Question

The number of bacteria increases in the geometric progression of 2^(n+1) per minute. If there are 2 bacteria to start with, find:
(i) the number of bacteria in the ninth minute
(ii) the total number of bacteria after 9 minutes.

Answer 1

Answer is (i) = 512 Nos. & (ii) = 2046 Nos.

Solution:
a = 2^(n+1)
& first term a1 = 2

For the 1st minute, I.e., n = 1
Therefore, second term a2 = 2^(1+1) = 2^2 = 4

similarly, when n = 2,
a3 = 2^(2+1) = 2^3 = 8

similarly, when n = 8,
a9 = 2^(8+1) = 2^9 = 512.

similarly, when n = 9,
a10 = 2^(9+1) = 2^10 = 1024

Therefore, no. of bacteria in the 9th minute
= a10 - a9
= 1024 - 512
= 512 Nos.


.
.
.
Since, common ratio r = a2/a1
Therefore, r = 4/2 = 2
Now, we know that sum of n terms is given by
Sn = {a1 × (1-r^n)} / (1-r)

Therefore, sum of the first 10 terms is,
S10 = {2 × (1-2^10)} / (1-2)
or S10 = (-2046)/(-1)
or S10 = 2046 Nos.

Hence, total no. of bacteria after 9 minutes are 2046 Nos.

Answer 2

Answer:

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