# Arithmetic progression problem

• Jan 8th 2010, 05:02 PM
mastermin346
Arithmetic 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?

• Jan 8th 2010, 05:29 PM
skeeter
Quote:

Originally Posted by mastermin346
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?

$\displaystyle S_n = \frac{n}{2}(t_1 + t_n) > 590$

where $\displaystyle t_n = t_1 + d(n-1)$

solve for the least value of $\displaystyle n$ that makes the sum greater than 590
• Jan 8th 2010, 08:41 PM
mastermin346
Quote:

Originally Posted by skeeter
$\displaystyle S_n = \frac{n}{2}(t_1 + t_n) > 590$

where $\displaystyle t_n = t_1 + d(n-1)$

solve for the least value of $\displaystyle n$ that makes the sum greater than 590

how?? i dont know how to do..
• Jan 8th 2010, 09:14 PM
Prove 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 PM
mastermin346
Quote:

Originally Posted by Prove 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...

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 PM
mr fantastic
Quote:

Originally Posted by mastermin346
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"???

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 $\displaystyle \frac{3n^2}{2} - \frac{n}{2} - 590 \geq 0$.