Sequence and SeriesMCQPYQ Nov. 20Question 1760 of 212
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Divide 69\displaystyle 69 into 3\displaystyle 3 parts which are in A.P. and are such that the product of first two parts is 483\displaystyle 483.

Options

A20,23,26\displaystyle 20, 23, 26
B21,23,25\displaystyle 21, 23, 25
C19,23,27\displaystyle 19, 23, 27
D22,23,24\displaystyle 22, 23, 24
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Correct Answer

Option b21,23,25\displaystyle 21, 23, 25

All Options:

  • A20,23,26\displaystyle 20, 23, 26
  • B21,23,25\displaystyle 21, 23, 25
  • C19,23,27\displaystyle 19, 23, 27
  • D22,23,24\displaystyle 22, 23, 24

Ad

Detailed Solution & Explanation

Let the three parts of 69\displaystyle 69 in A.P. be:
ad, a, a+da-d, \ a, \ a+d
Since their sum is 69\displaystyle 69:
(ad)+a+(a+d)=69(a-d) + a + (a+d) = 69
3a=69    a=233a = 69 \implies a = 23

Given that the product of the first two parts is 483\displaystyle 483:
(ad)a=483(a-d) \cdot a = 483
Substitute a=23\displaystyle a = 23:
(23d)23=483(23-d) \cdot 23 = 483
23d=21    d=223-d = 21 \implies d = 2

Thus, the three parts are:
ad=232=21a-d = 23-2 = 21
a=23a = 23
a+d=23+2=25a+d = 23+2 = 25
So the parts are 21,23,25\displaystyle 21, 23, 25.
Hence, **Option B** 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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