Hi
5/3 + 15/3 + 25/3 +....(10n-5)/3 = 125/3
solve for n
actually i know the what answer is
i wondering which way you prefer to solve for this
Thanks best regards
sum of an arithmetic series ...
$\displaystyle \sum_{i = 1}^n a_i = \frac{n}{2} (a_1 + a_n)$
$\displaystyle \sum_{i = 1}^n \frac{10i-5}{3} = \frac{n}{2} \left(\frac{5}{3} + \frac{10n-5}{3} \right) = \frac{125}{3}$
solve for $\displaystyle n$
Hello, blackjack21!
$\displaystyle \text{Solve for }n\!:\;\frac{5}{3} + \frac{15}{3} + \frac{25}{3} + \hdots + \frac{5(2n-1)}{3} \;=\;\frac{125}{3}$
Multiply by $\displaystyle \tfrac{3}{5}\!:\;\;1 + 3 + 5 + \hdots + (2n-1) \;=\;25$
The left side is an arithmetic series with first term $\displaystyle a=1$,
. . common difference $\displaystyle d = 2$, and $\displaystyle n$ terms.
Its sum is: .$\displaystyle S_n \:=\:\tfrac{n}{2}\big[2(1) + (n-1)2\big] \:=\:\tfrac{n}{2}\big[2 + 2n - 2] \:=\:n^2$
Therefore: .$\displaystyle n^2 \:=\:25 \quad\Rightarrow\quad n \:=\:5$