Algebra

Geometrik progressiya

murakkab 185 daqiqa geometrik progressiyageometric sequencecommon ratioqa_ngeometric meanfinite geometric seriesinfinite geometric seriesexponential growthcompound growthpartial sumsequence

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Geometrik progressiya doimiy foizli o‘sish va kamayishning asosiy diskret modelidir. Murakkab foiz, aholi o‘sishi, radioaktiv kamayish, rekursiv algoritmlar, masshtab va iterativ jarayonlarda har qadam oldingisining ma’lum ko‘paytuvchisiga teng bo‘ladi. AP absolute o‘zgarishni, GP esa relative/multiplicative o‘zgarishni modellashtiradi.

O'quv maqsadlari

Geometrik progressiyani doimiy nisbat orqali aniqlash
q umumiy nisbatni topish
a_n=a_1q^(n-1) formulasini qo‘llash
Ikki had orqali q va a_1 ni tiklash
Anchor term orqali a_n=a_m q^(n-m) formulasini ishlatish
Musbat GP da geometrik o‘rta xossasini qo‘llash
a_k^2=a_{k-1}a_{k+1} xossasini ishlatish
Teng indeks yig‘indili hadlar ko‘paytmasi invariantini tushuntirish
q>1 o‘suvchi GP ni tahlil qilish
0<q<1 kamayuvchi GP ni tahlil qilish
q<0 da ishora almashishini tahlil qilish
q=0 va q=1 maxsus holatlarini ajratish
Chekli GP yig‘indisi formulasini qo‘llash
q=1 maxsus yig‘indini alohida ishlatish
Partial sums orqali hadni tiklash
Geometrik partial-sum strukturani tanish
|q|<1 cheksiz GP yig‘indisini tushuntirish
Cheksiz yig‘indidan parametrlarni tiklash
Doimiy foizni q ko‘paytuvchiga aylantirish
Foizli o‘sish va kamayish masalalarini GP bilan model qilish
AP va GP modellarini farqlash
Real-life exponential growth/decay masalalarini yechish

Geometrik progressiyada qo‘shni hadlar orasidagi ayirma emas, nisbat doimiy: a_{n+1}/a_n=q. Shuning uchun har qadamda oldingi had q ga ko‘payadi va n-had a_n=a_1q^{n-1} ko‘rinishda yoziladi. q>1 bo‘lsa magnitude odatda o‘sadi, 0<q<1 bo‘lsa kamayadi, q<0 bo‘lsa ishora navbatma-navbat almashadi. GP ning eng kuchli symmetry xossasi additive emas, multiplicative: teng masofadagi hadlar ko‘paytmasi bir xil. Masalan a_k^2=a_{k-1}a_{k+1}. Bu geometrik o‘rta va ko‘plab tezkor test usullarining asosi. Chekli yig‘indida S_n=a_1(1-q^n)/(1-q), q≠1. Agar |q|<1 bo‘lsa q^n→0 va cheksiz yig‘indi S_∞=a_1/(1-q) ga yaqinlashadi. Shu sabab geometric sequence va geometric series tushunchalarini, shuningdek individual a_n va cumulative S_n ni qat’iy ajratish kerak.

Ta'riflar

Geometrik progressiya · Geometric progression

Nol bo‘lmagan ketma-ket hadlar uchun a_{n+1}/a_n=q doimiy bo‘lgan ketma-ketlik.

Har yangi had oldingisini bir xil q songa ko‘paytirish bilan olinadi.

Misol: $2,6,18,54,...$

Bu emas: $2,5,8,11,...$ AP, GP emas.

💡 Nol hadlar maxsus holatlarda alohida ko‘riladi.

Umumiy nisbat · Common ratio

GP da q=a_{n+1}/a_n doimiy qiymat.

Ketma-ket had nechaga ko‘payayotganini bildiradi.

Misol: $3,12,48$ da q=4.

Bu emas: $3,6,10$ da nisbat doimiy emas.

💡 a_n≠0 bo‘lgan standart ta’rifda.

Birinchi had · First term

GP ning boshlang‘ich hadi a_1.

Barcha hadlar shu qiymatdan q powers bilan olinadi.

Misol: $a_1=5$

Bu emas: q first term emas.

💡 Initial value sifatida talqin qilinadi.

n-had · nth term

GP ning n-indeksdagi hadi a_n=a_1q^{n-1}.

Birinchi haddan n-hadgacha n-1 marta q ga ko‘payamiz.

Misol: $a_5=2·3^4=162$

Bu emas: $a_n=a_1q^n$ off-by-one.

💡 n≥1.

Anchor formula · Tayanch had formulasi

a_n=a_mq^{n-m}.

Istalgan ma’lum hadni yangi boshlanish nuqtasi sifatida olish mumkin.

Misol: $a_{10}=a_7q^3$

Bu emas: $a_{10}=a_7q^{10}$ emas.

💡 Index difference ishlatiladi.

Geometrik o‘rta · Geometric mean

Musbat x,y uchun sqrt(xy); GP da o‘rta had qo‘shnilarining geometrik o‘rtasidir.

O‘rta hadning kvadrati yon hadlar ko‘paytmasiga teng.

Misol: $4,12,36$ da 12=sqrt(4·36).

Bu emas: Arithmetic mean (4+36)/2=20 boshqa tushuncha.

💡 Musbat qiymatlarda principal mean.

Geometrik qator · Geometric series

GP hadlarini qo‘shishdan hosil bo‘lgan yig‘indi.

Sequence hadlar ro‘yxati, series esa ularning yig‘indisi.

Misol: $1+2+4+8$

Bu emas: $1,2,4,8$ sequence.

💡 Finite va infinite turlari bor.

Chekli geometrik yig‘indi · Finite geometric sum

Birinchi n GP hadining yig‘indisi S_n.

Ma’lum n gacha jami qiymat.

Misol: $S_4=1+2+4+8=15$

Bu emas: $a_4=8$ total emas.

💡 q=1 alohida formula.

Cheksiz geometrik qator · Infinite geometric series

n cheksiz o‘sadigan GP hadlari yig‘indisining limiti, agar mavjud bo‘lsa.

Faqat hadlar yetarlicha tez 0 ga yaqinlashsa finite limit paydo bo‘ladi.

Misol: $1+1/2+1/4+...=2$

Bu emas: $1+2+4+...$ finite limitga ega emas.

💡 Real GP uchun |q|<1 convergence sharti.

Konvergentsiya · Convergence

Partial sums ma’lum finite limitga yaqinlashishi.

Yig‘indi bitta son tomon boradi.

Misol: $S_n→2$

Bu emas: S_n→∞ convergence emas.

💡 Infinite series uchun.

Divergensiya · Divergence

Partial sums finite limitga ega emasligi.

Yig‘indi barqaror finite qiymatga bormaydi.

Misol: $1+2+4+...$

Bu emas: 1+1/2+... convergent.

💡 |q|≥1 GP series odatda divergent.

O‘sish koeffitsiyenti · Growth factor

Har bosqichdagi multiplicative factor q>1.

Masalan +20% har safar ×1.2.

Misol: $q=1.2$

Bu emas: $d=0.2$ AP ayirma emas.

💡 Foizni decimal multiplierga aylantiring.

Kamayish koeffitsiyenti · Decay factor

0<q<1 bo‘lgan multiplicative factor.

Masalan 15% kamayish → ×0.85.

Misol: $q=0.85$

Bu emas: $q=-0.15$ noto‘g‘ri.

💡 Remaining fraction ishlatiladi.

Alternating GP · Ishora almashuvchi GP

q<0 bo‘lib hadlar ishorasi navbatma-navbat almashadigan GP.

Har qadam manfiy q ga ko‘paygani uchun sign almashadi.

Misol: $2,-6,18,-54,...$

Bu emas: $2,-6,-18,...$ nisbat doimiy emas.

💡 Magnitude |q| bo‘yicha o‘sishi yoki kamayishi mumkin.

Partial sum · Qisman yig‘indi

S_n=a_1+...+a_n.

Birinchi n hadgacha cumulative total.

Misol: $S_3=2+6+18=26$

Bu emas: $a_3=18$ individual term.

💡 Har sequence uchun umumiy tushuncha.

Compound growth · Murakkab o‘sish

Har davr oxirida oldingi qiymatga foiz qo‘llanadigan multiplicative model.

Foiz keyingi davrda yangi bazaga qo‘llanadi.

Misol: $P_n=P_0(1+r)^n$

Bu emas: $P_0+nrP_0$ simple linear model.

💡 Finance/population contexts.

Half-life style decay · Ulushli kamayish

Har bosqichda qiymatning doimiy ulushi qoladigan model.

Masalan har bosqichda yarmi qoladi → q=1/2.

Misol: $100,50,25,12.5,...$

Bu emas: $100,50,0,-50$ AP.

💡 Radioactive-like discrete model.

q=1 maxsus holat · Constant GP

Har had a_1 ga teng bo‘ladigan GP.

Har qadam ×1 qiymatni o‘zgartirmaydi.

Misol: $5,5,5,...$

Bu emas: $5,10,15$ q=1 emas.

💡 Finite sum S_n=na_1; fraction formula denominator 0 bo‘ladi.

q=0 maxsus holat

Recurrence a_{n+1}=qa_n nuqtai nazarida a_1 dan keyingi barcha hadlar 0.

Birinchi ko‘paytirishdayoq qiymat 0 bo‘ladi.

Misol: $7,0,0,0,...$

Bu emas: $7,0,1,0,...$ emas.

💡 Standard ratio keyingi 0/0 sabab alohida talqin qilinadi.

Foizdan q ga o‘tish · Percent-to-ratio

r foiz o‘sish uchun q=1+r, kamayish uchun q=1-r (r decimal).

Relative change multiplicative factor bilan ifodalanadi.

Misol: 8% o‘sish → q=1.08.

Bu emas: 8% o‘sish → q=0.08 emas.

💡 r=8%=0.08.

Fundamental tushunchalar

Multiplicative invariant

GP ni aniqlaydigan asosiy belgi consecutive ratio constant bo‘lishidir.

$a_{n+1}=q a_n$

q fixed.

Exponent counts transitions

a_1 dan a_n gacha n-1 transition bor.

$a_n=a_1q^{n-1}$

Off-by-one nazorat.

Anchor flexibility

Birinchi had shart emas; istalgan a_m dan a_n ga index difference orqali o‘tamiz.

$a_n=a_mq^{n-m}$

n>m yoki n<m bo‘lishi mumkin, q≠0 kerak bo‘lishi mumkin.

Multiplicative symmetry

Teng indeks yig‘indili hadlar ko‘paytmasi teng.

$a_i a_j=a_r a_s\quad(i+j=r+s)$

Explicit powers bir xil exponent sum beradi.

Three-term geometric mean

Uch consecutive GP had uchun middle squared = outer product.

$a_k^2=a_{k-1}a_{k+1}$

Real sign caveat; positive GP da middle positive geometric mean.

q magnitude controls size

|q| growth/decay magnitude’ni, q signi esa sign patternni boshqaradi.

$|q|$

|q|>1 magnitude grows; 0<|q|<1 shrinks.

Positive growth

q>1 va positive a_1 bo‘lsa hadlar o‘sadi.

$q>1$

Monotone positive growth.

Positive decay

0<q<1 va positive a_1 bo‘lsa hadlar kamayib 0 ga yaqinlashadi.

Positive monotone decay.

Alternating behavior

q<0 ishorani har qadam almashtiradi.

$(-1)^{n-1}|q|^{n-1}$

Magnitude |q| bo‘yicha.

Finite sum multiplier trick

S_n ni q ga ko‘paytirib originaldan ayirish middle termsni bekor qiladi.

$qS_n-S_n=a_1(q^n-1)$

q≠1.

Two equivalent finite-sum forms

Sign choice denominator bilan mos ravishda ikkita standard form bir xil.

$S_n=a_1\frac{1-q^n}{1-q}=a_1\frac{q^n-1}{q-1}$

q≠1.

q=1 branch

General fraction formula 0/0 beradi, ammo sequence constant.

$q=1\Rightarrow S_n=na_1$

Maxsus holat alohida.

Infinite-sum convergence gate

Geometric infinite sum finite bo‘lishi uchun va yetarli shart |q|<1.

$|q|<1$

Then q^n→0.

Infinite sum as finite-sum limit

Finite formula n→∞ limitida q^n yo‘qoladi.

$S_\infty=\frac{a_1}{1-q}$

|q|<1.

Partial sum recovers term

Har seriesda a_n=S_n-S_{n-1}.

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

GP uchun ham universal.

Partial sums of GP are affine in q^n

q≠1 da S_n=C(1-q^n) ko‘rinishida.

$S_n=\frac{a_1}{1-q}(1-q^n)$

Exponential remainder.

Remainder of convergent GP

Infinite totaldan first n sumni ayirsak tail ham geometric.

$R_n=S_\infty-S_n=\frac{a_1q^n}{1-q}$

|q|<1.

Percent growth is geometric

Constant percent change additive d emas, constant q yaratadi.

$q=1\pm r$

r decimal.

Doubling/halving time discrete

Har fixed periodda ×2 yoki ×1/2 bo‘lsa GP.

$q=2\text{ or }1/2$

Period index discrete.

AP vs GP diagnostic

AP: consecutive differences constant. GP: consecutive nonzero ratios constant.

Bir sequence ikkala bo‘lishi mumkin faqat constant nonzero sequence kabi special casesda.

Logarithmic inversion preview

a_n/a_1=q^{n-1} da n unknown bo‘lsa logarithm keyingi mavzularda natural vosita.

$n-1=\log_q(a_n/a_1)$

q>0,q≠1, ratio positive.

Model-domain check

Real-life modelda negative q, negative terms yoki fractional counts fizik ma’noga mos kelmasligi mumkin.

Algebraic GP va context restrictionsni ajrating.

Formula kutubxonasi

GP recurrence

$$a_{n+1}=qa_n$$

Har keyingi had oldingisining q baravari.

Shart: Standard GP recurrence

Xususiy holatlar: q=1 constant; q<0 alternating.

n-had formulasi

$$a_n=a_1q^{n-1}$$

Birinchi haddan n-hadga n-1 ko‘paytirish.

Shart: n≥1

Xususiy holatlar: q=0 recurrence maxsus talqin.

Anchor formula

$$a_n=a_mq^{n-m}$$

Istalgan ma’lum had orqali boshqa hadni topadi.

Shart: Relevant powers defined

Xususiy holatlar: n<m bo‘lsa negative exponent, q≠0.

Nisbatni ikki consecutive haddan

$$q=\frac{a_{n+1}}{a_n}$$

Common ratio hisoblanadi.

Shart: a_n≠0

Xususiy holatlar: Signs q ni aniqlaydi.

Nisbatni uzoq hadlardan

$$q^{n-m}=\frac{a_n}{a_m}$$

Index gap orqali q powerini topadi.

Shart: a_m≠0

Xususiy holatlar: Real q uchun parity bir nechta sign variant berishi mumkin.

Uch had invariant

$$a_k^2=a_{k-1}a_{k+1}$$

Middle term squared outer productga teng.

Shart: Consecutive GP terms

Xususiy holatlar: Positive GP da a_k=sqrt(product).

General product symmetry

$$a_i a_j=a_r a_s$$

Teng indeks yig‘indili hadlar producti teng.

Shart: i+j=r+s

Xususiy holatlar: a_k²=a_{k-h}a_{k+h}.

Chekli GP yig‘indisi

$$S_n=a_1\frac{1-q^n}{1-q}$$

Finite geometric series sum.

Shart: q≠1

Xususiy holatlar: q=1 alohida.

Chekli GP yig‘indisi alternativ

$$S_n=a_1\frac{q^n-1}{q-1}$$

Oldingi formula bilan aynan equivalent.

Shart: q≠1

Xususiy holatlar: Same value.

q=1 sum

$$q=1\Longrightarrow S_n=na_1$$

Constant sequence n marta qo‘shiladi.

Shart: q=1

Xususiy holatlar: a_n=a_1.

Partial sum recurrence

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

Har yangi had totalga qo‘shiladi.

Shart: n≥1,S_0=0

Xususiy holatlar: Universal sequence formula.

Partial sumsdan had

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

Cumulative difference termni beradi.

Shart: n≥1

Xususiy holatlar: Universal.

Cheksiz GP yig‘indisi

$$S_\infty=\frac{a_1}{1-q}$$

Convergent infinite GP finite limitga ega.

Shart: |q|<1

Xususiy holatlar: |q|≥1 da qo‘llanmaydi.

Infinite tail

$$R_n=S_\infty-S_n=\frac{a_1q^n}{1-q}$$

First n haddan keyingi infinite remainder.

Shart: |q|<1

Xususiy holatlar: R_0=S_infinity.

Tail first-term form

$$R_n=\frac{a_{n+1}}{1-q}$$

Tail o‘zi yangi infinite GP.

Shart: |q|<1

Xususiy holatlar: Equivalent to previous.

Foiz o‘sish factor

$$q=1+r$$

r relative increase multiplierga aylanadi.

Shart: r>-1 in positive contexts

Xususiy holatlar: 20% →1.2.

Foiz kamayish factor

$$q=1-r$$

Remaining fraction multiplier.

Shart: 0≤r≤1 typical

Xususiy holatlar: 15% →0.85.

Compound value after n periods

$$V_n=V_0q^n$$

Initial value after n multiplicative updates.

Shart: Discrete periods

Xususiy holatlar: If sequence indexed a_1=V_0 then index offset differs.

Doubling model

$$V_n=V_0 2^n$$

Har period ikki baravar.

Shart: q=2

Xususiy holatlar: Index conventionni tekshiring.

Halving model

$$V_n=V_0(1/2)^n$$

Har period yarmi qoladi.

Shart: q=1/2

Xususiy holatlar: Approaches 0, never negative.

Product around midpoint

$$a_{k-h}a_{k+h}=a_k^2$$

Multiplicative symmetryning centered formi.

Shart: Valid indices

Xususiy holatlar: h=1 gives neighbor identity.

Three positive GP terms from endpoints

$$b=\sqrt{ac}$$

Middle term endpoints geometric meanidir.

Shart: a,c>0 and positive GP

Xususiy holatlar: Negative middle possible in alternating contexts.

Infinite sumdan first term

$$a_1=S_\infty(1-q)$$

Total va ratio orqali first term.

Shart: |q|<1

Infinite sumdan q

$$q=1-\frac{a_1}{S_\infty}$$

First term va totaldan ratio candidate.

Shart: S_infinity≠0, |q|<1 verification

Xususiy holatlar: Candidate convergence sharti bilan tekshiriladi.

AP/GP diagnostic

$$a_{n+1}-a_n=d\quad\text{vs}\quad a_{n+1}/a_n=q$$

Additive va multiplicative patternni ajratadi.

Shart: Ratio denominator nonzero

Xususiy holatlar: Constant nonzero sequence d=0,q=1 ikkala model.

Teoremalar va isbotlar

📐 GP n-had teoremasi

Geometrik progressiyada a_n=a_1q^{n-1}.

n-hadga yetguncha n-1 marta ×q qilinadi.

Isbotni ko'rsatish

Berilgan: a_{n+1}=qa_n geometric recurrence.

Isbotlash kerak: a_n=a_1q^{n-1}.

a_2=a_1q.
a_3=a_2q=a_1q^2.
Har yangi indexda yana bitta q factor qo‘shiladi.
a_1 dan a_n gacha n-1 transition bor.
Shuning uchun a_n=a_1q^{n-1}.

n-had formula isbotlandi. ∎

📐 Multiplicative symmetry teoremasi

GP da i+j=r+s bo‘lsa a_i a_j=a_r a_s.

Exponentlar yig‘indisi bir xil bo‘lsa q ning umumiy poweri bir xil.

Isbotni ko'rsatish

Berilgan: a_k=a_1q^{k-1} GP va i+j=r+s.

Isbotlash kerak: a_i a_j=a_r a_s.

a_i a_j=a_1^2 q^{i-1+j-1}=a_1^2q^{i+j-2}.
a_r a_s=a_1^2q^{r+s-2}.
i+j=r+s bo‘lgani uchun exponentlar teng.
Demak productlar teng.

a_i a_j=a_r a_s. ∎

📐 Finite geometric series teoremasi

q≠1 bo‘lgan GP uchun S_n=a_1(1-q^n)/(1-q).

Yig‘indini q ga surib ayirganda ichki hadlar bekor bo‘ladi.

Isbotni ko'rsatish

Berilgan: S_n=a_1+a_1q+...+a_1q^{n-1}, q≠1.

Isbotlash kerak: S_n=a_1(1-q^n)/(1-q).

S_n=a_1+a_1q+...+a_1q^{n-1}.
qS_n=a_1q+a_1q^2+...+a_1q^n.
S_n-qS_n da barcha ichki hadlar bekor bo‘ladi.
(1-q)S_n=a_1-a_1q^n=a_1(1-q^n).
q≠1 sabab 1-q ga bo‘lamiz.

S_n=a_1(1-q^n)/(1-q). ∎

📐 Infinite geometric series convergence teoremasi

Real GP series ∑a_1q^{k} finite sumga konvergent iff |q|<1; bu holda S∞=a_1/(1-q).

Faqat |q|<1 da q^n 0 ga yo‘qoladi.

Isbotni ko'rsatish

Berilgan: Infinite geometric series a_1+a_1q+a_1q²+..., real q.

Isbotlash kerak: Convergent iff |q|<1 va sum a_1/(1-q).

q≠1 uchun S_n=a_1(1-q^n)/(1-q).
Agar |q|<1 bo‘lsa q^n→0, demak S_n→a_1/(1-q).
Agar |q|>1 bo‘lsa |q^n| o‘sadi va terms 0 ga bormaydi.
q=1 da terms a_1; q=-1 da nonzero a_1 uchun partial sums oscillate qiladi.
Shuning uchun nontrivial geometric series finite limitga aynan |q|<1 da ega.

|q|<1 iff convergence, va S∞=a_1/(1-q). ∎

📐 Partial-sum recovery teoremasi

Har qanday sequence uchun a_n=S_n-S_{n-1}; GP uchun shu difference’lar o‘zaro constant ratio q ga ega.

Cumulative data increments individual termsdir.

Isbotni ko'rsatish

Berilgan: S_n=a_1+...+a_n.

Isbotlash kerak: a_n=S_n-S_{n-1}; GP bo‘lsa recovered terms ratio q.

S_n=a_1+...+a_{n-1}+a_n.
S_{n-1}=a_1+...+a_{n-1}.
Ayirishda common terms bekor bo‘lib a_n qoladi.
Agar original sequence GP bo‘lsa a_{n+1}/a_n=q, demak recovered differences ham shu ratio ga ega.

Partial sums original GP termsni tiklaydi. ∎

Yechilgan misollar

oson $2,6,18,54,...$ uchun q ni toping.

💡 Maslahat: Ketma-ket hadlarni bo‘ling.

q=6/2=3.
18/6=3 va 54/18=3 bilan tekshiramiz.

✅ Javob: $q=3$

Nega bu usul ishlaydi: Consecutive ratio constant.

⚠️ Ayirmani 4 deb olib AP bilan adashtirmang.

oson $a_1=5,q=2$. $a_8$ ni toping.

💡 Maslahat: n-had formula.

a_8=5·2^{7}.
2^7=128.
a_8=640.

✅ Javob: $640$

Nega bu usul ishlaydi: 1-haddan 8-hadgacha 7 ko‘paytirish bor.

⚠️ Exponent 8 emas, n-1=7.

oson $a_1=81,q=1/3$. $a_5$ ni toping.

💡 Maslahat: Decay GP.

a_5=81(1/3)^4.
(1/3)^4=1/81.
a_5=1.

✅ Javob: $1$

Nega bu usul ishlaydi: Positive q<1 hadlarni kamaytiradi.

⚠️ 1/3 ni 3 deb yubormang.

oson $a_1=4,q=-2$. Dastlabki 5 hadni yozing.

💡 Maslahat: Har safar -2 ga ko‘paytiring.

a_1=4.
a_2=-8.
a_3=16.
a_4=-32.
a_5=64.

✅ Javob: $4,-8,16,-32,64$

Nega bu usul ishlaydi: Negative q signni navbatma-navbat almashtiradi.

⚠️ Magnitude va signni alohida kuzating.

oson $a_4=54,q=3$. $a_7$ ni toping.

💡 Maslahat: Anchor formula.

a_7=a_4q^{7-4}.
=54·3^3=54·27.
=1458.

✅ Javob: $1458$

Nega bu usul ishlaydi: Faqat 3 transition kerak.

⚠️ q^7 emas, q^{7-4}.

oson Musbat GP da $4,x,36$ consecutive hadlar. x ni toping.

💡 Maslahat: Geometric mean.

x²=4·36=144.
Musbat GP sabab x=12.

✅ Javob: $12$

Nega bu usul ishlaydi: Middle squared outer product.

Muqobil usul: sqrt(144)=12.

⚠️ Algebraik x=±12 dan positive-context x=12 tanlanadi.

oson $a_1=3,q=2,n=6$. $S_6$ ni toping.

💡 Maslahat: Finite geometric sum.

S_6=3(2^6-1)/(2-1).
=3(64-1).
=189.

✅ Javob: $189$

Nega bu usul ishlaydi: q>1 uchun alternative sum form qulay.

Muqobil usul: 3+6+12+24+48+96=189.

⚠️ Faqat a_6=96 ni javob qilmang.

oson $a_1=64,q=1/2,n=5$. $S_5$ ni toping.

💡 Maslahat: 1-q form qulay.

S_5=64[1-(1/2)^5]/(1-1/2).
=64(31/32)/(1/2).
=124.

✅ Javob: $124$

Nega bu usul ishlaydi: Finite decay series formula.

Muqobil usul: 64+32+16+8+4=124.

⚠️ Infinite sum 128 bilan aralashtirmang.

oson $5,5,5,...$ ning birinchi 20 hadi yig‘indisini toping.

💡 Maslahat: q=1 special case.

q=1.
S_{20}=20·5=100.

✅ Javob: $100$

Nega bu usul ishlaydi: General fraction formula q=1 da 0/0 beradi.

⚠️ Maxsus branchni tekshiring.

oson $1+1/2+1/4+1/8+...$ cheksiz yig‘indini toping.

💡 Maslahat: |q|<1.

a_1=1,q=1/2.
S∞=1/(1-1/2)=2.

✅ Javob: $2$

Nega bu usul ishlaydi: q powers 0 ga boradi.

⚠️ Finite 4-term sum 15/8 bilan cheksiz totalni adashtirmang.

ortacha $a_3=12,a_6=96$. q ni toping, q>0.

💡 Maslahat: Ratio gives q^3.

a_6/a_3=q^{3}.
96/12=8.
q^3=8 → q=2.

✅ Javob: $q=2$

Nega bu usul ishlaydi: Index gap 3.

⚠️ q=8 emas.

ortacha $a_2=6,a_5=162$, q>0. $a_1$ ni toping.

💡 Maslahat: Avval q, keyin a_1.

a_5/a_2=q^3=162/6=27.
q=3.
a_2=a_1q → a_1=6/3=2.

✅ Javob: $a_1=2$

Nega bu usul ishlaydi: Uzoq hadlar ratio q powerini beradi.

⚠️ Index gapni 5-2=3 deb oling.

ortacha GP da $a_3a_9=400$. $a_6^2$ ni toping.

💡 Maslahat: Indeks sumsni solishtiring.

3+9=12 va 6+6=12.
Product symmetry: a_3a_9=a_6².
a_6²=400.

✅ Javob: $400$

Nega bu usul ishlaydi: Equal index sum → equal product.

Muqobil usul: Positive GP bo‘lsa a_6=20.

⚠️ a_6 ning o‘zi so‘ralmagan.

ortacha Musbat GP da $a_4=10$ va $a_6=90$. $a_5$ ni toping.

💡 Maslahat: Middle geometric mean.

a_5²=a_4a_6=900.
Positive GP → a_5=30.

✅ Javob: $30$

Nega bu usul ishlaydi: Three-term invariant.

⚠️ ±30 dan positive branch.

ortacha $a_1=2,q=-3$. $S_5$ ni toping.

💡 Maslahat: Finite formula negative q bilan ham ishlaydi.

S_5=2[1-(-3)^5]/[1-(-3)].
(-3)^5=-243.
S_5=2(244)/4=122.

✅ Javob: $122$

Nega bu usul ishlaydi: Alternating signs cancellation formula ichida saqlanadi.

Muqobil usul: 2-6+18-54+162=122.

⚠️ q^5 signini yo‘qotmang.

ortacha $S_6=126$ va q=2$. $a_1$ ni toping.

💡 Maslahat: Finite sum formulani a_1 uchun yeching.

126=a_1(2^6-1)/(2-1).
126=63a_1.
a_1=2.

✅ Javob: $a_1=2$

Nega bu usul ishlaydi: Sum multiplier 63.

⚠️ 126/64 emas.

ortacha $a_1=3,q=2$. $S_n=93$ bo‘lsa n ni toping.

💡 Maslahat: Finite sum equation.

93=3(2^n-1).
31=2^n-1.
2^n=32.
n=5.

✅ Javob: $n=5$

Nega bu usul ishlaydi: Geometric total exponent equationga keladi.

⚠️ S_n formula ichidagi q^n ga e’tibor.

ortacha $S_n=2^{n+1}-2$. $a_n$ ni toping.

💡 Maslahat: Partial-sum difference.

a_n=S_n-S_{n-1}.
S_{n-1}=2^n-2.
a_n=2^{n+1}-2-(2^n-2)=2^n.

✅ Javob: $a_n=2^n$

Nega bu usul ishlaydi: Cumulative difference original term.

Muqobil usul: Sequence 2,4,8,... q=2.

⚠️ S_n ni a_n deb o‘qimang.

ortacha $S_n=3(2^n-1)$. GP ning a_1 va q sini toping.

💡 Maslahat: Difference yoki finite-sum pattern.

a_n=S_n-S_{n-1}.
=3(2^n-1)-3(2^{n-1}-1)=3·2^{n-1}.
Demak a_1=3,q=2.

✅ Javob: $a_1=3,\ q=2$

Nega bu usul ishlaydi: Partial-sum exponential form GP ni ochadi.

Muqobil usul: S_1=3, ratio differences=2.

⚠️ S_n coefficientini first term deb olish har doim tekshiruv talab qiladi.

ortacha $a_1=12,q=1/3$. $S_\infty$ ni toping.

💡 Maslahat: Convergence check first.

|q|=1/3<1.
S∞=12/(1-1/3).
=12/(2/3)=18.

✅ Javob: $18$

Nega bu usul ishlaydi: Infinite GP convergent.

⚠️ 12/(1+1/3) emas.

ortacha $S_\infty=20$ va q=0.2$. $a_1$ ni toping.

💡 Maslahat: a_1=S∞(1-q).

a_1=20(1-0.2).
=20·0.8=16.

✅ Javob: $16$

Nega bu usul ishlaydi: Infinite sum formula rearranged.

⚠️ q=20% =0.2.

ortacha $S_\infty=30$ va $a_1=12$. q ni toping.

💡 Maslahat: q=1-a_1/S∞.

q=1-12/30.
=1-0.4=0.6.
|q|<1 tekshiruvi o‘tadi.

✅ Javob: $q=0.6$

Nega bu usul ishlaydi: Inverse infinite sum.

⚠️ Convergence checkni unutmaslik kerak.

ortacha $a_1=8,q=1/2$. Dastlabki 4 haddan keyingi infinite tail yig‘indisini toping.

💡 Maslahat: Tail first term a_5.

a_5=8(1/2)^4=1/2.
R_4=a_5/(1-q).
=(1/2)/(1/2)=1.

✅ Javob: $1$

Nega bu usul ishlaydi: Tail o‘zi yangi infinite GP.

Muqobil usul: S∞=16,S_4=15 → remainder 1.

⚠️ a_4 ni tail first term deb olmang.

ortacha $1-1/2+1/4-1/8+...$ yig‘indini toping.

💡 Maslahat: q=-1/2 va |q|<1.

a_1=1,q=-1/2.
S∞=1/[1-(-1/2)].
=1/(3/2)=2/3.

✅ Javob: $\frac23$

Nega bu usul ishlaydi: Alternating convergent GP.

⚠️ Denominator 1-q=1+1/2.

ortacha 1000 so‘m qiymat har davr 10% oshadi. 5 davrdan keyingi qiymat?

💡 Maslahat: q=1.10.

V_5=1000(1.1)^5.
1.1^5=1.61051.
V_5=1610.51.

✅ Javob: $1610.51$ so‘m

Nega bu usul ishlaydi: Constant percent growth multiplicative.

⚠️ 1000+5·100 linear usul compound emas.

ortacha 800 birlik qiymat har davr 25% kamayadi. 4 davrdan keyin qancha qoladi?

💡 Maslahat: q=0.75.

V_4=800(0.75)^4.
0.75^4=0.31640625.
Natija 253.125.

✅ Javob: $253.125$

Nega bu usul ishlaydi: Har period qolgan qiymatning 75%i.

⚠️ q=0.25 emas.

ortacha Bakteriya soni har 3 soatda ikki baravar. Boshlang‘ich 500 bo‘lsa 12 soatdan keyin qancha?

💡 Maslahat: 12/3=4 doubling period.

n=4 period.
V=500·2^4.
=8000.

✅ Javob: $8000$

Nega bu usul ishlaydi: Fixed-period doubling GP.

⚠️ 12 ni exponent qilish noto‘g‘ri; period count 4.

ortacha Modda miqdori har 2 soatda yarmiga tushadi. Boshlang‘ich 160 mg. 8 soatdan keyin?

💡 Maslahat: 4 halving period.

n=8/2=4.
V=160(1/2)^4.
=10 mg.

✅ Javob: $10$ mg

Nega bu usul ishlaydi: Discrete half-life model.

⚠️ 8 exponent emas, 4 period.

murakkab GP da $a_2+a_4=30$ va $a_3=12$, q>0. q ni toping.

💡 Maslahat: a_2=a_3/q, a_4=a_3q.

12/q+12q=30.
2/q+2q=5.
2q²-5q+2=0.
(2q-1)(q-2)=0.
q=1/2 yoki q=2.

✅ Javob: $q=\frac12$ yoki $q=2$

Nega bu usul ishlaydi: Symmetric reciprocal ratios bir xil outer-sum berishi mumkin.

⚠️ Bitta javobga shoshilmang; q va 1/q symmetry bor.

murakkab Musbat GP da $a_1a_5=81$. $a_3$ ni toping.

💡 Maslahat: Index sums 1+5=3+3.

a_3²=a_1a_5=81.
Positive GP → a_3=9.

✅ Javob: $9$

Nega bu usul ishlaydi: Product symmetry.

⚠️ Negative middle musbat GP shartiga mos emas.

murakkab $a_1=1,q=3$. $S_n>1000$ bo‘ladigan eng kichik n ni toping.

💡 Maslahat: S_n=(3^n-1)/2.

(3^n-1)/2>1000 →3^n>2001.
3^6=729<2001.
3^7=2187>2001.
Eng kichik n=7.

✅ Javob: $7$

Nega bu usul ishlaydi: Monotone positive partial sums threshold.

⚠️ Real logarithmic estimate bo‘lsa ham integer check qiling.

murakkab $a_1=5,q=1/2$. Infinite totalning 99% iga yetish uchun eng kichik n ni toping.

💡 Maslahat: Tail/total ratio = q^n.

S∞=10.
S_n/S∞=1-q^n.
1-(1/2)^n≥0.99.
(1/2)^n≤0.01.
2^n≥100.
2^6=64<100,2^7=128≥100 → n=7.

✅ Javob: $7$

Nega bu usul ishlaydi: Relative remainder q^n.

Muqobil usul: Tail ≤1% total.

⚠️ 99% criterionni term size bilan aralashtirmang.

murakkab GP da $a_4=16$ va $a_8=256$, q>0. $S_8/S_4$ ni toping.

💡 Maslahat: q^4=16.

a_8/a_4=q^4=256/16=16 →q=2.
S_8/S_4=[2^8-1]/[2^4-1].
=255/15=17.

✅ Javob: $17$

Nega bu usul ishlaydi: a_1 factors cancel in sum ratio.

⚠️ 8/4=2 deb q olish xato.

murakkab Cheksiz GP da $S_\infty=24$ va ikkinchi had $a_2=6$. Musbat q<1 bo‘lsa q ni toping.

💡 Maslahat: a_1=6/q va S∞=(6/q)/(1-q).

24=6/[q(1-q)].
4q(1-q)=1.
4q²-4q+1=0.
(2q-1)²=0.
q=1/2.

✅ Javob: $q=\frac12$

Nega bu usul ishlaydi: Second term + total quadratic relation.

⚠️ a_2 ni a_1 deb olmang.

murakkab Cheksiz GP ning yig‘indisi 12, birinchi 3 had yig‘indisi 10.5. q ni toping, 0<q<1.

💡 Maslahat: S_3/S∞=1-q^3.

10.5/12=0.875=1-q^3.
q^3=0.125.
q=0.5.

✅ Javob: $q=\frac12$

Nega bu usul ishlaydi: Finite-to-infinite ratio q^n remainderni beradi.

⚠️ 1-q emas, 1-q^3.

murakkab $S_n=4(1-3^{-n})$. GP ning $a_1,q$ va $S_\infty$ sini toping.

💡 Maslahat: C(1-q^n) pattern.

Compare with a_1(1-q^n)/(1-q).
q=1/3.
a_1/(1-1/3)=4 →a_1=8/3.
S∞=4.

✅ Javob: $a_1=\frac83,\ q=\frac13,\ S_\infty=4$

Nega bu usul ishlaydi: Partial sum asymptote and exponential remainder directly reveal parameters.

Muqobil usul: a_n=S_n-S_{n-1} bilan ham.

⚠️ 4 ni first term deb olish noto‘g‘ri; u limit.

murakkab GP da $a_1=2,q=3$. $a_5+a_6+...+a_9$ ni toping.

💡 Maslahat: Sub-block o‘zi GP; yoki S_9-S_4.

S_9=2(3^9-1)/(3-1)=3^9-1=19682.
S_4=3^4-1=80.
Farq 19602.

✅ Javob: $19602$

Nega bu usul ishlaydi: Partial-sum difference universal.

Muqobil usul: a_5=162 dan 5-term GP sum ham mumkin.

⚠️ S_9-S_5 bo‘lsa a_5 yo‘qoladi.

murakkab GP da $a_3=8$ va q=-2. $a_1$ va $S_6$ ni toping.

💡 Maslahat: a_3=a_1q².

8=a_1·4 →a_1=2.
S_6=2[1-(-2)^6]/[1-(-2)].
=2(1-64)/3=-42.

✅ Javob: $a_1=2,\ S_6=-42$

Nega bu usul ishlaydi: Alternating GP total negative bo‘lishi mumkin.

Muqobil usul: Terms:2,-4,8,-16,32,-64 sum -42.

⚠️ Even power q^6 positive.

murakkab Bir investitsiya har yil 8% oshadi. 6 yil davomida har yil boshidagi qiymatlar $1000,1080,...$ bo‘lsa ushbu 6 qiymatning yig‘indisini toping.

💡 Maslahat: Finite geometric sum with q=1.08.

a_1=1000,q=1.08,n=6.
S_6=1000(1.08^6-1)/(0.08).
1.08^6≈1.586874322.
S_6≈7335.929.

✅ Javob: Taxminan $7335.93$

Nega bu usul ishlaydi: Repeated percent values finite GP series.

⚠️ “6 yil oxiridagi qiymat” bilan 6 annual snapshots totalini ajrating.

murakkab To‘p har sakrashda oldingi balandlikning 60% iga chiqadi. Dastlab 10 m balandlikdan tushsa, keyingi sakrash cho‘qqilarining cheksiz yig‘indisi?

💡 Maslahat: First bounce height 6, q=.6.

a_1=10·0.6=6.
q=0.6.
Sum=6/(1-0.6)=15.

✅ Javob: $15$ m

Nega bu usul ishlaydi: Bounce peak heights geometric decay.

Muqobil usul: Agar total travelled distance so‘ralsa tushish/ko‘tarilishlar alohida hisoblanadi.

⚠️ Initial 10 m bu “keyingi sakrash cho‘qqilari” seriesiga kirmaydi.

murakkab $0.272727\ldots$ ni kasrga aylantiring.

💡 Maslahat: 0.27+0.0027+... geometric series sifatida yozing.

0.272727...=27/100+27/10000+... .
a_1=27/100, q=1/100.
S=(27/100)/(1-1/100).
=(27/100)/(99/100)=27/99=3/11.

✅ Javob: $\frac3{11}$

Nega bu usul ishlaydi: Repeating decimal convergent geometric seriesdir.

Muqobil usul: x=0.2727... deb 100x-x usuli ham mumkin.

⚠️ q=0.01; 0.27 emas.

murakkab Kvadratning yuzi 100. Har bosqichda yangi shakl oldingi yuzaning 1/4 qismiga teng. Barcha bosqichlar yuzalarining cheksiz yig‘indisini toping.

💡 Maslahat: a_1=100,q=1/4.

|q|=1/4<1.
S∞=100/(1-1/4).
=100/(3/4)=400/3.

✅ Javob: $\frac{400}{3}$

Nega bu usul ishlaydi: Self-similar areas geometric decay beradi.

⚠️ Faqat qo‘shimcha shakllar so‘ralsa first term boshqacha bo‘lishi mumkin.

murakkab GP da $S_3=21$ va $S_6=189$, q>0. q ni toping.

💡 Maslahat: S_6=S_3+q^3S_3 relationni ko‘ring.

First 3 terms sum S_3.
Next 3 terms har biriga q^3 factor berilgan, ularning sum’i q^3S_3.
S_6=S_3(1+q^3).
189=21(1+q^3) →9=1+q^3.
q^3=8 →q=2.

✅ Javob: $q=2$

Nega bu usul ishlaydi: Equal-length consecutive blocks geometric factor q^blocklength bilan bog‘langan.

Muqobil usul: Finite-sum formulas ratio orqali ham.

⚠️ S_6/S_3=q^3 emas; 1+q^3.

murakkab GP da $a_1=3,q=2$. $a_6+a_7+\cdots+a_{10}$ ni toping.

💡 Maslahat: 5-term sub-block first term a_6.

a_6=3·2^5=96.
Block 5 had, ratio 2.
Sum=96(2^5-1)/(2-1).
=96·31=2976.

✅ Javob: $2976$

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

Muqobil usul: $S_{10}-S_5$ ham 2976 beradi.

⚠️ S_10-S_6 bo‘lsa a_6 yo‘qoladi.

murakkab Bir kompaniya daromadi 1-yil 100 mln, har yil 12% oshadi. Dastlabki 5 yilning jami daromadini toping.

💡 Maslahat: Yearly revenues finite GP, q=1.12.

a_1=100,q=1.12,n=5 (mln).
S_5=100(1.12^5-1)/(0.12).
1.12^5≈1.7623416832.
S_5≈100·0.7623416832/0.12≈635.284736.

✅ Javob: Taxminan $635.285$ mln

Nega bu usul ishlaydi: Har yilgi qiymatlar constant percent bilan geometric; “jami” finite series.

Muqobil usul: Har yilni alohida hisoblab qo‘shish mumkin.

⚠️ 5-yil qiymati bilan 5 yillik totalni adashtirmang.

Umumiy xatolar

❌ GP da ayirmalarni tekshirish

GP ning defining invariant’i ratio, difference emas.

✅ Consecutive nonzero hadlar nisbatini tekshiring.

2,6,18 da differences 4,12; ratios 3,3.

❌ a_n=a_1q^n yozish

1-haddan n-hadgacha n-1 transition bor.

✅ a_n=a_1q^{n-1}.

a_1 formula n=1 da q^0=1 berishi kerak.

❌ Uzoq hadlarda q=a_n/a_m olish

Bu ratio q^{n-m}, q ning o‘zi emas.

✅ a_n/a_m=q^{n-m}.

a_6/a_3=q^3.

❌ q<0 bo‘lsa barcha hadlar manfiy deb o‘ylash

Negative ratio signni alternatsiya qiladi.

✅ Ishora (-1) power bilan almashishini kuzating.

2,-6,18,-54.

❌ a_k=(a_{k-1}+a_{k+1})/2 ni GP ga qo‘llash

Bu AP arithmetic mean xossasi.

✅ Musbat GP da a_k=sqrt(a_{k-1}a_{k+1}).

4,12,36.

❌ Geometrik o‘rtada ± rootni avtomatik olish

Geometric mean musbat sonlar uchun principal positive root.

✅ Context/sign patternni tekshiring.

Positive GP middle term 12, -12 emas.

❌ Finite sumda q^n o‘rniga q^{n-1}

Yig‘indi multiplier trick endpointda q^n beradi.

✅ S_n=a_1(1-q^n)/(1-q).

n=1 da formula a_1 berishini tekshiring.

❌ q=1 ni umumiy fraction formulaga qo‘yish

Denominator 1-q=0.

✅ q=1 da alohida S_n=na_1.

5+5+... 20 had =100.

❌ Infinite sum formulasini |q|≥1 da ishlatish

q^n 0 ga bormaydi; partial sums finite limitga yaqinlashmaydi.

✅ Avval |q|<1 convergence gate.

1+2+4+... divergent.

❌ q=-1/2 da denominatorni 1-1/2 olish

Formula 1-q; negative q bo‘lsa minus negative plusga aylanadi.

✅ 1-(-1/2)=3/2.

1-1/2+1/4-...=2/3.

❌ Cheksiz qatorning first termini initial pre-period value bilan adashtirish

Context series qaysi bosqichdan boshlanishiga bog‘liq.

✅ Seriesdagi haqiqiy birinchi hadni alohida yozing.

Bounce peaksda first bounce 0.6h, initial drop h emas.

❌ 20% o‘sishda q=0.20 olish

0.20 faqat qo‘shiladigan ulush; new total 120%.

✅ q=1.20.

100→120.

❌ 20% kamayishda q=-0.20 olish

Qiymatning 80%i qoladi; sign almashmaydi.

✅ q=0.80.

100→80.

❌ Har period foizni original bazaga qo‘llash

Compound model yangi qiymat ustida ishlaydi.

✅ Repeated multiplier q^n.

1000 at 10%: 1000·1.1^n.

❌ S_n bilan a_n ni bir xil deb olish

S_n cumulative, a_n individual term.

✅ a_n=S_n-S_{n-1}.

S_n=2^{n+1}-2 → a_n=2^n.

❌ Tailni a_n dan boshlash

First n termsdan keyingi tail a_{n+1} bilan boshlanadi.

✅ R_n=a_{n+1}/(1-q).

First 4 terms after tail starts at a_5.

❌ Product symmetryda indekslar ayirmasini tenglashtirish

Invariant exponent sumga bog‘liq.

✅ i+j=r+s bo‘lsa products equal.

a_3a_9=a_6².

❌ q va 1/q symmetryni unutish

Uzoq/symmetric constraints ba’zan reciprocal ikkita positive ratio beradi.

✅ Barcha admissible rootsni tekshiring.

a_2+a_4 constraint q=2 yoki 1/2 berishi mumkin.

❌ Period uzunligini exponent bilan adashtirish

Exponent update soni, vaqt birligi emas.

✅ n=total time / period length.

12 soat, doubling every3h → n=4.

❌ AP va GP modelini signal so‘zlar bilan ko‘r-ko‘rona tanlash

“oshadi”ning absolute yoki percent ekanini ajratish kerak.

✅ Constant amount → AP; constant factor/percent → GP.

Har oy +50 vs +5%.

Noto'g'ri tasavvurlar

Geometrik progressiyada hadlar doim o‘sadi.

0<q<1 da kamayadi, q<0 da sign almashadi, q=1 da constant.

GP faqat musbat sonlardan iborat.

Negative a_1 yoki q<0 bilan manfiy/alternating GP lar ham mavjud.

q common difference bilan bir xil tushuncha.

d additive change; q multiplicative ratio.

Geometrik o‘rta oddiy arithmetic average.

Positive x,y geometric mean sqrt(xy), arithmetic mean (x+y)/2.

Finite geometric sum uchun |q|<1 shart.

Finite sum q≠1 formula bilan istalgan real q da (defined terms) ishlaydi; |q|<1 faqat infinite convergence uchun.

q=1 GP emas.

Constant nonzero sequence q=1 bo‘lgan GP va d=0 bo‘lgan AP ham.

Cheksiz seriesda terms 0 ga borsa har doim sum convergent.

Term→0 necessary, lekin umumiy serieslarda sufficient emas; geometric seriesda |q|<1 bilan sufficient.

Negative q bo‘lsa infinite sum albatta divergent.

Masalan q=-1/2 da |q|<1 va series convergent.

Doimiy foiz o‘sish chiziqli model.

Foiz oldingi qiymatga nisbatan olinadi, shuning uchun exponential/geometric.

Partial sumsning o‘zi ham GP bo‘ladi.

Odatda yo‘q; S_n=C(1-q^n) affine-exponential formda, ratio constant emas.

Amaliy qo'llanilishi

Moliya

Murakkab foiz, investitsiya va qarz qoldig‘i har davr multiplicative factor bilan o‘zgaradi.

Aholi va biologiya

Doimiy foizli population growth va cell doubling geometric/exponential model beradi.

Radioaktiv kamayish

Har fixed intervalda doimiy ulush qolsa q<1 geometric decay modeli paydo bo‘ladi.

Fizika — sakrash va aks-sado

Har bounce/amplitude oldingisining doimiy ulushi bo‘lsa finite/infinite geometric series ishlatiladi.

Kompyuter fanlari

Recursive branching, repeated scaling va ayrim algorithm work levels geometric counts beradi.

Grafika va masshtab

Har iterationda obyekt o‘lchami bir xil scale factor bilan o‘zgaradi.

Marketing va network growth

Har participant doimiy o‘rtacha multiplier bilan yangi participant olib kelsa level counts geometric modelga yaqinlashadi.

Error approximation

Convergent geometric series tail formulasi truncation errorni aniq baholaydi.

Geometrik progressiya: ratio, term va series decision map

GP masalalarida avval constant ratio aniqlanadi. So‘ng savol bitta had, chekli yig‘indi yoki cheksiz yig‘indi ekaniga qarab alohida engine tanlanadi; infinite branchda |q|<1 sharti majburiy.

Xulosa

Cheat sheet: a_{n+1}=q a_n; a_n=a_1q^{n-1}; a_n=a_m q^{n-m}. Musbat GP da a_k^2=a_{k-1}a_{k+1}. Chekli yig‘indi: S_n=a_1(1-q^n)/(1-q)=a_1(q^n-1)/(q-1), q≠1; q=1 da S_n=na_1. |q|<1 bo‘lsa S_∞=a_1/(1-q). Partial sumsda a_n=S_n-S_{n-1}. Foiz o‘sishi r bo‘lsa q=1+r; kamayish r bo‘lsa q=1-r.

Keyingi “Matnli masalalar” va ko‘rsatkichli tenglamalarda geometrik o‘sish modellari, compound growth va q^n tipidagi noma’lum ko‘rsatkichlar chuqurlashadi.

Bog'liq mavzular

Oldin bilishingiz kerak: Daraja va uning xossalari, darajali ifodalar, Arifmetik progressiya yig'indisi

Bog'liq mavzular: Funksiya

Keyingi mavzular: Matnli masalalar, Ko'rsatkichli tenglama, Logarifmlar: hisoblashga doir masalalar

Manbalar

OpenStax Intermediate Algebra 2e — 12.3 Geometric Sequences and Series — Geometric sequence, finite sum, infinite convergence and sum formulas.
OpenStax Algebra and Trigonometry 2e — Sequences and Series — Broader sequence/series context and notation.
OpenStax Intermediate Algebra — 12.3 Geometric Sequences and Series — Worked finite and infinite geometric-series examples.
Khan Academy — Precalculus Series: Geometric series — Finite geometric series, summation notation and word-problem practice.
Khan Academy — Infinite geometric series practice — Convergent/divergent infinite geometric-series practice.
MathTest.uz — Arifmetik progressiya yig‘indisi — Contrast between additive arithmetic accumulation and multiplicative geometric accumulation.

Ro'yxatdan o'tib, mashq qilishni boshlang