IPMAT Indore 2023 - QA (SA) Q5 Explanation

Explanation:

3034 

(3034) Given f(1) = 1

and f(n) = 3n - f(n - 1) | for all integers n > 1

Put n = 2 in the given functional equation we get,

f(2) = 3 × 2 - f(1) = 6 - 1 = 5.

Put n = 3, we get,

f (3) = 3 × 3 - f(2) = 9 - 5 = 4

Put n = 4, we get,

f(4) = 3 × 4 - f(3) = 12 - 4 = 8

Similarly, f(5) = 3 × 5 - f(4) = 15 - 8 = 7 and so on.

f(1), f(2), f(3) f(4) etc.

Arrange these values in a sequence, we observe…them as terms of 2 ways Arithmetic Progressions, each with common difference = 3.

As we have to find value of f(2023), let's take the first AP.

f(1), f(3), f(5) ……. … .. where n is odd.

⇒1, 4, 7 …. .. …… ..

The position of f(1023) in this AP will be 1012 (2023+1 2 = 1012)

Using the formula of general term a𝑛=𝑎+(𝑛−1)𝑑, where ‘a’ is the first term & ‘d’ the common difference.’

f(2023) = 𝑎1012 = 1 + (1012−1)3 = 1 + 1011 × 3

= 1 + 3033

f(2023) = 𝑎1012 = 3034. Ans