1 ) The sum of all natural numbers from 100 to 300 which are divisible by 4 or 5??
2) If x = 1 + 1/3 + 1/3² +.............∞
y = 1 + 1/4 + 1/ 4² +............∞ Find xy
a)2
b)1
c)8/9
d)1/2
pls solve this....
thanks in advance..
Anjana (ipcc) (46 Points)
03 June 20111 ) The sum of all natural numbers from 100 to 300 which are divisible by 4 or 5??
2) If x = 1 + 1/3 + 1/3² +.............∞
y = 1 + 1/4 + 1/ 4² +............∞ Find xy
a)2
b)1
c)8/9
d)1/2
pls solve this....
thanks in advance..
ROSHAN PAUL JOSEPH
(CA Final student)
(311 Points)
Replied 03 June 2011
1 ) The sum of all natural numbers from 100 to 300 which are divisible by 4 or 5??
ROSHAN PAUL JOSEPH
(CA Final student)
(311 Points)
Replied 03 June 2011
100 /4= 25
300/4=75
100/5=20
300/5=60
100/20=5
300/20=15
Sum,S = 4*(25+26+…….+75)+ 5*(20+21+…….+60)-20*(5+6+…+15)
S=4a+5b-20c
Sum of n terms of A.P= (n/2) * {2a + (n-1)d}
a={(75-24)/2}*[2*25+(51-1)*1]
a= (51/2) * (50+50)
a=2550
b={(60-19)/2}*[2*20+(41-1)*1]
b= (41/2) * (40+40)
b=1640
c=={(15-4)/2}*[2*5+(11-1)*1]
c= (11/2) * (10+10)
c=110
S= 2550*4 + 1640*5 – 110*20
=16200
ROSHAN PAUL JOSEPH
(CA Final student)
(311 Points)
Replied 03 June 2011
2) If x = 1 + 1/3 + 1/3² +.............∞
y = 1 + 1/4 + 1/ 4² +............∞
xy=2
ROSHAN PAUL JOSEPH
(CA Final student)
(311 Points)
Replied 03 June 2011
Sum of n terms of GP when common ratio is less than 1;
S= a (1- rn) / (1-r)
x = 1* [1- (1/3n)] / [ 1- (1/3)]
y= 1* [1- (1/4n)] / [ 1- (1/4)]
xy= 1*[1- (1/3n)] / (2/3) * 1 *[1- (1/4n)]/ (3/4)
=1*1* (3/2) * (4/3) * [1- (1/3n)] * [1- (1/4n)]
=2*[ 1 - (1/3n) - (1/4n) + (1/12n)]
=2
(1/3n) (1/4n) (1/12n) are approximately equal to 0
Mukesh Kumar Singh
(CA-FINAL)
(4094 Points)
Replied 21 June 2011
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hitesh
(CA Final Student)
(147 Points)
Replied 21 June 2011
 Originally posted by : mukesh kumar singh |
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> what's my name > what's my name > what's my name > My name is Sheela > Sheela ki jawani > I'm just s*xy for you > Main tere haath na aani > Na na na sheela > Sheela ki jawani > I'm just s*xy for you > Main tere haath na aani |
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