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CAT 2019 Slot 1 — QA Question 21

Geometric ProgressionEasy

If the population of a town is p in the beginning of any year then it becomes 3 + 2p in the beginning of the next year. If the population in the beginning of 2019 is 1000, then the population in the beginning of 2034 will be

Answer & solution

  • A

    (1003)15 + 6

  • B

    (997)15 - 3

  • (1003)215 - 3

  • D

    (997)214 + 3

Solution

Easy

The recurrence pn+1=2pn+3p_{n+1}=2p_n+3 is linear. Iterate a few times to spot the pattern: the pp-term picks up a factor of 2 each year while the constant 3 accumulates as a geometric series. After 15 years, sum that series.

1

Iterate the recurrence. Start from pp (in 2019) and apply →2(⋅)+3\to 2(\cdot)+3.

yr 1: 2p+3yr 2: 2(2p+3)+3=22p+2⋅3+3yr 3: 23p+22⋅3+2⋅3+3\begin{aligned} &\text{yr }1:\ 2p+3\\ &\text{yr }2:\ 2(2p+3)+3=2^2p+2\cdot3+3\\ &\text{yr }3:\ 2^3p+2^2\cdot3+2\cdot3+3 \end{aligned}
2

General term after nn years. The pattern from step 1 generalises.

pn=2np+3(2n−1+2n−2+⋯+1)\begin{aligned} &p_n=2^np+3\big(2^{n-1}+2^{n-2}+\cdots+1\big) \end{aligned}
3

Plug in n=15n=15 (2034) and sum the geometric series. The bracket is 215−12^{15}-1.

p15=215p+3 (215−1)(2034 is 15 yrs after 2019)⇒ p15=215(1000)+3⋅215−3(p=1000)⇒ p15=215(1000+3)−3=(1003) 215−3\begin{aligned} &p_{15}=2^{15}p+3\,(2^{15}-1)\quad\text{(2034 is 15 yrs after 2019)}\\ &\Rightarrow\ p_{15}=2^{15}(1000)+3\cdot2^{15}-3\quad\text{(}p=1000\text{)}\\ &\Rightarrow\ p_{15}=2^{15}(1000+3)-3=(1003)\,2^{15}-3 \end{aligned}
Population in 2034=(1003) 215−3\text{Population in 2034}=(1003)\,2^{15}-3
CAT 2019 Slot 1 QA Q21: If the population of a town is p in the beginning of any year then it becomes 3 + 2p in the beginning of the n — Solution | TheCATExam