# sorry asking the same question

• October 26th 2012, 07:04 AM
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 ?

thanks..............
• October 26th 2012, 07:20 AM
Salahuddin559
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
• October 26th 2012, 08:11 AM
TheEmptySet
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 ?

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

An arithmetic sequence can always be written as

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

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

So the sequence has the form

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

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

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

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

Now just plug everything in.

Also note that

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

Using the formula for the sum of the first n integers

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