Sequence and SeriesMCQMTP June 24 Series IQuestion 1815 of 213
All Questions

Find the sum of all natural numbers betn $250\displaystyle and $1000 which are exactly divisible by 3\displaystyle 3:

Select Your Answer (Keyboard: A, B, C, D)
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Correct Answer

✅ Option a — $1,56,375$

All Options:

  • A$1,56,375$
  • B$1,56,357$
  • C$1,65,375$
  • D$1,65,357$

Detailed Solution & Explanation

The natural numbers between $250\displaystyle and $1000 that are divisible by 3\displaystyle 3 form an A.P. series:
252,255,258,…,999252, 255, 258, \dots, 999
Here:
First term a=252\displaystyle a = 252
Common difference d=3\displaystyle d = 3
Last term l=999\displaystyle l = 999.

Using the last term formula:
l=a+(n−1)dl = a + (n-1)d
999=252+(n−1)3999 = 252 + (n-1)3
747=(n−1)3747 = (n-1)3
n−1=249  ⟹  n=250n-1 = 249 \implies n = 250

Now, find the sum of these $250$ terms:
S250=2502[a+l]S_{250} = 2502\displaystyle \frac{250}{2} [a + l]
S250=125[252+999]S_{250} = 125 [252 + 999]
S250=125[1251]=156375S_{250} = 125 [1251] = 156375
Hence, **Option A** is the correct answer.

About This Chapter: Sequence and Series

Paper

Paper 3: Quantitative Aptitude

Weightage

4-6 Marks

Key Topics

Arithmetic & Geometric Progressions

This chapter covers Arithmetic Progressions (AP) and Geometric Progressions (GP). Students learn how to find the nth term, sum of n terms, arithmetic/geometric means, and sum to infinity of a GP.

View Official ICAI Syllabus

Exam Strategy Tip

For complex 'sum of series' questions, a great hack is to substitute n = 1 and n = 2 into the question and the options to see which one matches, completely bypassing the formula.

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