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NDA 2017
Exam
NDA
Year
2017
Subject
Mathematics
Chapter
Sequences and Series

NDA 2017 · Mathematics · Sequences and Series

If S_ n =nP+ n(n-1) Q 2 , where S_ n denotes the sum of the first n...

If Sn=nP+n(n−1)Q2\mathrm{S}_{\mathrm{n}}=\mathrm{nP}+\frac{\mathrm{n}(\mathrm{n}-1) \mathrm{Q}}{2}, where Sn\mathrm{S}_{\mathrm{n}} denotes the sum of the first n terms of an AP , then the common difference is

Options

  1. AP+Q\mathrm{P}+\mathrm{Q}
  2. B2P+3Q2 \mathrm{P}+3 \mathrm{Q}
  3. C2Q2 Q
  4. DQQ

Correct answer

D. QQ

Step-by-step solution

Sn=np+n(n−1)Q2\mathrm{S}_{\mathrm{n}}=\mathrm{np}+\frac{\mathrm{n}(\mathrm{n}-1) \mathrm{Q}}{2} We know, T1=S1\mathrm{T}_1=\mathrm{S}_1 and T2=S2−S1\mathrm{T}_2=\mathrm{S}_2-\mathrm{S}_1 Common difference (d)=T2−T1∴S1=(1)P+1(1−1)Q2=P+0=PS2=(2)P+2(2−1)Q2=2P+2Q2=2P+Q(\mathrm{d})=\mathrm{T}_2-\mathrm{T}_1\therefore \mathrm{S}_1=(1) \mathrm{P}+\frac{1(1-1) \mathrm{Q}}{2}=\mathrm{P}+0=\mathrm{P}\mathrm{S}_2=(2) \mathrm{P}+\frac{2(2-1) \mathrm{Q}}{2}=2 \mathrm{P}+\frac{2 \mathrm{Q}}{2}=2 \mathrm{P}+\mathrm{Q}. ∴T1=P;T2=2P+Q−P=P+Q∴ Common difference (d)=T2−T1\begin{aligned} & \therefore \mathrm{T}_1=\mathrm{P} ; \mathrm{T}_2=2 \mathrm{P}+\mathrm{Q}-\mathrm{P}=\mathrm{P}+\mathrm{Q} \\ & \therefore \text{ Common difference }(\mathrm{d})=\mathrm{T}_2-\mathrm{T}_1\end{aligned}=\mathrm{P}+\mathrm{Q}-\mathrm{P}=\mathrm{Q}$
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