Airthmetic Progression

Nice and long.

$5+13+21.....+181$

Introducing the sum formula: ${ a }_{ n }={ a }_{ 1 }+(n-1)d$

Introducing the arithmetic series formula: ${ S }_{ n }=\frac { n({ a }_{ 1 }+{ a }_{ n }) }{ 2 }$

where: D=common difference n=Term number ${ a }_{ 1 }$ =first term ${ a }_{ n }$ =Term n ${ S }_{ n }$ =Sum to n

$1=5+(n-1)8$

$n=23$

${ S }_{ 23 }=\frac { 23(5+181) }{ 2 } =2139$

The sum $S$ of an AP of $n$ terms starting with $a$ and end with $l$ is given by:

$S = \dfrac {n}{2} (a+l)$

It can be seen that the $i^{th}$ tern of the series $\{5, 13, 21, ... 181\}$ is $a_i = 8i-3$ for $i = 1,2,3,...$

The number of terms of the series, $n = \dfrac {181+3} {8} = 23$

Therefore,

$S = 5 + 13 + 21 + ... + 181 = \dfrac {n}{2} (a+l) = \dfrac {23}{2} (5+181) = \boxed{2139}$

Common Difference = 8 No. of terms = 23 Sum = 12.5 (5+181) = 2139