sorry asking the same question

Hello friends ,

I am solving this question:

The sum of how may tersm of the series 6 + 12 + 18 + 24 + ... is 1800 ?

where I came across this equation n^2+n-600=0

got the values n=24 or n=25

but i dident understood how to put these values in the eqn as the answer is

ie sum of the terms = only 24

and the question is

sum of how may tersm of the series 6 + 12 + 18 + 24 + ... is 1800?

so how to calcullare sum of the terms ?

so please help me

thanks..............

Re: sorry asking the same question

Hi, it is simple arithmetic series sum. The values are not 24 and 25, those are 24 and -25 (minus 25). So, the answer is 24.

Salahuddin

Maths online

Re: sorry asking the same question

Quote:

Originally Posted by

**manishamishra** Hello friends ,

I am solving this question:

The sum of how may tersm of the series 6 + 12 + 18 + 24 + ... is 1800 ?

where I came across this equation n^2+n-600=0

got the values n=24 or n=25

but i dident understood how to put these values in the eqn as the answer is

ie sum of the terms = only 24

and the question is

sum of how may tersm of the series 6 + 12 + 18 + 24 + ... is 1800?

so how to calcullare sum of the terms ?

so please help me

thanks..............

An arithmetic sequence can always be written as

$\displaystyle a_n=a_1+d(n-1)$

Where $\displaystyle a_1$ is the first term and $\displaystyle d$ is the common difference.

So the sequence has the form

$\displaystyle a_n=6+6(n-1)=6n$

So to find out how many terms are in the sequence set $\displaystyle a_n=1800 \iff 6n=1800 \iff n=300$

The sum of the first $\displaystyle n$ terms of an arithmetic sequence is given by

$\displaystyle S_n=\frac{n(a_1+a_n)}{2}$

Now just plug everything in.

Also note that

$\displaystyle \sum_{n=1}^{300}6n=6\sum_{n=1}^{300}n=6 \cdot \frac{300\cdot 300}{2}$

Using the formula for the sum of the first n integers

$\displaystyle \sum_{n=1}^{k}=\frac{k(k+1)}{2}$