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NDA 2019
Exam
NDA
Year
2019
Subject
Mathematics
Chapter
Sequences and Series

NDA 2019 · Mathematics · Sequences and Series

The sum of (p+q)^ th and (p-q)^ th terms of an AP is equal to

The sum of (p+q)th (p+q)^{\text{th }} and (p−q)th (p-q)^{\text{th }} terms of an AP is equal to

Options

  1. A(2p)th (2 \mathrm{p})^{\text{th }} term
  2. B(2q)th (2 q)^{\text{th }} term
  3. CTwice the pth \mathrm{p}^{\text{th }} term
  4. DTwice the qth q^{\text{th }} term

Correct answer

C. Twice the pth \mathrm{p}^{\text{th }} term

Step-by-step solution

Let the first term of AP=aLet\mathrm{AP}=\mathrm{a} Let the commondifferenceofcommon difference of\mathrm{AP}=\mathrm{d}(p+q)th  term =Tp+q=a+(p+q−1)d(p−q)th  term =Tp−q=a+(p−q−1)dTp+q+Tp−q=a+(p+q−1)d+a+(p−q−1)d=2a+(p+q−1+p−q−1)d=2a+(2p−2)d=2[a+(p−1)d]=2Tp\begin{aligned} & (\mathrm{p}+\mathrm{q})^{\text{th }} \text{ term }=\mathrm{T}_{\mathrm{p}+\mathrm{q}}=\mathrm{a}+(\mathrm{p}+\mathrm{q}-1) \mathrm{d} \\ & (\mathrm{p}-\mathrm{q})^{\text{th }} \text{ term }=T_{p-\mathrm{q}}=\mathrm{a}+(\mathrm{p}-\mathrm{q}-1) \mathrm{d} \\ & \mathrm{T}_{\mathrm{p}+\mathrm{q}}+T_{p-q}=\mathrm{a}+(\mathrm{p}+\mathrm{q}-1) \mathrm{d}+\mathrm{a}+(\mathrm{p}-\mathrm{q}-1) \mathrm{d} \\ & =2 \mathrm{a}+(\mathrm{p}+\mathrm{q}-1+\mathrm{p}-\mathrm{q}-1) \mathrm{d} \\ & =2 \mathrm{a}+(2 \mathrm{p}-2) \mathrm{d}=2[\mathrm{a}+(\mathrm{p}-1) \mathrm{d}]=2 \mathrm{ T}_{\mathrm{p}}\end{aligned}
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