yig‘indi farqi · Difference of partial sums

S_q-S_{p-1}=a_p+a_{p+1}+...+a_q.

Oldingi jamlanmani kattaroq jamlanmadan ayirib kerakli interval qoladi.

Misol: $a_5+...+a_{12}=S_{12}-S_4$

Bu emas: $S_{12}-S_5$ bo‘lsa a_5 ham yo‘qoladi.

💡 p=1 da S_0=0 convention ishlatilishi mumkin.

S_0 · Empty sum

Hech qanday had qo‘shilmagan yig‘indi uchun qulay convention S_0=0.

Index-range formulalarini yagona ko‘rinishda yozishga yordam beradi.

Misol: $S_3-S_0=S_3$

Bu emas: $S_0=a_0$ emas.

💡 Bu algebraik convention.

Yig‘indidan hadni tiklash · Recovering a term

a_n=S_n-S_{n-1}.

Jamlangan totalning bir qadamdagi ortishi aynan navbatdagi had.

Misol: $S_8-S_7=a_8$

Bu emas: $S_8-S_6=a_8$ emas; a_7+a_8 chiqadi.

💡 Partial sums sequence bilan original sequence bog‘lanishi.

Kvadratik partial-sum modeli · Quadratic partial sum

AP ning S_n formulasi n bo‘yicha kvadratik polinom bo‘ladi.

Hadlar chiziqli bo‘lsa, ularning jamlanmasi quadratic o‘sadi.

Misol: $a_n=3n+1$ bo‘lsa S_n quadratic.

Bu emas: Har quadratic S_n avtomatik AP emas; S_n-S_{n-1} affine bo‘lishi tekshiriladi.

💡 Discrete accumulation g‘oyasi.

Triangular number · Uchburchak son

T_n=1+2+...+n=n(n+1)/2.

Eng oddiy AP yig‘indisi.

Misol: $T_{10}=55$

Bu emas: $1+2+4+...$ triangular sum emas.

💡 Ko‘p formulalarning canonical misoli.

Arithmetic-series word model · Amaliy yig‘indi modeli

Har bosqichdagi miqdor AP bo‘lib, savol jami miqdorni so‘rasa arithmetic-series model ishlatiladi.

“Har kuni 3 taga ko‘proq, jami nechta?” kabi.

Misol: 1-kun 10, har kuni +2, 7 kunlik jami.

Bu emas: Faqat 7-kun qiymatini so‘rasa series emas, nth term.

💡 Savolda “jami” signal so‘z.

Formula tanlash · Formula selection

Berilgan ma’lumotlarga mos equivalent S_n formulasini tanlash jarayoni.

a_n ma’lum bo‘lsa endpoint formula, d ma’lum bo‘lsa difference formula qulay.

Misol: $a_1,a_n,n$ → n(a_1+a_n)/2.

Bu emas: Har masalada avval a_n ni topish majburiy emas.

💡 Hisobni qisqartiradi.

Fundamental tushunchalar

Had va yig‘indi farqi

a_n bitta qiymat; S_n esa a_1 dan a_n gacha barcha hadlar jamlanmasi.

$a_n\neq S_n$

S_1=a_1 bo‘lsa ham n>1 da ma’nolari keskin farq qiladi.

Gauss juftlash invariant

AP da boshidan va oxiridan teng masofadagi hadlar yig‘indisi bir xil.

$a_k+a_{n+1-k}=a_1+a_n$

Indekslar yig‘indisi n+1.

Average × count prinsipi

Chekli AP yig‘indisi hadlar average’ining hadlar soniga ko‘paytmasidir.

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

Endpoint average barcha hadlar average’iga teng.

Ikki asosiy formula ekvivalent

a_n ni nth-term formula bilan almashtirish ikki mashhur S_n formulasini bog‘laydi.

$S_n=\frac n2(a_1+a_n)=\frac n2[2a_1+(n-1)d]$

Topic49 formula Topic50 formulaga ulanadi.

n toq: markaziy had

Toq sondagi AP hadlarining average’i markaziy hadning o‘zi.

$S_n=n a_{(n+1)/2}$

n toq.

n juft: markaziy juft

Juft n da ikkita markaziy had bor; ularning average’i butun AP average’iga teng.

$S_n=\frac n2(a_{n/2}+a_{n/2+1})$

n juft.

Partial-sum recurrence

Har yangi total oldingi totalga navbatdagi had qo‘shish bilan olinadi.

$S_n=S_{n-1}+a_n$

n≥1, S_0=0.

Difference recovers term

Partial-sum sequence first difference’i original sequence hadidir.

$a_n=S_n-S_{n-1}$

n≥1.

AP → quadratic S_n

a_n affine bo‘lgani uchun S_n n bo‘yicha quadratic bo‘ladi.

$a_n=pn+q\Rightarrow S_n=\frac p2n(n+1)+qn$

n musbat butun.

Quadratic S_n → AP

Agar S_n=An^2+Bn va S_0=0 bo‘lsa a_n=S_n-S_{n-1}=2An+(B-A), ya’ni AP.

$S_n=An^2+Bn$

Constant term 0 empty-sum convention bilan mos.

Oraliq yig‘indi

Ketma-ket indekslar p dan q gacha bo‘lsa, oldingi partial sumni ayiramiz.

$\sum_{k=p}^{q}a_k=S_q-S_{p-1}$

1≤p≤q.

Triangular numbers as base model

1,2,...,n AP ning yig‘indisi n(n+1)/2; ko‘p chiziqli summalar shu modelga kamayadi.

$1+2+\cdots+n=\frac{n(n+1)}2$

n∈N.

Constant sequence sum

d=0 bo‘lsa barcha hadlar a_1 va yig‘indi n a_1.

$d=0\Rightarrow S_n=na_1$

AP maxsus holati.

Negative difference allowed

Kamayuvchi AP yig‘indisi formulalari o‘zgarmaydi; faqat hadlar va total signi kontekstga bog‘liq.

$d<0$

Formula aynan ishlaydi.

Zero-crossing and sum

Hadlar musbatdan manfiyga o‘tsa ham algebraik yig‘indi signed total hisoblanadi.

$S_n=\sum a_k$

Kontekstda “jami miqdor” manfiy qiymatni fizik talqin qilishni talab qilishi mumkin.

Unknown n requires discreteness

S_n tenglamadan n topilgach n musbat butun bo‘lishi shart.

$n\in\mathbb N$

Quadratic algebraic rootlar filtrlanadi.

Symmetric-pair speed trick

Agar yig‘indi to‘liq AP block bo‘lsa har juft endpoint sum bir xil; ko‘p individual hadlarni hisoblash shart emas.

$a_p+a_q=a_{p+1}+a_{q-1}$

p+q fixed.

Sigma linearity

Konstant coefficient yig‘indidan tashqariga chiqadi va yig‘indilar ajratiladi.

$\sum(ck+b)=c\sum k+\sum b$

Bir xil index bounds.

Arithmetic mean connection

AP ning finite mean qiymati endpoint meaniga teng.

$\bar a=\frac{S_n}{n}=\frac{a_1+a_n}{2}$

n>0.

Cumulative real-world interpretation

a_n bosqichdagi miqdor, S_n esa shu vaqtgacha jami miqdor.

$S_n=S_{n-1}+a_n$

Units: a_n va S_n bir xil physical unit, lekin semantic scope boshqa.

Formula selection by known data

a_1,a_n,n ma’lum → endpoint formula; a_1,d,n ma’lum → difference formula; S_n/S_{n-1} ma’lum → term difference.

Ma’lumotni avval inventory qiling.

Scope boundary: finite vs infinite series

AP d≠0 bo‘lsa hadlar 0 ga yaqinlashmaydi, shuning uchun odatiy infinite sum convergent emas; bu mavzu finite sumsga qaratilgan.

Cheksiz qator konvergentsiyasi alohida mavzu.

Formula kutubxonasi

Partial sum ta’rifi

$$S_n=a_1+a_2+\cdots+a_n$$

Birinchi n AP hadining to‘g‘ridan-to‘g‘ri yig‘indisi.

Shart: n≥1

Xususiy holatlar: S_1=a_1.

Endpoint formula

$$S_n=\frac{n(a_1+a_n)}{2}$$

Hadlar soni × endpointlar average’i.

Shart: n musbat butun

Xususiy holatlar: n juft/toqdan qat’i nazar.

Difference formula

$$S_n=\frac{n}{2}[2a_1+(n-1)d]$$

Oxirgi hadni alohida topmasdan yig‘indi beradi.

Shart: n musbat butun

Xususiy holatlar: d=0 → S_n=na_1.

Oxirgi had orqali

$$a_n=a_1+(n-1)d$$

Topic49 nth-term formulasi yig‘indi formulasining prerequisite’i.

Shart: n≥1

Xususiy holatlar: d=0 maxsus holat.

Average-term form

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

Chekli AP average’i endpointlar average’iga teng.

Shart: n>0

Xususiy holatlar: n toq bo‘lsa middle termga teng.

Toq n markaziy had

$$S_n=n a_{(n+1)/2}$$

Yig‘indi = hadlar soni × markaziy had.

Shart: n toq musbat butun

Xususiy holatlar: Masalan S_9=9a_5.

Juft n markaziy juft

$$S_n=\frac n2(a_{n/2}+a_{n/2+1})$$

n/2 juftning har biri markaziy juft yig‘indisiga teng.

Shart: n juft musbat butun

Xususiy holatlar: Masalan S_10=5(a_5+a_6).

Partial-sum recurrence

$$S_n=S_{n-1}+a_n$$

Jamlangan total bir yangi hadga ortadi.

Shart: n≥1, S_0=0

Xususiy holatlar: S_1=S_0+a_1=a_1.

Haddni partial sumsdan tiklash

$$a_n=S_n-S_{n-1}$$

Ikki consecutive total farqi navbatdagi had.

Shart: n≥1

Xususiy holatlar: a_1=S_1-S_0.

Indeks oralig‘i yig‘indisi

$$a_p+a_{p+1}+\cdots+a_q=S_q-S_{p-1}$$

Katta partial sumdan oldingi qismni olib tashlaydi.

Shart: 1≤p≤q

Xususiy holatlar: p=1 → S_q.

Sigma notation

$$\sum_{k=1}^{n}a_k=S_n$$

Partial sumni sigma yozuvida ifodalaydi.

Shart: n musbat butun

Xususiy holatlar: Index nomi k,j,r bo‘lishi mumkin.

Triangular number

$$1+2+\cdots+n=\frac{n(n+1)}{2}$$

1 dan n gacha natural sonlar yig‘indisi.

Shart: n∈N

Xususiy holatlar: T_n notation ishlatiladi.

Birinchi n toq son yig‘indisi

$$1+3+5+\cdots+(2n-1)=n^2$$

Birinchi n toq son yig‘indisi perfect square.

Shart: n∈N

Xususiy holatlar: n=1 da 1.

Birinchi n juft son yig‘indisi

$$2+4+\cdots+2n=n(n+1)$$

Birinchi n juft son yig‘indisi.

Shart: n∈N

Xususiy holatlar: 2T_n.

Affine hadlar yig‘indisi

$$\sum_{k=1}^{n}(pk+q)=\frac{p n(n+1)}{2}+qn$$

Chiziqli hadlar yig‘indisini quadratic formulaga keltiradi.

Shart: n∈N

Xususiy holatlar: p=d, q=a_1-d.

AP partial sum quadratic form

$$S_n=\frac d2n^2+\left(a_1-\frac d2\right)n$$

S_n ning n bo‘yicha quadratic ekanini ko‘rsatadi.

Shart: n∈N

Xususiy holatlar: d=0 da linear na_1.

Quadratic S_n dan a_n

$$S_n=An^2+Bn\Longrightarrow a_n=2An+(B-A)$$

Quadratic partial sumdan original affine term olinadi.

Shart: S_0=0

Xususiy holatlar: Common difference d=2A.

Quadratic S_n dan d

$$S_n=An^2+Bn\Longrightarrow d=2A$$

Partial-sum quadratic curvature common difference’ning yarmi bilan bog‘liq.

Shart: S_0=0

Xususiy holatlar: A=0 → d=0.

Quadratic S_n dan a_1

$$S_n=An^2+Bn\Longrightarrow a_1=A+B$$

S_1=A+B bo‘lgani uchun first term shu.

Shart: S_0=0

Xususiy holatlar: A+B signed bo‘lishi mumkin.

Unknown n equation

$$S_n=T\Longrightarrow n[2a_1+(n-1)d]=2T$$

Berilgan totalga nechta had kerakligini quadratic tenglamaga aylantiradi.

Shart: n musbat butun

Xususiy holatlar: Algebraik rootsdan faqat positive integer qabul qilinadi.

Unknown d

$$d=\frac{2S_n/n-2a_1}{n-1}$$

Total, first term va countdan common difference.

Shart: n>1

Xususiy holatlar: n=1 da d aniqlanmaydi.

Unknown first term

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

Average’dan yarim total driftni ayirish.

Shart: n≥1

Xususiy holatlar: d=0 → a_1=S_n/n.

Endpoint sum invariant

$$a_k+a_{n+1-k}=a_1+a_n$$

Har simmetrik juft endpoint sumiga teng.

Shart: Valid symmetric indices

Xususiy holatlar: n toq bo‘lsa middle term o‘zi bilan juftlanadi.

Finite AP mean

$$\bar a=\frac{1}{n}\sum_{k=1}^{n}a_k=\frac{a_1+a_n}{2}$$

Finite AP mean endpoint midpointidir.

Shart: n>0

Xususiy holatlar: Toq n da mean=middle term.

Constant AP sum

$$d=0\Longrightarrow S_n=na_1$$

Bir xil n ta sonning yig‘indisi.

Shart: d=0

Xususiy holatlar: a_n=a_1.

Teoremalar va isbotlar

📐 Arifmetik progressiya yig‘indisi teoremasi

Birinchi hadi a_1, n-hadi a_n bo‘lgan chekli AP uchun S_n=n(a_1+a_n)/2.

Boshidan va oxiridan juftlasak barcha juftlar bir xil sum beradi.

Isbotni ko'rsatish

Berilgan: a_1,...,a_n arifmetik progressiya.

Isbotlash kerak: S_n=n(a_1+a_n)/2.

  1. S_n=a_1+a_2+...+a_n deb yozamiz.
  2. Teskari tartibda ham S_n=a_n+a_{n-1}+...+a_1.
  3. Ustunma-ust qo‘shilganda har juftlik a_1+a_n ga teng.
  4. Jami n ta shunday ustun bor: 2S_n=n(a_1+a_n).
  5. Ikki tomonni 2 ga bo‘lib formula olinadi.

S_n=n(a_1+a_n)/2. ∎

📐 Difference-form teoremasi

a_1,d,n ma’lum bo‘lsa S_n=n[2a_1+(n-1)d]/2.

AP ning endpointi first termdan n-1 qadam uzoqda.

Isbotni ko'rsatish

Berilgan: a_1,d,n ma’lum AP.

Isbotlash kerak: S_n=n[2a_1+(n-1)d]/2.

  1. Topic49 dan a_n=a_1+(n-1)d.
  2. Asosiy formula S_n=n(a_1+a_n)/2.
  3. a_n o‘rniga a_1+(n-1)d qo‘yamiz.
  4. Ichkarida 2a_1+(n-1)d hosil bo‘ladi.
  5. Natija S_n=n[2a_1+(n-1)d]/2.

Difference form isbotlandi. ∎

📐 Partial-sum difference teoremasi

Har qanday ketma-ketlik uchun a_n=S_n-S_{n-1}, S_0=0.

Jamlangan hisoblagichning bir qadamdagi ortishi shu qadamdagi miqdor.

Isbotni ko'rsatish

Berilgan: S_n=a_1+...+a_n va S_{n-1}=a_1+...+a_{n-1}.

Isbotlash kerak: a_n=S_n-S_{n-1}.

  1. S_n=a_1+a_2+...+a_{n-1}+a_n.
  2. S_{n-1}=a_1+a_2+...+a_{n-1}.
  3. Birinchisidan ikkinchisini ayiramiz.
  4. Barcha umumiy hadlar bekor bo‘ladi va a_n qoladi.

a_n=S_n-S_{n-1}. ∎

📐 Markaziy had teoremasi

Toq n li finite AP uchun S_n=n a_{(n+1)/2}.

Simmetrik hadlar markaz atrofida balance qiladi.

Isbotni ko'rsatish

Berilgan: n toq va a_1,...,a_n AP.

Isbotlash kerak: S_n=n a_{(n+1)/2}.

  1. n toq bo‘lgani uchun m=(n+1)/2 integer markaziy indeks.
  2. AP simmetriyasi bo‘yicha a_1+a_n=2a_m.
  3. Asosiy formula S_n=n(a_1+a_n)/2.
  4. a_1+a_n=2a_m ni qo‘ysak S_n=na_m.

S_n=n a_{(n+1)/2}. ∎

📐 Quadratic partial-sum characterization

S_0=0 bo‘lgan ketma-ketlikda S_n=An²+Bn bo‘lsa, a_n=2An+(B-A) arifmetik progressiya va d=2A; aksincha har AP partial sum shu quadratic ko‘rinishga ega.

Discrete derivative quadraticni linearga tushiradi, discrete summation lineardan quadraticga ko‘taradi.

Isbotni ko'rsatish

Berilgan: S_0=0 va S_n=An²+Bn yoki a_n AP.

Isbotlash kerak: Quadratic partial sum iff arithmetic term sequence.

  1. Agar S_n=An²+Bn bo‘lsa a_n=S_n-S_{n-1}.
  2. Hisob: A[n²-(n-1)²]+B=A(2n-1)+B=2An+(B-A), affine formula.
  3. Shuning uchun common difference 2A.
  4. Aksincha AP uchun S_n=n[2a_1+(n-1)d]/2.
  5. Yoyilganda S_n=(d/2)n²+(a_1-d/2)n, ya’ni quadratic va constant term 0.

AP ↔ zero-constant quadratic partial sum characterization isbotlandi. ∎

Yechilgan misollar

oson $3+7+11+15+19$ yig‘indini toping.

💡 Maslahat: Bu 5 hadli AP; endpoint formulani ishlating.

  1. a_1=3, a_5=19, n=5.
  2. S_5=5(3+19)/2.
  3. =5·22/2=55.

✅ Javob: $55$

Nega bu usul ishlaydi: Endpointlar average’i 11, 5 ta had bor.

Muqobil usul: Bevosita qo‘shish ham mumkin.

⚠️ Kichik misolda ham formula strukturasini tanish muhim.

oson $a_1=5,d=3,n=10$. $S_{10}$ ni toping.

💡 Maslahat: Difference-form formuladan foydalaning.

  1. S_{10}=10[2·5+9·3]/2.
  2. Ichki qiymat 10+27=37.
  3. S_{10}=5·37=185.

✅ Javob: $185$

Nega bu usul ishlaydi: a_1,d,n bevosita difference-formga mos.

Muqobil usul: a_{10}=32 ni topib 10(5+32)/2 qilish mumkin.

⚠️ n-1=9 ni unutish off-by-one xato.

oson $a_1=40,d=-4,n=8$. $S_8$ ni toping.

💡 Maslahat: Kamayuvchi AP uchun ham formula o‘zgarmaydi.

  1. a_8=40+7(-4)=12.
  2. S_8=8(40+12)/2.
  3. =4·52=208.

✅ Javob: $208$

Nega bu usul ishlaydi: Negative d faqat endpointni kamaytiradi.

Muqobil usul: Difference form: 8[80-28]/2.

⚠️ d<0 bo‘lsa yig‘indi formulasi ishlamaydi degan fikr noto‘g‘ri.

oson $2+4+6+\cdots+100$ ni hisoblang.

💡 Maslahat: Bu birinchi 50 juft son.

  1. a_1=2,a_n=100,d=2.
  2. n=1+(100-2)/2=50.
  3. S=50(2+100)/2=25·102=2550.

✅ Javob: $2550$

Nega bu usul ishlaydi: Avval hadlar sonini aniqlab, endpoint formulani ishlatamiz.

Muqobil usul: $2(1+2+...+50)=2·1275$.

⚠️ 100 ni hadlar soni deb olish xato.

oson $1+3+5+\cdots+99$ ni hisoblang.

💡 Maslahat: 99=2n-1 dan n ni toping.

  1. 2n-1=99 → n=50.
  2. Birinchi n toq son yig‘indisi n².
  3. S=50²=2500.

✅ Javob: $2500$

Nega bu usul ishlaydi: Odd-number identity AP sum formulasidan keladi.

Muqobil usul: 50(1+99)/2=2500.

⚠️ 99 ta had emas; 50 ta toq son bor.

oson $1+2+3+\cdots+200$ ni hisoblang.

💡 Maslahat: Triangular number formulasidan foydalaning.

  1. S=200·201/2.
  2. 100·201=20100.

✅ Javob: $20100$

Nega bu usul ishlaydi: Natural sonlar AP: a_1=1,d=1.

Muqobil usul: Gauss pairing: 100 ta juft ×201.

⚠️ 200·200/2 deb yozish endpoint +1 ni yo‘qotadi.

oson $a_1=7,a_{15}=49$. $S_{15}$ ni toping.

💡 Maslahat: Endpoint formula eng qisqa.

  1. S_{15}=15(7+49)/2.
  2. =15·56/2=15·28.
  3. =420.

✅ Javob: $420$

Nega bu usul ishlaydi: d ni topish shart emas.

Muqobil usul: d=3 dan difference form ham ishlaydi.

⚠️ Keraksiz parametr topish hisobni cho‘zadi.

oson 9 ta hadli AP ning markaziy hadi $a_5=17$. $S_9$ ni toping.

💡 Maslahat: n toq: S_n=n·middle term.

  1. n=9 va middle index (9+1)/2=5.
  2. S_9=9a_5.
  3. =9·17=153.

✅ Javob: $153$

Nega bu usul ishlaydi: Simmetrik juftlar average’i middle termga teng.

Muqobil usul: Endpointlarni bilsak ham shu natija keladi.

⚠️ d yoki a_1 ni topish shart emas.

oson 10 ta hadli AP da $a_5=14$ va $a_6=18$. $S_{10}$ ni toping.

💡 Maslahat: Juft n uchun markaziy juft formulasi.

  1. n/2=5.
  2. S_{10}=5(a_5+a_6).
  3. =5(14+18)=160.

✅ Javob: $160$

Nega bu usul ishlaydi: Har symmetric pair 32 ga teng va 5 ta pair bor.

Muqobil usul: d=4 ni topib endpointlarni tiklash mumkin.

⚠️ Markaziy ikki hadning o‘rtachasi AP average’i.

ortacha $a_1=4,d=5$ bo‘lgan AP da $S_n=329$ bo‘lsa n ni toping.

💡 Maslahat: S_n formulasidan quadratic tenglama tuzing.

  1. 329=n[8+5(n-1)]/2.
  2. 658=n(5n+3).
  3. 5n²+3n-658=0.
  4. Diskriminant 9+13160=13169=?? Bu perfect square emasligini ko‘ramiz; berilgan total mos integer n bermaydi.
  5. n musbat butun bo‘lishi kerak, demak bunday n yo‘q.

✅ Javob: Musbat butun $n$ mavjud emas

Nega bu usul ishlaydi: Partial sums discrete; arbitrary total AP partial sum bo‘lavermaydi.

Muqobil usul: Yaqin S_11=319, S_12=378 ni tekshirish mumkin.

⚠️ Quadratic real ildizni yaxlitlab olish mumkin emas.

ortacha $a_1=2,d=3$ bo‘lgan AP da $S_n=155$ bo‘lsa n ni toping.

💡 Maslahat: Quadratic tenglamani factorlang.

  1. 155=n[4+3(n-1)]/2.
  2. 310=n(3n+1).
  3. 3n²+n-310=0.
  4. (3n+31)(n-10)=0.
  5. Musbat integer root n=10.

✅ Javob: $n=10$

Nega bu usul ishlaydi: Partial-sum equation exact integer root beradi.

Muqobil usul: S_10=10(2+29)/2=155 bilan tekshiring.

⚠️ Negative algebraic root -31/3 qabul qilinmaydi.

ortacha $S_{12}=300$ va $d=4$. $a_1$ ni toping.

💡 Maslahat: Unknown-first-term formulasidan foydalaning.

  1. a_1=S_n/n-(n-1)d/2.
  2. =300/12-11·4/2.
  3. =25-22=3.

✅ Javob: $a_1=3$

Nega bu usul ishlaydi: Average term 25; first va last endpoint midpointi 25.

Muqobil usul: Difference formni a_1 uchun yechish mumkin.

⚠️ 300/12 ni a_1 deb olish d=0 holatidagina to‘g‘ri.

ortacha $S_{20}=860$ va $a_1=5$. d ni toping.

💡 Maslahat: Unknown-d formuladan foydalaning.

  1. 2S_n/n=2·860/20=86.
  2. 86=2a_1+(n-1)d=10+19d.
  3. 19d=76.
  4. d=4.

✅ Javob: $d=4$

Nega bu usul ishlaydi: Endpoint average total/n bilan bog‘langan.

Muqobil usul: a_{20}=2S/n-a_1=86-5=81; d=(81-5)/19=4.

⚠️ n emas, n-1 ga bo‘linadi.

ortacha $S_n=3n^2+2n$. $a_n$ formulasini va d ni toping.

💡 Maslahat: a_n=S_n-S_{n-1}.

  1. S_{n-1}=3(n-1)²+2(n-1).
  2. a_n=3n²+2n-[3(n-1)²+2(n-1)].
  3. =6n-1.
  4. d=6.

✅ Javob: $a_n=6n-1,\ d=6$

Nega bu usul ishlaydi: Quadratic partial sumning discrete difference’i affine.

Muqobil usul: A=3,B=2 formula bo‘yicha a_n=2An+(B-A)=6n-1.

⚠️ S_n formulasidagi n² koeffitsiyentni bevosita d deb olish xato; d=2A.

ortacha $S_n=2n^2-5n$. $a_1$ va d ni toping.

💡 Maslahat: Quadratic partial-sum coefficientlaridan foydalaning.

  1. A=2,B=-5.
  2. a_1=A+B=-3.
  3. d=2A=4.
  4. Tekshiruv: a_n=4n-7, a_1=-3.

✅ Javob: $a_1=-3,\ d=4$

Nega bu usul ishlaydi: S_1=a_1 va curvature 2A common difference beradi.

Muqobil usul: a_n=S_n-S_{n-1} ni to‘liq hisoblash mumkin.

⚠️ B=-5 ni a_1 deb olish noto‘g‘ri.

ortacha $S_{15}=420$ va $S_{14}=371$. $a_{15}$ ni toping.

💡 Maslahat: Consecutive partial sums farqi.

  1. a_{15}=S_{15}-S_{14}.
  2. =420-371=49.

✅ Javob: $a_{15}=49$

Nega bu usul ishlaydi: Oxirgi qo‘shilgan had total farqidir.

Muqobil usul: Agar AP parametrlari ma’lum bo‘lsa nth term bilan ham topiladi.

⚠️ S_15/S_14 nisbatini ishlatish kerak emas.

ortacha $S_{20}=650$ va $S_{12}=258$. $a_{13}+a_{14}+\cdots+a_{20}$ ni toping.

💡 Maslahat: Partial-sum difference.

  1. Kerakli block indekslari 13 dan20 gacha.
  2. Yig‘indi S_{20}-S_{12}.
  3. 650-258=392.

✅ Javob: $392$

Nega bu usul ishlaydi: S_12 birinchi 12 hadni olib tashlaydi.

Muqobil usul: 8 ta hadni alohida topish shart emas.

⚠️ S_20-S_13 bo‘lsa a_13 ham olib tashlanadi.

ortacha $a_n=4n+1$. Birinchi 30 had yig‘indisini toping.

💡 Maslahat: Affine sequence: a_1 va a_30 ni toping yoki sigma ishlating.

  1. a_1=5.
  2. a_{30}=121.
  3. S_{30}=30(5+121)/2=15·126=1890.

✅ Javob: $1890$

Nega bu usul ishlaydi: Affine formula AP beradi; endpoint formula tez.

Muqobil usul: $4·30·31/2+30=1860+30=1890$.

⚠️ a_1=1 deb olish interceptni first term bilan adashtirish.

ortacha $\sum_{k=1}^{40}(3k-2)$ ni hisoblang.

💡 Maslahat: Sigma linearity yoki AP endpoint formula.

  1. a_1=1, a_{40}=118.
  2. S=40(1+118)/2.
  3. =20·119=2380.

✅ Javob: $2380$

Nega bu usul ishlaydi: 3k-2 affine termlar AP.

Muqobil usul: $3·40·41/2-2·40=2460-80$.

⚠️ Upper bound 40 hadlar soni, chunki lower bound 1.

ortacha $10+13+16+\cdots+100$ ni hisoblang.

💡 Maslahat: Avval n ni toping.

  1. d=3.
  2. 100=10+(n-1)3.
  3. 90=3(n-1) → n=31.
  4. S=31(10+100)/2=31·55=1705.

✅ Javob: $1705$

Nega bu usul ishlaydi: Endpoint va count aniqlangach sum bevosita.

Muqobil usul: Difference form ham bir xil natija beradi.

⚠️ 100 ni n deb olish xato.

ortacha AP da $a_4=11$ va $a_{10}=35$. $a_4+a_5+\cdots+a_{10}$ ni toping.

💡 Maslahat: Bu 4 dan10 gacha 7 ta hadli block.

  1. Blockning birinchi hadi 11, oxirgi hadi 35.
  2. Hadlar soni 10-4+1=7.
  3. Yig‘indi 7(11+35)/2.
  4. =7·23=161.

✅ Javob: $161$

Nega bu usul ishlaydi: Har consecutive AP blockning o‘zi ham AP.

Muqobil usul: $S_{10}-S_3$ orqali ham topish mumkin.

⚠️ Hadlar soni q-p emas, q-p+1.

ortacha AP da $a_5+a_{15}=60$. $a_6+a_7+\cdots+a_{14}$ ni toping.

💡 Maslahat: 5 va15 ning midpointi 10; block 6..14 markazi ham 10.

  1. a_5+a_15=2a_{10}=60, demak a_{10}=30.
  2. 6 dan14 gacha 9 ta had bor.
  3. Bu blockning markaziy hadi a_{10}=30.
  4. Yig‘indi 9·30=270.

✅ Javob: $270$

Nega bu usul ishlaydi: Odd-length AP block sum = count × middle term.

Muqobil usul: Endpoint a_6+a_14=2a_10=60; 9·60/2=270.

⚠️ 60 ni middle term deb olish ikki marta katta xato.

ortacha AP da $S_9=180$. $a_5$ ni toping.

💡 Maslahat: 9 toq; S_9=9a_5.

  1. a_5=S_9/9.
  2. =180/9=20.

✅ Javob: $a_5=20$

Nega bu usul ishlaydi: Finite AP average middle termga teng.

Muqobil usul: Endpoint average ham 20.

⚠️ Bu xossa faqat 5 middle index bo‘lgan 9-term block uchun.

ortacha AP da $S_{10}=250$. $a_5+a_6$ ni toping.

💡 Maslahat: S_10=5(a_5+a_6).

  1. 250=5(a_5+a_6).
  2. a_5+a_6=50.

✅ Javob: $50$

Nega bu usul ishlaydi: 10 ta had 5 ta symmetric pairga bo‘linadi.

Muqobil usul: Endpoint sum a_1+a_10 ham 50.

⚠️ 250/10=25 average; middle pair sum esa 50.

ortacha AP da $a_1=6$ va $S_{16}=696$. $a_{16}$ ni toping.

💡 Maslahat: Endpoint formula orqali a_n ni yeching.

  1. 696=16(6+a_{16})/2.
  2. 696=8(6+a_{16}).
  3. 87=6+a_{16}.
  4. a_{16}=81.

✅ Javob: $81$

Nega bu usul ishlaydi: Total/n average=43.5; endpointlar average’i shu.

Muqobil usul: a_16=2S/n-a_1=87-6.

⚠️ d ni topishga hojat yo‘q.

ortacha AP da $a_{20}=73$ va $S_{20}=800$. $a_1$ ni toping.

💡 Maslahat: Endpoint formulani a_1 uchun yeching.

  1. 800=20(a_1+73)/2.
  2. 80=a_1+73.
  3. a_1=7.

✅ Javob: $7$

Nega bu usul ishlaydi: Endpoint sum 2S/n ga teng.

Muqobil usul: Average 40, demak first endpoint 40-(73-40)=7.

⚠️ 800/20=40 ni a_1 deb olish xato.

murakkab AP da $S_n=4n^2+n$. $S_{20}-S_{10}$ ni toping va uni hadlar blocki sifatida talqin qiling.

💡 Maslahat: Formulaga n=20 va10 qo‘ying.

  1. S_{20}=4·400+20=1620.
  2. S_{10}=4·100+10=410.
  3. Farq 1210.
  4. Bu a_{11}+a_{12}+...+a_{20} yig‘indisi.

✅ Javob: $1210$

Nega bu usul ishlaydi: Partial-sum farqi indeks oralig‘idagi totalni beradi.

Muqobil usul: a_n=8n-3 ni topib 10 hadli blockni sum qilish mumkin.

⚠️ S_20-S_10 a_10 ni emas, a_11 dan boshlaydi.

murakkab $S_n=5n^2-2n$ bo‘lsa $a_{25}$ ni toping.

💡 Maslahat: Discrete difference formulasidan foydalaning.

  1. A=5,B=-2.
  2. a_n=2An+(B-A)=10n-7.
  3. a_{25}=250-7=243.

✅ Javob: $243$

Nega bu usul ishlaydi: Quadratic cumulative formula original AP ni affine qiladi.

Muqobil usul: $S_{25}-S_{24}$ ni bevosita hisoblash ham mumkin.

⚠️ a_n ni S_n formulasiga n=25 qo‘yib 3075 deb olish xato; u total.

murakkab AP da $a_1=3,d=2$. Qaysi n uchun $S_n=168$?

💡 Maslahat: Quadratic sum equation.

  1. 168=n[6+2(n-1)]/2.
  2. 168=n(n+2).
  3. n²+2n-168=0.
  4. (n-12)(n+14)=0.
  5. Musbat integer n=12.

✅ Javob: $n=12$

Nega bu usul ishlaydi: Discrete root filtering bilan totalga mos count topiladi.

Muqobil usul: S_12=12(3+25)/2=168.

⚠️ -14 algebraik root indeks bo‘la olmaydi.

murakkab AP da $a_1=8,d=5$. $S_n>1000$ bo‘ladigan eng kichik n ni toping.

💡 Maslahat: S_n quadratic inequality; yaqin integerlarni tekshiring.

  1. S_n=n[16+5(n-1)]/2=n(5n+11)/2.
  2. n(5n+11)>2000.
  3. n=18: 18·101=1818<2000.
  4. n=19: 19·106=2014>2000.
  5. Eng kichik n=19.

✅ Javob: $19$

Nega bu usul ishlaydi: Partial sums increasing, chunki hadlar musbat; threshold integer tekshiruv bilan topiladi.

Muqobil usul: Quadratic positive root atrofidagi integersni tekshirish mumkin.

⚠️ Real thresholdni ceiling qilishdan oldin monotonicity va exact check qiling.

ortacha Teatrning 1-qatorida 20 o‘rin, har keyingi qatorda 3 tadan ko‘p. 25 qatorning jami o‘rindiqlar sonini toping.

💡 Maslahat: Qatorlardagi o‘rinlar AP; jami series.

  1. a_1=20,d=3,n=25.
  2. a_{25}=20+24·3=92.
  3. S_{25}=25(20+92)/2.
  4. =25·56=1400.

✅ Javob: $1400$

Nega bu usul ishlaydi: Har qator individual term, barcha qatorlar jami partial sum.

Muqobil usul: Difference form bilan bevosita 25[40+72]/2.

⚠️ Faqat a_25=92 ni javob qilish nth-term/sum chalkashligi.

ortacha Sportchi 1-kuni 12 ta mashq qiladi va har kuni 2 tadan ko‘paytiradi. 30 kunda jami nechta mashq?

💡 Maslahat: Daily counts AP.

  1. a_1=12,d=2,n=30.
  2. a_{30}=12+29·2=70.
  3. S_{30}=30(12+70)/2=15·82=1230.

✅ Javob: $1230$

Nega bu usul ishlaydi: Doimiy absolute increment arithmetic series beradi.

⚠️ “30-kuni nechta?” bilan “30 kunda jami?”ni ajrating.

ortacha Ishlab chiqarish 1-kun 100 dona, har kuni 15 donaga oshadi. 14 kunlik jami ishlab chiqarish?

💡 Maslahat: AP sum.

  1. a_1=100,d=15,n=14.
  2. a_{14}=100+13·15=295.
  3. S_{14}=14(100+295)/2.
  4. =7·395=2765.

✅ Javob: $2765$

Nega bu usul ishlaydi: Daily output AP, cumulative output series.

Muqobil usul: Difference-form bevosita ishlaydi.

⚠️ 14·295 qilish barcha kunlar 295 degani bo‘lib qoladi.

ortacha O‘quvchi 1-hafta 40 ta masala, har hafta 5 tadan ko‘proq ishlaydi. 12 haftada jami nechta?

💡 Maslahat: Week counts AP.

  1. a_1=40,d=5,n=12.
  2. a_{12}=95.
  3. S_{12}=12(40+95)/2=6·135=810.

✅ Javob: $810$

Nega bu usul ishlaydi: Har hafta bir xil +5.

⚠️ 12-haftadagi 95 va 12 haftalik 810 boshqa miqdorlar.

ortacha Bir bino zinapoyasining 1-qatorida 4 ta bezak, har keyingi qatorda 2 tadan ko‘p. 18 qatorga jami nechta bezak?

💡 Maslahat: AP sum.

  1. a_1=4,d=2,n=18.
  2. a_{18}=38.
  3. S_{18}=18(4+38)/2=9·42=378.

✅ Javob: $378$

Nega bu usul ishlaydi: Row pattern AP.

⚠️ 18·38 maksimum qatorni barcha qatorlarga ko‘paytirish xato.

murakkab Jamg‘arma rejasi: 1-oy 200 ming so‘m, har keyingi oy oldingidan 50 ming so‘m ko‘proq. 1 yil davomida jami qancha?

💡 Maslahat: 12 oy AP sum.

  1. a_1=200,d=50,n=12 (ming so‘m).
  2. a_{12}=200+11·50=750.
  3. S_{12}=12(200+750)/2.
  4. =6·950=5700 ming so‘m.

✅ Javob: $5\,700\,000$ so‘m

Nega bu usul ishlaydi: Oylik badallar linear o‘sadi; total arithmetic series.

⚠️ Birlikni ming so‘mdan so‘mga qaytarishni unutmang.

murakkab Bir korxona 1-oy 500 birlik ishlab chiqaradi, har oy 20 birlik kamayadi. 12 oy jami ishlab chiqarish?

💡 Maslahat: Kamayuvchi AP sum.

  1. a_1=500,d=-20,n=12.
  2. a_{12}=500-220=280.
  3. S_{12}=12(500+280)/2.
  4. =6·780=4680.

✅ Javob: $4680$

Nega bu usul ishlaydi: Negative difference bilan finite sum normal ishlaydi.

⚠️ Oxirgi had musbat ekanini kontekst uchun tekshirish yaxshi.

murakkab AP da $a_1=10,d=4$. 8-haddan 20-hadgacha yig‘indini toping.

💡 Maslahat: Block endpoint yoki partial-sum difference.

  1. a_8=10+7·4=38.
  2. a_{20}=10+19·4=86.
  3. Hadlar soni 20-8+1=13.
  4. Block sum=13(38+86)/2=13·62=806.

✅ Javob: $806$

Nega bu usul ishlaydi: Consecutive sub-blockning o‘zi AP.

Muqobil usul: $S_{20}-S_7$ bilan ham topiladi.

⚠️ S_20-S_8 emas; u 9-haddan boshlaydi.

murakkab $\sum_{k=7}^{25}(2k+3)$ ni hisoblang.

💡 Maslahat: Lower bound 7; terms count 19.

  1. Birinchi term k=7: 17.
  2. Oxirgi term k=25: 53.
  3. Hadlar soni 25-7+1=19.
  4. Sum=19(17+53)/2=19·35=665.

✅ Javob: $665$

Nega bu usul ishlaydi: Affine summand consecutive k lar bo‘yicha AP.

Muqobil usul: [S_25 for 2k+3]-[S_6] orqali ham.

⚠️ 25-7=18 emas, inclusive count 19.

murakkab AP da $S_5=45$ va $S_{10}=165$. $S_{15}$ ni toping.

💡 Maslahat: Avval block averages yoki a_1,d ni tiklang.

  1. S_5=5[2a_1+4d]/2 → 2a_1+4d=18 → a_1+2d=9.
  2. S_{10}=5[2a_1+9d]=165 → 2a_1+9d=33.
  3. Birinchi tenglamani 2 ga ko‘paytirib 2a_1+4d=18.
  4. Ayirib 5d=15 → d=3.
  5. a_1+6=9 → a_1=3.
  6. S_{15}=15[6+14·3]/2=15·48/2=360.

✅ Javob: $360$

Nega bu usul ishlaydi: Ikki partial sum AP parametrlarini aniqlaydi.

Muqobil usul: Quadratic S_n=An²+Bn coefficientlarini S_5,S_10 dan topish mumkin.

⚠️ S_15 ni S_10+S_5 deb olish noto‘g‘ri; keyingi 5 had first 5 hadga teng emas.

murakkab AP da $S_n=n^2+4n$. $a_{30}+a_{31}$ ni toping.

💡 Maslahat: Avval a_n ni discrete difference orqali toping.

  1. A=1,B=4, shuning uchun a_n=2n+(4-1)=2n+3.
  2. a_{30}=63.
  3. a_{31}=65.
  4. Yig‘indi 128.

✅ Javob: $128$

Nega bu usul ishlaydi: Quadratic partial sum affine term sequence beradi.

Muqobil usul: a_{30}+a_{31}=S_{31}-S_{29} deb ham topish mumkin.

⚠️ S_30+S_31 ni olish kerak emas.

murakkab AP ning birinchi 21 hadi yig‘indisi 1050. $a_{11}$ ni toping.

💡 Maslahat: 21 toq; middle term average.

  1. S_{21}=21a_{11}.
  2. a_{11}=1050/21=50.

✅ Javob: $50$

Nega bu usul ishlaydi: Odd-length AP total count × middle term.

Muqobil usul: Endpoint sum a_1+a_21=100 ham keladi.

⚠️ d yoki endpointlarni topishga ma’lumot yetmasa ham middle term aniq.

murakkab AP ning birinchi 20 hadi yig‘indisi 760. $a_{10}+a_{11}$ ni toping.

💡 Maslahat: Even-middle pair formula.

  1. S_{20}=10(a_{10}+a_{11}).
  2. 760=10(a_{10}+a_{11}).
  3. a_{10}+a_{11}=76.

✅ Javob: $76$

Nega bu usul ishlaydi: 20 had 10 symmetric pairdan iborat.

⚠️ 760/20=38 average, middle pair sum 76.

murakkab AP da $a_3+a_8=42$. Birinchi 10 had yig‘indisini toping.

💡 Maslahat: Indeks yig‘indisi 11; a_3+a_8 endpoint pair a_1+a_10 ga teng.

  1. 3+8=11=1+10.
  2. AP pair-sum invariant bo‘yicha a_1+a_{10}=a_3+a_8=42.
  3. S_{10}=10·42/2=210.

✅ Javob: $210$

Nega bu usul ishlaydi: Teng indeks yig‘indili juftlar hadlar yig‘indisi teng.

Muqobil usul: a_3+a_8=2a_{5.5} formal average g‘oyasi bilan ham tushuniladi.

⚠️ a_3 va a_8 ni alohida aniqlash shart emas.

murakkab AP da $S_{12}=216$ va $a_1=4$. $S_6$ ni toping.

💡 Maslahat: Avval d ni toping, keyin S_6.

  1. 216=12[8+11d]/2=6(8+11d).
  2. 36=8+11d → 11d=28 → d=28/11.
  3. S_6=6[8+5·28/11]/2.
  4. =3(88/11+140/11)=3·228/11=684/11.

✅ Javob: $\frac{684}{11}$

Nega bu usul ishlaydi: Partial sums arbitrary integer bo‘lishi shart emas; AP rational d ga ega bo‘lishi mumkin.

Muqobil usul: a_12=2S_12/12-a_1=36-4=32; d=(32-4)/11.

⚠️ 216 ning yarmi S_6 emas; second half terms kattaroq.

Umumiy xatolar

❌ a_n bilan S_n ni bir xil deb olish

a_n bitta had, S_n esa birinchi n hadning jami.

✅ Savolda “n-had”mi yoki “birinchi n had yig‘indisi”mi aniqlang.

a_10=32 bo‘lishi S_10=32 degani emas.

❌ S_n=a_1+nd formulasini ishlatish

Bu yig‘indi formula emas va nth-term formula ham off-by-one.

✅ S_n=n(a_1+a_n)/2 yoki n[2a_1+(n-1)d]/2.

a_1=2,d=3,n=10.

❌ Hadlar sonini oxirgi had qiymati bilan adashtirish

n indeks/count, a_n esa qiymat.

✅ Avval n ni nth-term tenglamadan toping.

2+4+...+100 da n=50, 100 emas.

❌ Endpoint formulada 2 ga bo‘lishni unutish

n(a_1+a_n) ikki marta yig‘indi beradi.

✅ S_n=n(a_1+a_n)/2.

1+...+100=5050, 10100 emas.

❌ n-1 o‘rniga n d ishlatish

Birinchi haddan n-hadgacha n-1 qadam.

✅ a_n=a_1+(n-1)d va sum difference formda ham n-1 ishlating.

a_10= a_1+9d.

❌ Kamayuvchi AP uchun sum formulasi ishlamaydi deb o‘ylash

Formula d ishorasidan mustaqil.

✅ Negative d ni aynan formulaga qo‘ying.

40,36,...,12 yig‘indisi normal hisoblanadi.

❌ S_q-S_p bilan p dan q gacha block olish

S_p ichida a_p ham bor, shuning uchun u ham ayiriladi.

✅ a_p+...+a_q=S_q-S_{p-1}.

a_5+...+a_12=S_12-S_4.

❌ S_n-S_{n-2}=a_n deb olish

Ikki qadam farqda ikki had qoladi.

✅ a_n=S_n-S_{n-1}.

S_8-S_6=a_7+a_8.

❌ Quadratic S_n ni a_n deb o‘qish

S_n cumulative total.

✅ a_n=S_n-S_{n-1}.

S_n=3n²+2n → a_n=6n-1.

❌ S_n=An²+Bn da d=A deb olish

Discrete difference slope 2A.

✅ d=2A.

A=3 → d=6.

❌ Unknown n uchun real rootni yaxlitlash

n discrete musbat butun indeks.

✅ Quadratic rootsdan faqat musbat integer rootni qabul qiling.

n≈10.4 bo‘lsa “10” yoki “11” deb taxmin qilinmaydi.

❌ Toq n markaziy had formulani juft n ga qo‘llash

Juft n da yagona middle term yo‘q.

✅ Juft n uchun S_n=(n/2)(a_{n/2}+a_{n/2+1}).

S_10=5(a_5+a_6).

❌ Juft n da middle pair average’ini total deb olish

Middle pair sum faqat bitta symmetric-pair sum.

✅ S_n=(n/2)×middle-pair sum.

a_5+a_6=32 bo‘lsa S_10=160.

❌ Sigma upper-lower differenceini hadlar soni deb olish

Inclusive bounds sabab +1 kerak.

✅ count=q-p+1.

k=7..25 → 19 had.

❌ Real-life “jami” savolida faqat oxirgi bosqich qiymatini topish

a_n final period, S_n cumulative total.

✅ Final termni topgach sum formulasiga o‘ting.

30-kun 70 mashq, 30 kun jami 1230.

❌ Doimiy foizli o‘sishni arithmetic series deb olish

Doimiy foiz multiplicative pattern, AP emas.

✅ AP faqat absolute increment constant bo‘lsa.

Har oy +50 ming → AP; +5% → GP.

❌ d=0 holatda unknown-d formulasini 0 ga bo‘lib ishlatish

Ba’zi rearranged formulalarda n-1 yoki d denominator bo‘ladi.

✅ Maxsus holatni alohida tekshiring.

Constant AP: S_n=na_1.

❌ S_n ni S_p+S_q kabi chiziqli indeks bo‘yicha qo‘shish

Partial sum n bo‘yicha odatda quadratic; S_{p+q}≠S_p+S_q.

✅ Kerakli blockni partial-sum difference bilan yozing.

S_15≠S_10+S_5 umumiy holda.

❌ Pairingda tasodifiy hadlarni juftlash

Teng pair sum faqat simmetrik indekslarda guaranteed.

✅ Indekslar yig‘indisi n+1 bo‘lgan juftlarni tanlang.

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

❌ Birliklarni yo‘qotish

Real-life total birlik bilan javob berilishi kerak.

✅ Hisob davomida ming so‘m/dona/o‘rin kabi birlikni saqlang.

5700 ming so‘m = 5 700 000 so‘m.

Noto'g'ri tasavvurlar

Arifmetik progressiya yig‘indisi faqat musbat hadlarda ishlaydi.

Formula manfiy va aralash ishorali hadlarda ham algebraik yig‘indi uchun ishlaydi.

S_n va a_n faqat notation farqi, qiymat bir xil.

S_n cumulative total, a_n individual term; faqat n=1 da S_1=a_1.

Yig‘indi formulasini ishlatish uchun d albatta ma’lum bo‘lishi kerak.

a_1,a_n,n ma’lum bo‘lsa d siz endpoint formula yetarli.

Yig‘indi formulasini ishlatish uchun a_n albatta topilishi kerak.

a_1,d,n ma’lum bo‘lsa difference form bevosita ishlaydi.

Odd n da markaziy had yig‘indiga qo‘shimcha ravishda alohida qo‘shiladi.

S_n=n·middle term formulasi markaziy hadni allaqachon to‘liq hisobga oladi.

Partial sums ham arifmetik progressiya bo‘ladi.

AP d≠0 bo‘lsa S_n quadratic va consecutive differences a_n o‘zgaradi; odatda AP emas.

Sigma belgisi yangi matematik operatsiya bo‘lib, oddiy qo‘shishdan farq qiladi.

Sigma faqat indekslangan ko‘p hadli yig‘indining ixcham yozuvidir.

Birinchi n toq son yig‘indisi n(n+1)/2.

Bu natural sonlar yig‘indisi; birinchi n toq son yig‘indisi n².

S_n quadratic bo‘lsa original ketma-ketlik ham quadratic.

a_n=S_n-S_{n-1}; quadratic partial sumdan affine/arithmetic terms chiqadi.

Cheksiz arifmetik progressiyaning ham shu formula bilan finite totalini topish mumkin.

Bu formulalar finite n uchun. d≠0 AP ning infinite seriesi odatda konvergent emas.

Amaliy qo'llanilishi

Teatr va stadion

Qatorlardagi o‘rindiqlar doimiy miqdorda oshsa, barcha qatorlardagi jami o‘rindiqlar arithmetic series bilan topiladi.

Ishlab chiqarish

Kunlik ishlab chiqarish har kuni bir xil dona miqdoriga oshsa yoki kamaysa, davr bo‘yicha jami ishlab chiqarish S_n bilan hisoblanadi.

Ta’lim va mashq rejasi

Har hafta bajariladigan masalalar yoki takrorlar soni doimiy oshirilsa, umumiy workload AP sumidir.

Jamg‘arma rejasi

Oylik badal har oy bir xil absolute summa bilan oshirilsa, davr oxirigacha kiritilgan jami mablag‘ arithmetic series.

Qurilish va dizayn

Pog‘onalar, qatorlar yoki parallel elementlar uzunligi/soni teng qadam bilan o‘zgarsa, jami material miqdori AP sum formulasiga keladi.

Dasturlash

Loopda har iteratsiyadagi workload chiziqli oshsa, total operation count arithmetic-series formulasi bilan baholanadi.

Ma’lumotlar tahlili

Diskret chiziqli trendning cumulative total qiymati quadratic partial-sum model beradi.

Algoritmik optimizatsiya

1+2+...+n yoki pk+q tipidagi loop xarajatlarini yopiq formula bilan O(1) hisoblash mumkin.

Arifmetik progressiya yig‘indisi: juftlash va cumulative workflow

AP yig‘indisi — symmetry → average → totala₁, a₂, … , aₙ₋₁, aₙa₁+aₙ = a₂+aₙ₋₁ = …Sₙ = n(a₁+aₙ)/2KNOWN ENDPOINTSa₁, aₙ, n→ endpoint formulaaverage=(a₁+aₙ)/2KNOWN DIFFERENCEa₁, d, n→ 2a₁+(n−1)d→ SₙPARTIAL SUMSaₙ=Sₙ−Sₙ₋₁p..q = S_q−S_{p−1}quadratic S → linear aFINAL QAaₙ ≠ Sₙ • count inclusive • n∈N • units • correct partial-sum boundary

Arifmetik series masalalarida avval berilgan ma’lumot turini tanlang: endpointlar, common difference yoki partial sums. Yakunda had/yig‘indi notationi va inclusive indeks chegaralarini tekshiring.

Xulosa

Cheat sheet: S_n=a_1+...+a_n. Asosiy formulalar: S_n=n(a_1+a_n)/2 va S_n=n[2a_1+(n-1)d]/2. AP hadlarining o‘rtacha qiymati (a_1+a_n)/2. Partial sums uchun a_n=S_n-S_{n-1}. a_p+...+a_q=S_q-S_{p-1}. n toq bo‘lsa S_n=n·a_{(n+1)/2}. Har doim a_n bilan S_n ni farqlang va n musbat butunligini tekshiring.

Keyingi “Geometrik progressiya” mavzusida doimiy ayirma o‘rniga doimiy nisbat paydo bo‘ladi. AP yig‘indisidagi juftlash g‘oyasidan farqli ravishda GP yig‘indisi ko‘paytirish-ayirish orqali olinadi.

Bog'liq mavzular

Oldin bilishingiz kerak: Arifmetik progressiya, Chiziqli tenglamalar

Bog'liq mavzular: Funksiya, Daraja va uning xossalari, darajali ifodalar

Keyingi mavzular: Geometrik progressiya, Matnli masalalar

Manbalar

Ro'yxatdan o'tib, mashq qilishni boshlang