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Ask iit jee aieee pet cbse icse state board community Community Discussion Question: Answer it......(rates assured)
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[Post New]21 Dec 2007 13:40:00 IST
    Subject: Answer it......(rates assured)
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nishant_delhi (0)

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Q.  Find the sum to n terms of the sequence , 8, 88, 888, 8888.........
(plz.. explain briefly)
    
[Post New]21 Dec 2007 14:09:42 IST
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srujana (3220)

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8, 88, 888, 8888.........
 
can be written as
 
8(1+11+111+.......upto n terms)
multiplying and dividing by 9
 
8/9 (9+99+999+.....upto n terms)
 
8/9 (10+100+1000+.......upto n terms  - (1+1+......n))
 
8/9 (10/9 (10n -1) - n )) 
 
 [ sum upto n terms for a progression in G.P.  a/(r-1) * (rn-1) ]
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[Post New]25 Dec 2007 23:22:34 IST
    Subject: Re:Answer it......(rates assured)
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karthik_18049200 (52)

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there is  a formula
a+aa+aaa+aaaa+..................nterms =a/9(10+102+103+.......10n-(1+1+.......n))
=a/9(10(10n-1)-9n/9)
=a/81(10n+1-9n-10)
this is applicable for all a belongs to R
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[Post New]25 Dec 2007 23:29:39 IST
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chimanshu_007 (11437)

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yeah....correct

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