Q154 / 270 · NDA 2017 (Phase 2) · Mathematics

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

  • A
    $\mathrm{P}+\mathrm{Q}$
  • B
    $2 \mathrm{P}+3 \mathrm{Q}$
  • C
    $2 Q$
  • Correct answer
    D
    $Q$

Solution

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

Hence, the correct answer is option: 'D'.

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