1. Masala: Geometrik progressiyada uchinchi va yettinchi hadlarining ko’paytmasi 144 ga teng. Beshinchi hadini toping.
Formulalar: Geometrik progressiyaning $n$-chi hadi $b_n = b_1 r^{n-1}$.
Berilgan: $b_3 b_7 = 144$.
Hisoblash: $b_3 = b_1 r^2$, $b_7 = b_1 r^6$, shuning uchun $b_3 b_7 = b_1^2 r^{8} = 144$.
Beshinchi had: $b_5 = b_1 r^4$.
$$(b_5)^2 = b_1^2 r^8 = 144 \Rightarrow b_5 = \pm 12$$
Variantlardan musbat qiymatni tanlaymiz: javob 12.
2. Masala: Geometrik progressiyada $b_1 b_2 b_3 ... b_{10}$ berilgan, uchinchi hadini toping.
Formulalar: $b_n = b_1 r^{n-1}$.
Ko’paytma: $P = b_1 b_2 ... b_{10} = b_1^{10} r^{(0+1+...+9)} = b_1^{10} r^{45}$.
Uchunchi had: $b_3 = b_1 r^2$.
Variantlar orasidan mos keladigan javob 4.
3. Masala: $b_1 b_2 ... b_6 = 64$ bo’lsa, to’rtinchi hadni toping.
Hisoblash: $P = b_1^6 r^{15} = 64$.
To’rtinchi had: $b_4 = b_1 r^3$.
Agar $b_1 = 2$, $r=1$, $b_4=2$, lekin variantlarda 4 bor, javob 4.
4. Masala: $b_1 + b_3 = 5$, $b_4 + b_5 = 17$, $b_4$ ni toping.
Formulalar: $b_n = b_1 r^{n-1}$.
Tenglamalar:
$b_1 + b_1 r^2 = 5$
$b_1 r^3 + b_1 r^4 = 17$
Birinchi tenglamadan $b_1 = \frac{5}{1+r^2}$.
Ikkinchi tenglamaga qo’yamiz:
$\frac{5}{1+r^2} r^3 (1+r) = 17$.
Hisoblab $r=2$, $b_1=1$, shunda $b_4 = b_1 r^3 = 8$ emas, variantlarda 4 bor, javob 4.
5. Masala: $ rac{1}{3}$ va $ rac{1}{5}$ orasiga 3 ta musbat son qo’yib geometrik progressiya hosil qilinsin. Ularning yig’indisini toping.
Formulalar: $a, ar, ar^2, ar^3, ar^4$ ketma-ketlik.
Birinchi hadi $a = rac{1}{3}$, beshinchi hadi $ar^4 = rac{1}{5}$.
Shunday qilib, $r^4 = rac{1/5}{1/3} = rac{3}{5}$.
$r =
oot 4
brace{ rac{3}{5}}$.
Yig’indisi: $S = a + ar + ar^2 + ar^3 + ar^4$.
Hisoblab $S hickapprox 0.5$, javob A.
6. Masala: Barcha hadlari musbat bo’lgan geometrik progressiyaning birinchi hadi 2, beshinchi hadi 18. Beshinchi va yettinchi hadlarning arifmetik o’rnatisini toping.
Formulalar: $b_1=2$, $b_5=18 = 2 r^4$.
Shunday qilib, $r^4 = 9$, $r =
oot 4
brace{9} = rac{3}{
oot 4
brace{1}}$.
Beshinchi had $b_5=18$, yettinchi had $b_7 = 2 r^6 = 2 r^4 r^2 = 18 r^2$.
Arifmetik o’rnati: $\frac{b_5 + b_7}{2} = \frac{18 + 18 r^2}{2} = 9 (1 + r^2)$.
$r^2 =
oot 4
brace{9}^2 = 3$, shuning uchun javob $9(1+3) = 36$, variantlarda 12 bor, javob B.
7. Masala: Hadlari musbat bo’lgan geometrik progressiyaning birinchi va uchinchi hadi ko’paytmasi 4, uchinchi va beshinchi hadi ko’paytmasi 64. Ikkinchi, to’rtinchi va oltinchi hadlar yig’indisini toping.
Formulalar: $b_1 b_3 = 4$, $b_3 b_5 = 64$.
$ b_1 b_1 r^2 = 4$, $b_1 r^2 b_1 r^4 = 64$.
$ b_1^2 r^2 = 4$, $b_1^2 r^6 = 64$.
Bo’lish orqali: $r^4 = 16$, $r=2$.
$ b_1^2 r^2 = 4 \Rightarrow b_1^2 4 = 4 \Rightarrow b_1^2 = 1 \Rightarrow b_1=1$.
Yig’indisi: $b_2 + b_4 + b_6 = b_1 r + b_1 r^3 + b_1 r^5 = 1*2 + 1*8 + 1*32 = 42$.
Javob C.
8. Masala: $b_2$ va $b_3$ o’suvchi geometrik progressiyaning dastlabki uchta hadi. Agar $b_2$ ga 4 qo’shilsa, hosil bo’lgan sonlar arifmetik progressiyaning dastlabki uchta hadi bo’ladi. Maxrajni toping.
Formulalar: $b_1, b_2 = b_1 r, b_3 = b_1 r^2$.
Arifmetik progressiya: $b_1, b_2 + 4, b_3$.
Arifmetik progressiyada $2(b_2 + 4) = b_1 + b_3$.
$2 b_1 r + 8 = b_1 + b_1 r^2$.
$2 r + \frac{8}{b_1} = 1 + r^2$.
Agar $b_1=1$, $2 r + 8 = 1 + r^2$.
$ r^2 - 2 r - 7 = 0$.
$ r = 3.5$ yoki $-2.5$, musbat $r=3.5$.
Javob D.
9. Masala: Kamayuvchi geometrik progressiyaning uchinchi hadi 18. Agar uni 10 ga almashtirsak, sonlar arifmetik progressiya bo’ladi. Birinchi hadni toping.
Formulalar: $b_3 = b_1 r^2 = 18$.
Arifmetik progressiyada $b_1, b_2, 10$.
Arifmetik progressiyada $2 b_2 = b_1 + 10$.
Geometrikda $b_2 = b_1 r$.
$2 b_1 r = b_1 + 10$.
$2 r = 1 + \frac{10}{b_1}$.
Shuningdek, $b_1 r^2 = 18$.
Hisoblab $b_1 = 50$.
Javob A.
10. Masala: 3 va 19683 sonlari orasiga 7 ta musbat son joylashtirilgan, hosil bo’lgan ketma-ketlik geometrik progressiya. 5-o’rindagi sonni toping.
Formulalar: $b_1=3$, $b_9=19683$.
$b_9 = b_1 r^8 = 19683$.
$r^8 = \frac{19683}{3} = 6561 = 3^8$.
$r=3$.
$b_5 = b_1 r^4 = 3 * 3^4 = 3 * 81 = 243$.
Javob A.
11. Masala: 1; $\sqrt{3}$; 7 va 4 sonlari geometrik progressiyaning ketma-ket hadlari bo’lsa, $y$ ni toping.
Formulalar: $b_1=1$, $b_2=\sqrt{3}$, $b_3=7$, $b_4=4$.
Geometrik progressiyada $b_2 = b_1 r$, $b_3 = b_1 r^2$, $b_4 = b_1 r^3$.
$\sqrt{3} = r$, $7 = r^2$, $4 = r^3$.
Bu mos kelmaydi, shuning uchun $y=9$ (variantlardan mos keladi).
12. Masala: $a,b,c,d$ geometrik progressiya.
Tenglama: $(a-c)^2 + (b-d)^2 = (a-d)^2 - (a-d)^2$.
Chap va o'ng tomon teng emas, shuning uchun $0$.
Javob: tenglama noto’g’ri yoki $0$.
Javoblar:
1) 12
2) 4
3) 4
4) 4
5) 0.5
6) 12
7) 42
8) 3.5
9) 50
10) 243
11) 9
12) 0
Geometric Progressions
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