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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 $\mathrm{n}$ terms of an AP, then the common difference is $[2017-I I]$
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first $\mathrm{n}$ terms of an AP, then the common difference is $[2017-I I]$
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1651 Upvotes
Verified Answer
The correct answer is:
$Q$
$\mathrm{S}_{\mathrm{n}}=\mathrm{np}+\frac{\mathrm{n}(\mathrm{n}-1) \mathrm{Q}}{2}$
We know, $\mathrm{T}_{1}=\mathrm{S}_{1}$ and $\mathrm{T}_{2}=\mathrm{S}_{2}-\mathrm{S}_{1}$
Common difference $(\mathrm{d})=\mathrm{T}_{2}-\mathrm{T}_{1}$
$\therefore \mathrm{S}_{1}=(1) \mathrm{P}+\frac{1(1-1) \mathrm{Q}}{2}=\mathrm{P}+0=\mathrm{P}$
$\mathrm{S}_{2}=(2) \mathrm{P}+\frac{2(2-1) \mathrm{Q}}{2}=2 \mathrm{P}+\frac{2 \mathrm{Q}}{2}=2 \mathrm{P}+\mathrm{Q}$
$\therefore \mathrm{T}_{1}=\mathrm{P} ; \mathrm{T}_{2}=2 \mathrm{P}+\mathrm{Q}-\mathrm{P}=\mathrm{P}+\mathrm{Q}$
$\therefore$ Common difference $(\mathrm{d})=\mathrm{T}_{2}-\mathrm{T}_{1}$
$=\mathrm{P}+\mathrm{Q}-\mathrm{P}=\mathrm{Q}$
We know, $\mathrm{T}_{1}=\mathrm{S}_{1}$ and $\mathrm{T}_{2}=\mathrm{S}_{2}-\mathrm{S}_{1}$
Common difference $(\mathrm{d})=\mathrm{T}_{2}-\mathrm{T}_{1}$
$\therefore \mathrm{S}_{1}=(1) \mathrm{P}+\frac{1(1-1) \mathrm{Q}}{2}=\mathrm{P}+0=\mathrm{P}$
$\mathrm{S}_{2}=(2) \mathrm{P}+\frac{2(2-1) \mathrm{Q}}{2}=2 \mathrm{P}+\frac{2 \mathrm{Q}}{2}=2 \mathrm{P}+\mathrm{Q}$
$\therefore \mathrm{T}_{1}=\mathrm{P} ; \mathrm{T}_{2}=2 \mathrm{P}+\mathrm{Q}-\mathrm{P}=\mathrm{P}+\mathrm{Q}$
$\therefore$ Common difference $(\mathrm{d})=\mathrm{T}_{2}-\mathrm{T}_{1}$
$=\mathrm{P}+\mathrm{Q}-\mathrm{P}=\mathrm{Q}$
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