What is the maximum value of the sum of the numbers 36, 33, 30, 27, 24, ...?

Given:

The sequence is: 36, 33, 30, 27, 24, ...

We need to find the maximum value of the sum of the numbers.

This is an arithmetic progression (AP) where:

First term (a) = 36

Common difference (d) = 33 - 36 = -3

The sum of n terms of an AP is given by:

Formula used:

Sum (S_n) = \(\dfrac{n}{2}[2a + (n-1)d]\)

Calculation:

We need to find the maximum value of the sum. The AP ends when the last term becomes ≥ 0.

Last term (l) = a + (n-1)d

For the last term to be ≥ 0:

⇒ 36 + (n-1)(-3) ≥ 0

⇒ 36 - 3(n-1) ≥ 0

⇒ 36 - 3n + 3 ≥ 0

⇒ 39 - 3n ≥ 0

⇒ 3n ≤ 39

⇒ n ≤ 13

Thus, the sequence has a maximum of n = 13 terms.

Now, calculate the sum of these 13 terms:

S₁₃ = \(\dfrac{13}{2}[2(36) + (13-1)(-3)]\)

⇒ S₁₃ = \(\dfrac{13}{2}[72 + 12(-3)]\)

⇒ S₁₃ = \(\dfrac{13}{2}[72 - 36]\)

⇒ S₁₃ = \(\dfrac{13}{2}[36]\)

⇒ S₁₃ = 13 × 18

⇒ S₁₃ = 234

∴ The correct answer is option (3).