Thread: 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?

please help me

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

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..

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...

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 that is greater than what you found"???

Originally Posted by mastermin346

what do you mean by "Choose the first whole value of 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 $\displaystyle \frac{3n^2}{2} - \frac{n}{2} - 590 \geq 0$.