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NDA 2024
Exam
NDA
Year
2024
Subject
Mathematics
Chapter
Sequences and Series

NDA 2024 · Mathematics · Sequences and Series

If p times the p^ th term of an AP is equal to q times the q^ th...

If pp times the pth p^{\text{th }} term of an AP is equal to qq times the qth q^{\text{th }} term (p≠q)(p \neq q), then what is the (p+q)th (p+q)^{\text{th }} term equal to?

Options

  1. A00
  2. Bp+qp+q
  3. Cpqp q
  4. Dpq(p+q)p q(p+q)

Correct answer

A. 00

Step-by-step solution

Suppose 'a' and 'd' be the first term and common difference of AP respectively. ∵nth  term =Tn=a+(n−1)⋅d∴(p+q)th  term =Tp+q=a+(p+q−1)d....(i)\begin{aligned} & \because \mathrm{n}^{\text{th }} \text{ term }=\mathrm{T}_{\mathrm{n}}=\mathrm{a}+(\mathrm{n}-1) \cdot \mathrm{d}\\ & \therefore \quad(\mathrm{p}+\mathrm{q})^{\text{th }} \text{ term }=\mathrm{T}_{\mathrm{p}+\mathrm{q}}=\mathrm{a}+(\mathrm{p}+\mathrm{q}-1) \mathrm{d}....(i) \end{aligned} Given, p⋅Tp=q⋅Tq\mathrm{p} \cdot \mathrm{T}_{\mathrm{p}}=\mathrm{q} \cdot \mathrm{T}_{\mathrm{q}}⇒p⋅[a+(p−1)d]=q⋅[a+(q−1)d]⇒(p−q)[a+(p+q−1)d]=0⇒[a+(p+q−1)d]=0{∵p≠q}⇒Tp+q=0{ from (i)}\begin{aligned} & \Rightarrow \mathrm{p} \cdot[\mathrm{a}+(\mathrm{p}-1) \mathrm{d}]=\mathrm{q} \cdot[\mathrm{a}+(\mathrm{q}-1) \mathrm{d}] \\ & \Rightarrow(\mathrm{p}-\mathrm{q})[\mathrm{a}+(\mathrm{p}+\mathrm{q}-1) \mathrm{d}]=0 \\ & \Rightarrow[\mathrm{a}+(\mathrm{p}+\mathrm{q}-1) \mathrm{d}]=0 \quad\{\because \mathrm{p} \neq \mathrm{q}\} \\ & \Rightarrow \mathrm{T}_{\mathrm{p}+\mathrm{q}}=0 \quad\{\text{ from }(\mathrm{i})\} \end{aligned}$$
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