GATE Question of the Day - General Aptitude Question



The sum of n terms of the series 4+44+444+------ is


Please give the reason for your answer. All the best

15 comments:

  1. VamshiSeptember 1, 2012 at 10:20 PM

    option C....just take 4 as common and take the g.p of the terms ... we will get the answer..

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  2. AnonymousSeptember 1, 2012 at 10:32 PM

    It's C
    Its just as easy as substituting value of 'n' and checking options.

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  3. VamshiSeptember 1, 2012 at 10:33 PM

    if it is right post me...

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  4. VamshiSeptember 1, 2012 at 10:34 PM

    yaaa checking is easy... bt we can do it through process also

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  5. SouravManiSeptember 2, 2012 at 8:38 AM

    ya from my knowledge answer is "c"! but i have doubt on this?

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  6. VamshiSeptember 2, 2012 at 9:00 AM

    dude may i know what is that?

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  7. UnknownSeptember 2, 2012 at 12:20 PM

    yah it is C because it is in the geometric progression

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  8. SouravManiSeptember 2, 2012 at 11:29 PM

    hey after a stong confirmation i found the right answer the same option "C"

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  9. UnknownSeptember 3, 2012 at 10:45 AM

    option c......
    simple trick is that put only n=o,then we see that all term become zero.which only in case option c.so don't waste your time use short trick.

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  10. AnonymousSeptember 4, 2012 at 12:35 PM

    Option C

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  11. UnknownSeptember 5, 2012 at 2:35 PM

    yaa... put n=o.. den only in c ol terms wil be zero... so ans. is "C"

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  12. AnonymousSeptember 5, 2012 at 11:52 PM

    ans iz c

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  13. ExamsavvySeptember 6, 2012 at 2:44 PM

    Answer to this question is option C

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  14. luckySeptember 17, 2012 at 9:32 PM

    this question is very easy.put n=1 then u will get the 4...so option c

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  15. AnonymousSeptember 27, 2012 at 10:29 AM

    sum= 4+44+444+.........+tn
    sum= 4+ 44+.........+tn-1+tn

    subtracting above two equations we get

    0=4+40+400+.............+(tn-tn-1)-tn
    tn=(4+40+400+...n terms)
    tn=4*(10^(n)-1)/(10-1) -- G.P summation
    sum=sigma of tn
    sum=4/9*(sigma of 10^(n)-sigma of 1)
    sum=4/9*[10*(10^(n)-1)/(10-1)-n]
    sum=option 'C'

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