Sequence and SeriesMCQPYQ Jun 23Question 1773 of 212
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If pth\displaystyle p^{th} term of an AP is q\displaystyle q and its qth\displaystyle q^{th} term is p\displaystyle p, then what will be the value of (p+q)th\displaystyle (p+q)^{th} term?

Options

A0\displaystyle 0
B1\displaystyle 1
Cp+q1\displaystyle p+q-1
D2(p+q1)\displaystyle 2(p+q-1)
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Correct Answer

Option a0\displaystyle 0

All Options:

  • A0\displaystyle 0
  • B1\displaystyle 1
  • Cp+q1\displaystyle p+q-1
  • D2(p+q1)\displaystyle 2(p+q-1)

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Detailed Solution & Explanation

Let the first term be a\displaystyle a and common difference be d\displaystyle d.
We have:
tp=a+(p1)d=qt_p = a + (p-1)d = q
tq=a+(q1)d=pt_q = a + (q-1)d = p
Subtracting the second equation from the first:
(pq)d=qp    d=1(p-q)d = q-p \implies d = -1
Substitute d=1\displaystyle d = -1:
ap+1=q    a=p+q1a - p + 1 = q \implies a = p + q - 1

Now, find the (p+q)th\displaystyle (p+q)^{\text{th}} term:
tp+q=a+(p+q1)d=(p+q1)+(p+q1)(1)=0t_{p+q} = a + (p+q-1)d = (p+q-1) + (p+q-1)(-1) = 0
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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