The sum of the first [m]n[/m] positive integers is given by

Question

The sum of the first  n  positive integers is  \frac{n(n+1)}{2} . For each positive integer  n , a sequence  a_1, a_2, a_3, \ldots  is defined as follows:

a_n = -n   if  n  is a multiple of 3, and  a_n = n   otherwise

What is the sum of the first  33  terms of this sequence?

A. 99
B. 165
C. 260
D. 363
E. 561

rajdeep212003 · Answer

Hlo, I’m not fully getting what the sequence definition is trying to say here.

Especially the part where aₙ is defined — a bit confusing how the values are being assigned when n is a multiple of 3 vs otherwise.

Could you please explain the question once in simpler terms or maybe walk through a small example?

That would really help.

kevincan · Answer

Sorry, I was editing the question as you were posting. Hopefully it’s clearer now.

rajdeep212003 · Answer

Thank you for it, and here is the solution.

Using the given formula:

Sum of first n integers = n(n+1)/2

So for n = 33:

Total sum (1 to 33) = 33×34/2 = 561

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Now consider multiples of 3:

3, 6, 9, ..., 33

We can write them as:
3 × (1, 2, 3, ..., 11)

---

Sum of (1 to 11):

= 11×12/2 = 66

So sum of multiples of 3:

= 3 × 66 = 198

---

Now important idea:

These multiples were counted as +ve in 561,
but in the sequence they are −ve.

So we need to subtract them twice:

Final sum = 561 − 2×198
= 561 − 396
= 165

---

Final Answer: 165 (Option B)

---

Key shortcut insight:

Instead of building the whole sequence,
→ take normal sum
→ subtract twice the multiples of 3
I feel that is the fastest solution .

---

– Rajdeep

The sum of the first [m]n[/m] positive integers is given by

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The sum of the first [m]n[/m] positive integers is given by

Prep Toolkit

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