Results 1 to 3 of 3

Math Help - sorry asking the same question

  1. #1
    Newbie
    Joined
    Oct 2012
    From
    Pune
    Posts
    3

    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..............
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Junior Member
    Joined
    Oct 2012
    From
    India
    Posts
    61
    Thanks
    3

    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
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
    Joined
    Feb 2008
    From
    Yuma, AZ, USA
    Posts
    3,764
    Thanks
    78

    Re: sorry asking the same question

    Quote Originally Posted by manishamishra View Post
    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

    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}
    Follow Math Help Forum on Facebook and Google+

Search Tags


/mathhelpforum @mathhelpforum