IPMAT Indore 2023 · Quantitative Ability · Question 5

Step 1 — generate the first few terms.

f\left( 1 \right)=1

f\left( 2 \right)=6-1=5

f\left( 3 \right)=9-5=4

f\left( 4 \right)=12-4=8

f\left( 5 \right)=15-8=7

f\left( 6 \right)=18-7=11

f\left( 7 \right)=21-11=10

Step 2 — look at the odd-indexed terms only (since 2023 is odd):

f\left( 1 \right)=1, f\left( 3 \right)=4, f\left( 5 \right)=7, f\left( 7 \right)=10, \ldots

These form an arithmetic progression with common difference 3.

Step 3 — justify the pattern. Add two consecutive instances of the recurrence:

f\left( n \right)=3n-f\left( n-1 \right) \text{and}f\left( n-1 \right)=3\left( n-1 \right)-f\left( n-2 \right)

Substituting the second into the first:

f\left( n \right)=3n-3\left( n-1 \right)+f\left( n-2 \right)=f\left( n-2 \right)+3

So every term exceeds the one two places before it by exactly 3 — confirming the AP.

Step 4 — evaluate. Writing 2023=1+2k gives k=1011, and

f\left( 2023 \right)=f\left( 1 \right)+3k=1+3\left( 1011 \right)=1+3033=3034

Answer: 3034

Shortcut. Turn the recurrence into a two-step one. From f\left( n \right)+f\left( n-1 \right)=3n and f\left( n-1 \right)+f\left( n-2 \right)=3\left( n-1 \right), subtracting gives

f\left( n \right)-f\left( n-2 \right)=3

So the odd terms are an AP with difference 3 starting at f\left( 1 \right)=1, and

f\left( 2023 \right)=1+3\cdot \frac{2023-1}{2}=1+3033=3034

The reusable move: whenever a recurrence has the form f\left( n \right)+f\left( n-1 \right)=\left( \text{linear in}n \right), subtract consecutive instances. The alternating sign cancels and you are left with a clean arithmetic progression on every second term.