27,28,30 and 31 provide answers for those numbers

Mathematics

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3 parts
28
1
4
30
2
3
31
175

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Haya, hebu tutatue maswali hayo.

Jibu la Swali la 27:
Ujazo wa frustum ni:

46629.52 cm3\boxed{\text{46629.52 cm}^3}

Part 28· 3 steps

Jumla ya n-vigezo vya kwanza vya A.P. inatolewa na Sn=2n2−nS_n = 2n^2 - n. Tafuta kigezo cha kwanza na tofauti ya kawaida.

  1. Tafuta kigezo cha kwanza (a1a_1).

    Kigezo cha kwanza ni jumla ya kigezo kimoja cha kwanza (S1S_1).

    S1=2(1)2−1S_1 = 2(1)^2 - 1 S1=2(1)−1S_1 = 2(1) - 1 S1=2−1S_1 = 2 - 1 S1=1S_1 = 1

    Hivyo, kigezo cha kwanza a1=1a_1 = 1.

  2. Tafuta kigezo cha pili (a2a_2).

    Jumla ya vigezo viwili vya kwanza ni S2=a1+a2S_2 = a_1 + a_2.

    S2=2(2)2−2S_2 = 2(2)^2 - 2 S2=2(4)−2S_2 = 2(4) - 2 S2=8−2S_2 = 8 - 2 S2=6S_2 = 6

    Kwa kuwa S2=a1+a2S_2 = a_1 + a_2, tunapata:

    6=1+a26 = 1 + a_2 a2=6−1a_2 = 6 - 1 a2=5a_2 = 5
  3. Tafuta tofauti ya kawaida (dd).

    Tofauti ya kawaida ni d=a2−a1d = a_2 - a_1.

    d=5−1d = 5 - 1 d=4d = 4

    Kigezo cha kwanza ni 1\boxed{1} na tofauti ya kawaida ni 4\boxed{4}.


Part 302

Kigezo cha 3 cha G.P. ni 18 na kigezo cha 6 ni 486. Tafuta kigezo cha kwanza na uwiano wa kawaida.

Fomula ya kigezo cha n cha G.P. ni an=arn−1a_n = ar^{n-1}.

  1. Andika milinganyo kutokana na taarifa uliyopewa.

    Kigezo cha 3: a3=ar3−1=ar2=18a_3 = ar^{3-1} = ar^2 = 18 (Mlinganyo 1)
    Kigezo cha 6: a6=ar6−1=ar5=486a_6 = ar^{6-1} = ar^5 = 486 (Mlinganyo 2)

  2. Tatua kwa uwiano wa kawaida (rr).

    Gawanya Mlinganyo 2 kwa Mlinganyo 1:

    ar5ar2=48618\frac{ar^5}{ar^2} = \frac{486}{18} r5−2=27r^{5-2} = 27 r3=27r^3 = 27 r=273r = \sqrt[3]{27} r=3r = 3
  3. Tatua kwa kigezo cha kwanza (aa).

    Badilisha r=3r=3 katika Mlinganyo 1:

    a(3)2=18a(3)^2 = 18 9a=189a = 18 a=189a = \frac{18}{9} a=2a = 2

    Kigezo cha kwanza ni 2\boxed{2} na uwiano wa kawaida ni 3\boxed{3}.


Part 31175

Kigezo cha 4 cha A.P. ni 13 na kigezo cha 7 ni 22. Tafuta jumla ya vigezo 10 vya kwanza.

Fomula ya kigezo cha n cha A.P. ni an=a+(n−1)da_n = a + (n-1)d.
Fomula ya jumla ya vigezo n vya kwanza vya A.P. ni Sn=n2[2a+(n−1)d]S_n = \frac{n}{2}[2a + (n-1)d].

  1. Andika milinganyo kutokana na taarifa uliyopewa.

    Kigezo cha 4: a4=a+(4−1)d=a+3d=13a_4 = a + (4-1)d = a + 3d = 13 (Mlinganyo 1)
    Kigezo cha 7: a7=a+(7−1)d=a+6d=22a_7 = a + (7-1)d = a + 6d = 22 (Mlinganyo 2)

  2. Tatua kwa tofauti ya kawaida (dd).

    Toa Mlinganyo 1 kutoka Mlinganyo 2:

    (a+6d)−(a+3d)=22−13(a + 6d) - (a + 3d) = 22 - 13 3d=93d = 9 d=93d = \frac{9}{3} d=3d = 3
  3. Tatua kwa kigezo cha kwanza (aa).

    Badilisha d=3d=3 katika Mlinganyo 1:

    a+3(3)=13a + 3(3) = 13 a+9=13a + 9 = 13 a=13−9a = 13 - 9 a=4a = 4
  4. Tafuta jumla ya vigezo 10 vya kwanza (S10S_{10}).

    Tumia fomula Sn=n2[2a+(n−1)d]S_n = \frac{n}{2}[2a + (n-1)d] na n=10n=10, a=4a=4, na d=3d=3.

    S10=102[2(4)+(10−1)3]S_{10} = \frac{10}{2}[2(4) + (10-1)3] S10=5[8+(9)3]S_{10} = 5[8 + (9)3] S10=5[8+27]S_{10} = 5[8 + 27] S10=5[35]S_{10} = 5[35] S10=175S_{10} = 175

    Jumla ya vigezo 10 vya kwanza ni 175\boxed{175}.

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