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The first term of an arithmetic progression is 4 and the last term is 20. The sum of the terms is 252. Calculate the number of terms and the common difference of the progression

Mathematics

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2 parts
5

45\frac{4}{5}

0.8
6

80 dakika80 \text{ dakika}

Step-by-step solution

5 steps

Haya, nitasuluhisha maswali yote mawili.

Part 5· 2 steps

Tunapewa:
Nambari ya kwanza (aa) = 4
Nambari ya mwisho (ll) = 20
Jumla ya nambari (SnS_n) = 252

Tunahitaji kukokotoa idadi ya nambari (nn) na tofauti ya kawaida (dd).

  1. Tafuta idadi ya nambari (nn) kwa kutumia fomula ya jumla ya mfuatano wa hesabu.

    Fomula ya jumla ni Sn=n2(a+l)S_n = \frac{n}{2}(a + l).

    252=n2(4+20)252 = \frac{n}{2}(4 + 20) 252=n2(24)252 = \frac{n}{2}(24) 252=12n252 = 12n

    Kugawanya pande zote mbili kwa 12:

    n=25212n = \frac{252}{12} n=21n = 21

    Idadi ya nambari ni 21.

  2. Tafuta tofauti ya kawaida (dd) kwa kutumia fomula ya nambari ya mwisho.

    Fomula ya nambari ya mwisho ni l=a+(n−1)dl = a + (n-1)d.

    20=4+(21−1)d20 = 4 + (21-1)d 20=4+20d20 = 4 + 20d

    Toa 4 kutoka pande zote mbili:

    20−4=20d20 - 4 = 20d 16=20d16 = 20d

    Gawanya pande zote mbili kwa 20:

    d=1620d = \frac{16}{20}

    Kurahisisha sehemu:

    d=45d = \frac{4}{5}

    Tofauti ya kawaida ni 45\boxed{\frac{4}{5}} au 0.8\boxed{0.8}.

Part 6

80 dakika80 \text{ dakika}

Tunapewa:
Idadi ya awali ya bakteria (P0P_0) = 2500
Idadi huongezeka kwa 60% kila baada ya dakika 20.
Idadi inayolengwa (PtP_t) = 16384 bakteria

Tunahitaji kujua itachukua muda gani kufikia idadi inayolengwa.

  1. Tambua kiwango cha ukuaji na uanzishe fomula.

    Kiwango cha ukuaji kwa kila dakika 20 ni 1+0.60=1.61 + 0.60 = 1.6.
    Fomula ya ukuaji wa kielelezo ni Pt=P0×(kiwango_cha_ukuaji)kP_t = P_0 \times (kiwango\_cha\_ukuaji)^k, ambapo kk ni idadi ya vipindi vya dakika 20.

    16384=2500×(1.6)k16384 = 2500 \times (1.6)^k
  2. Tatua kwa kk.

    Gawanya pande zote mbili kwa 2500:

    163842500=(1.6)k\frac{16384}{2500} = (1.6)^k 6.5536=(1.6)k6.5536 = (1.6)^k

    Ili kupata kk, tunaweza kutumia logariti au kujaribu nguvu za 1.6:
    1.61=1.61.6^1 = 1.6
    1.62=2.561.6^2 = 2.56
    1.63=4.0961.6^3 = 4.096
    1.64=6.55361.6^4 = 6.5536
    Hivyo, k=4k = 4.

  3. Kokotoa muda wote.

    Kwa kuwa kk ni idadi ya vipindi vya dakika 20, muda wote ni k×20k \times 20 dakika.
    Muda = 4×20=804 \times 20 = 80 dakika.
    Itachukua 80 dakika\boxed{80 \text{ dakika}} kufikia idadi ya bakteria 16384.

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