Sequence and SeriesMCQMTP June 24 Series IQuestion 1815 of 212
All Questions

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

Options

A1,56,375\displaystyle 1,56,375
B1,56,357\displaystyle 1,56,357
C1,65,375\displaystyle 1,65,375
D1,65,357\displaystyle 1,65,357
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Correct Answer

Option a1,56,375\displaystyle 1,56,375

All Options:

  • A1,56,375\displaystyle 1,56,375
  • B1,56,357\displaystyle 1,56,357
  • C1,65,375\displaystyle 1,65,375
  • D1,65,357\displaystyle 1,65,357

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Detailed Solution & Explanation

The natural numbers between 250\displaystyle 250 and 1000\displaystyle 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+(n1)dl = a + (n-1)d
999=252+(n1)3999 = 252 + (n-1)3
747=(n1)3747 = (n-1)3
n1=249    n=250n-1 = 249 \implies n = 250

Now, find the sum of these 250\displaystyle 250 terms:
S250=2502[a+l]S_{250} = \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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