Question 36

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Jambmaths question: 

The sum of first n positive is

Option A: 

$\tfrac{1}{2}n(n-1)$

Option B: 

$n(n+1)$

Option C: 

$n(n-1)$

Option D: 

$\tfrac{1}{2}n(n+1)$

Jamb Maths Solution: 

$\begin{align}  & 1+2+3+4+---+n \\ & {{T}_{n}}=l=n \\ & a=1,\text{ }d=2-1=1 \\ & {{S}_{n}}=\tfrac{n}{2}\left[ a+l \right]=\tfrac{n}{2}\left[ 1+n \right] \\ & {{S}_{n}}=\tfrac{n}{2}\left[ n+1 \right] \\\end{align}$

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