32,33,34,35 provide answers to those

Mathematics

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4 parts
32
64
33
384
34
10

103\frac{10}{3}

35
765

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11 steps

Haya, hebu tutatue maswali hayo.

Part 32· 3 steps

Kigezo cha 5 cha A.P. ni 19 na kigezo cha 12 ni 40. Tafuta kigezo cha 20.

Fomula ya kigezo cha n cha A.P. ni an=a+(n−1)da_n = a + (n-1)d.

  1. Andika milinganyo kutokana na taarifa uliyopewa.

    Kigezo cha 5: a5=a+(5−1)d=a+4d=19a_5 = a + (5-1)d = a + 4d = 19 (Mlinganyo 1)
    Kigezo cha 12: a12=a+(12−1)d=a+11d=40a_{12} = a + (12-1)d = a + 11d = 40 (Mlinganyo 2)

  2. Tatua kwa tofauti ya kawaida (dd) na kigezo cha kwanza (aa).

    Toa Mlinganyo 1 kutoka Mlinganyo 2:

    (a+11d)−(a+4d)=40−19(a + 11d) - (a + 4d) = 40 - 19 7d=217d = 21 d=217d = \frac{21}{7} d=3d = 3

    Badilisha d=3d=3 katika Mlinganyo 1:

    a+4(3)=19a + 4(3) = 19 a+12=19a + 12 = 19 a=19−12a = 19 - 12 a=7a = 7
  3. Tafuta kigezo cha 20 (a20a_{20}).

    a20=a+(20−1)da_{20} = a + (20-1)d a20=7+(19)3a_{20} = 7 + (19)3 a20=7+57a_{20} = 7 + 57 a20=64a_{20} = 64

    Kigezo cha 20 ni 64\boxed{64}.


Part 33384

Kigezo cha 3 cha G.P. ni 12 na kigezo cha 6 ni 96. Tafuta kigezo cha 8.

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=12a_3 = ar^{3-1} = ar^2 = 12 (Mlinganyo 1)
    Kigezo cha 6: a6=ar6−1=ar5=96a_6 = ar^{6-1} = ar^5 = 96 (Mlinganyo 2)

  2. Tatua kwa uwiano wa kawaida (rr) na kigezo cha kwanza (aa).

    Gawanya Mlinganyo 2 kwa Mlinganyo 1:

    ar5ar2=9612\frac{ar^5}{ar^2} = \frac{96}{12} r3=8r^3 = 8 r=83r = \sqrt[3]{8} r=2r = 2

    Badilisha r=2r=2 katika Mlinganyo 1:

    a(2)2=12a(2)^2 = 12 4a=124a = 12 a=124a = \frac{12}{4} a=3a = 3
  3. Tafuta kigezo cha 8 (a8a_8).

    a8=ar8−1a_8 = ar^{8-1} a8=3×(2)7a_8 = 3 \times (2)^7 a8=3×128a_8 = 3 \times 128 a8=384a_8 = 384

    Kigezo cha 8 ni 384\boxed{384}.


Part 3410

Jumla ya vigezo vya kwanza vya A.P. ni 200. Ikiwa kigezo cha kwanza ni 5 na kigezo cha mwisho ni 35, tafuta idadi ya vigezo na tofauti ya kawaida.

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

  1. Tafuta idadi ya vigezo (nn).

    Tunapewa Sn=200S_n = 200, a=5a = 5, na l=35l = 35.

    200=n2(5+35)200 = \frac{n}{2}(5 + 35) 200=n2(40)200 = \frac{n}{2}(40) 200=20n200 = 20n n=20020n = \frac{200}{20} n=10n = 10
  2. Tafuta tofauti ya kawaida (dd).

    Tunajua an=l=35a_n = l = 35, a=5a = 5, na n=10n = 10.

    35=5+(10−1)d35 = 5 + (10-1)d 35=5+9d35 = 5 + 9d 35−5=9d35 - 5 = 9d 30=9d30 = 9d d=309d = \frac{30}{9} d=103d = \frac{10}{3}

    Idadi ya vigezo ni 10\boxed{10} na tofauti ya kawaida ni 103\boxed{\frac{10}{3}}.


Part 35765

Kigezo cha 4 cha G.P. ni 24 na kigezo cha 7 ni 192. Tafuta jumla ya vigezo 8 vya kwanza.

Fomula ya kigezo cha n cha G.P. ni an=arn−1a_n = ar^{n-1}.
Fomula ya jumla ya vigezo n vya kwanza vya G.P. ni Sn=a(rn−1)r−1S_n = \frac{a(r^n - 1)}{r - 1} (kwa r≠1r \neq 1).

  1. Andika milinganyo kutokana na taarifa uliyopewa.

    Kigezo cha 4: a4=ar4−1=ar3=24a_4 = ar^{4-1} = ar^3 = 24 (Mlinganyo 1)
    Kigezo cha 7: a7=ar7−1=ar6=192a_7 = ar^{7-1} = ar^6 = 192 (Mlinganyo 2)

  2. Tatua kwa uwiano wa kawaida (rr) na kigezo cha kwanza (aa).

    Gawanya Mlinganyo 2 kwa Mlinganyo 1:

    ar6ar3=19224\frac{ar^6}{ar^3} = \frac{192}{24} r3=8r^3 = 8 r=83r = \sqrt[3]{8} r=2r = 2

    Badilisha r=2r=2 katika Mlinganyo 1:

    a(2)3=24a(2)^3 = 24 8a=248a = 24 a=248a = \frac{24}{8} a=3a = 3
  3. Tafuta jumla ya vigezo 8 vya kwanza (S8S_8).

    Tumia fomula Sn=a(rn−1)r−1S_n = \frac{a(r^n - 1)}{r - 1} na n=8n=8, a=3a=3, na r=2r=2.

    S8=3(28−1)2−1S_8 = \frac{3(2^8 - 1)}{2 - 1} S8=3(256−1)1S_8 = \frac{3(256 - 1)}{1} S8=3(255)S_8 = 3(255) S8=765S_8 = 765

    Jumla ya vigezo 8 vya kwanza ni 765\boxed{765}.

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