CAT 2023 Slot 1 Solved Paper

Question : 66 of 66

Marks: +1, -0

A lab experiment measures the number of organisms at 8 am every day. Starting with 2 organisms on
the first day, the number of organisms on any day is equal to 3 more than twice the number on the
previous day. If the number of organisms on the nth day exceeds one million, then the lowest possible value
of n is

Your Answer:

Validate

Solution:

Given on day-1, there are 2 organisms.
On day- 2 , there are

2 * 2+3=7

and on day -3 , there are

2 * 7+3=17

..
Let us try to form a pattern:

\;2=2+0(n=1)

\;7=4+3(n=2)

\;17=8+9 [8+3^{⋆} 3] (n=3)

\;37=16+21 [16+3^{⋆} 7] n=4

\;T(n)=2^n+3 (2^{n-1}-1)

We know that

2^{20}=2^{10} × 2^{10}=1024 × 1024

, which is more than 1 million.
Let us check for

{\n}=19

2^{19}+3 (2^{18}-1) =2^{19}+3 ⋅ 2^{18}-3=2 ⋅ 2^{19}+2^{18}-3=2^{20}+2^{18}-3

which is more than 1 million.
Let us check for

n=18

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