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

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- Nov 4th 2012, 03:42 PMblackjack21Basic sum of terms equation
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 - Nov 4th 2012, 04:07 PMskeeterRe: Basic sum of terms equation
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 $\displaystyle n$ - Nov 4th 2012, 04:21 PMpickslidesRe: Basic sum of terms equation
the second term is 15/13 or 15/3 ?

- Nov 4th 2012, 06:49 PMSorobanRe: Basic sum of terms equation
Hello, blackjack21!

Quote:

$\displaystyle \text{Solve for }n\!:\;\frac{5}{3} + \frac{15}{3} + \frac{25}{3} + \hdots + \frac{5(2n-1)}{3} \;=\;\frac{125}{3}$

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: .$\displaystyle n^2 \:=\:25 \quad\Rightarrow\quad n \:=\:5$

- Nov 5th 2012, 01:57 AMblackjack21Re: Basic sum of terms equation
- Nov 5th 2012, 01:59 AMblackjack21Re: Basic sum of terms equation
thanx for the two diffrent solution better for me

Thanks and best regards