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2 parts
5

45\frac{4}{5}

0.8
6

80 minutes80 \text{ minutes}

Step-by-step solution

5 steps

Okay, I will solve both questions in English.

Part 5· 2 steps

Given:
First term (aa) = 4
Last term (ll) = 20
Sum of terms (SnS_n) = 252

We need to calculate the number of terms (nn) and the common difference (dd).

  1. Find the number of terms (nn) using the formula for the sum of an arithmetic progression.

    The sum formula is 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

    Divide both sides by 12:

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

    The number of terms is 21.

  2. Find the common difference (dd) using the formula for the last term.

    The formula for the last term is l=a+(n−1)dl = a + (n-1)d.

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

    Subtract 4 from both sides:

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

    Divide both sides by 20:

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

    Simplify the fraction:

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

    The common difference is 45\boxed{\frac{4}{5}} or 0.8\boxed{0.8}.

Part 6

80 minutes80 \text{ minutes}

Given:
Initial number of bacteria (P0P_0) = 2500
The number increases by 60% every 20 minutes.
Target number (PtP_t) = 16384 bacteria

We need to find out how long it will take to reach the target number.

  1. Determine the growth factor and set up the formula.

    The growth factor per 20-minute period is 1+0.60=1.61 + 0.60 = 1.6.
    The exponential growth formula is Pt=P0×(growth factor)kP_t = P_0 \times (\text{growth factor})^k, where kk is the number of 20-minute periods.

    16384=2500×(1.6)k16384 = 2500 \times (1.6)^k
  2. Solve for kk.

    Divide both sides by 2500:

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

    To find kk, we can test powers of 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
    Thus, k=4k = 4.

  3. Calculate the total time.

    Since kk is the number of 20-minute periods, the total time is k×20k \times 20 minutes.
    Time = 4×20=804 \times 20 = 80 minutes.
    It will take 80 minutes\boxed{80 \text{ minutes}} to reach 16384 bacteria.

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