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Question on A.P. formula

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knightwalker Just gettin' started! Default Avatar
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Question on A.P. formula Post Mon Aug 31, 2009 7:24 am
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  • Lap #[LAPCOUNT] ([LAPTIME])
    In an increasing sequence of 10 consecutive integers the sum of the first 5 integers in 560. What is the sum of the last 5 integers in the sequence?

    now the approach given in the answer is fairly straightforward, saying that sum of first five integers from x,x+1,x+2,x+3 & x+4 gives us the equation 5x+10=560,thus 5x=550 and x=110.

    However when i tried to do it on my own i applied the formula for the sum of an Arithmetic Progression which is:

    (2x+(n-1)d)*n/d

    where x is the first term of the A.P., n is the number of terms, and d is the difference b/w terms.
    when applying this formula you end up with,
    (2x+(5-1)1)*5/1=560
    -> 2x+4=112
    -> x=108/2 which equals 54

    am i making some really silly mistake here? is the formula wrong? (i rechecked it in a couple of places)... does it not apply for consecutive numbers? Conceptually the answer is obviously wrong as 5 consecutive numbers adding to over 500 means each must be greater than 100 but i can't seem to see the mistake i'm making...

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    sreak1089 GMAT Destroyer! Default Avatar
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    Post Mon Aug 31, 2009 7:38 am
    Hi knightwalker,

    Your formula seems to be wrong. Correct formula for the AP is:

    (n/2)[(2a + (n-1)d]

    => (5/2)[2x + 4] = 560
    => 10x + 20 = 560 * 2
    => 2x + 4 = 224
    => x = 220/2 = 110

    Hope this clarifies.

    Thanked by: knightwalker
    knightwalker Just gettin' started! Default Avatar
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    Post Mon Aug 31, 2009 7:47 am
    thanks a ton for the clarification sreak

    bpgen Really wants to Beat The GMAT!
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    Post Mon Mar 22, 2010 7:50 pm
    another approach:

    Take average of consecutive numbers=> (first+last)/2
    Applying this above formula, we get, (n+(n+4))/2=560/5 => n=110

    now, sum of last five consecutive numbers would be (((n+5)+(n+9))/2) *5 (i.e average*5)

    Therefore after substituting n, we get (((n+5)+(n+9))/2) *5 =>585

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    kstv GMAT Destroyer! Default Avatar
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    Post Mon Mar 22, 2010 8:55 pm
    The sum of the first 5 is 560
    Let us say the average of 1st 5 nos be x ( the actual value is 560/5, but that's not imp)
    this is the 3rd no.
    the average of 6 - 10 is the 8th no
    i.e x+5
    so the total 6 to 10 will be 25 more than the average of 1st to 5th
    i.e 560+25 = 585
    I have a firm belief that all AP sums in GMAT are solvable without memorising any formula just the concepts.
    Please correct me if it is otherwise or better still, prove me wrong ('ll appreciate it!).

    suhaibsyed Just gettin' started! Default Avatar
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    Post Tue Jun 19, 2012 6:31 am
    thanks

    jcnasia Just gettin' started!
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    Post Sat Jun 23, 2012 4:57 am
    If a, b, c, d, and e are consecutive integers and a+b+c+d+e=560, then the next five integers are...
    a+5, b+5, c+5, d+5, and e+5, and they add up to...

    a+b+c+d+e+(5*5) = 560 + 25 = 585

    Thanked by: theCEO

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