How many terms at least of the arithmetic progression 1,4,7,10,.............are needed to give a sum greater than 590,starting from the first term of the arithmetic progression?

please help me(Crying)

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- Jan 8th 2010, 05:02 PMmastermin346Arithmetic progression problem
How many terms at least of the arithmetic progression 1,4,7,10,.............are needed to give a sum greater than 590,starting from the first term of the arithmetic progression?

please help me(Crying) - Jan 8th 2010, 05:29 PMskeeter
- Jan 8th 2010, 08:41 PMmastermin346
- Jan 8th 2010, 09:14 PMProve It
Do as you have been instructed.

$\displaystyle S_n = \frac{n}{2}(t_1 + t_n)$

$\displaystyle = \frac{n}{2}[t_1 + t_1 + (n - 1)d]$

$\displaystyle = \frac{n}{2}[2t_1 + (n - 1)d]$

You should be able to see that $\displaystyle t_1 = 1$ and $\displaystyle d = 3$

So $\displaystyle S_n = \frac{n}{2}[2(1) + 3(n - 1)]$

$\displaystyle = \frac{n}{2}(2 + 3n - 3)$

$\displaystyle = \frac{n}{2}(3n - 1)$

$\displaystyle = \frac{3n^2}{2} - \frac{n}{2}$.

You know that $\displaystyle S_n > 590$

So $\displaystyle \frac{3n^2}{2} - \frac{n}{2} > 590$

Solve for $\displaystyle n$.

Choose the first whole value of $\displaystyle n$ that is greater than what you found... - Jan 8th 2010, 10:00 PMmastermin346
what do you mean by "Choose the first whole value of http://www.mathhelpforum.com/math-he...b31363a1-1.gif that is greater than what you found"???

- Jan 8th 2010, 10:23 PMmr fantastic
You have been given almost the entire solution. Your job is to read and process what you've been told and then finish off the problem.

Solve $\displaystyle \frac{3n^2}{2} - \frac{n}{2} = 590 \Rightarrow \frac{3n^2}{2} - \frac{n}{2} - 590 = 0$ for n. Choose the value of n that is positive and satisfies $\displaystyle \frac{3n^2}{2} - \frac{n}{2} - 590 \geq 0$.