Sequence and SeriesPYQ Sept 25Question 4130 of 150
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The sum of all natural numbers between 200 and 600 those are divisible by 13 is

Options

A12493
B14493
C16493
D18493
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Correct Answer

✅ Option a — 12493

All Options:

  • A12493
  • B14493
  • C16493
  • D18493

Detailed Solution & Explanation

We need to find the sum of all natural numbers in the open interval (200,600)\displaystyle (200, 600) that are divisible by 13.
First, find the smallest multiple of 13 greater than 200: 200÷13≈15.38  ⟹  First multiple a=13×16=208200 \div 13 \approx 15.38 \implies \text{First multiple } a = 13 \times 16 = 208
Next, find the largest multiple of 13 less than 600: 600÷13≈46.15  ⟹  Last multiple l=13×46=598600 \div 13 \approx 46.15 \implies \text{Last multiple } l = 13 \times 46 = 598
This forms an arithmetic progression: 208,221,234,…,598\displaystyle 208, 221, 234, \dots, 598. Let n\displaystyle n be the number of terms. Using the formula l=a+(n−1)d\displaystyle l = a + (n-1)d where d=13\displaystyle d = 13: 598=208+(n−1)13598 = 208 + (n-1)13 390=13(n−1)390 = 13(n-1) n−1=39013=30  ⟹  n=31n-1 = \frac{390}{13} = 30 \implies n = 31
The sum of an AP is given by: Sn=n2(a+l)S_n = \frac{n}{2}(a + l) S31=312(208+598)=312(806)=31×403S_{31} = \frac{31}{2}(208 + 598) = \frac{31}{2}(806) = 31 \times 403 S31=12,493S_{31} = 12,493
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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