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Pls help me out...

CPT 856 views 7 replies

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..

Replies (7)

1 ) The sum of all natural numbers from 100 to 300 which are divisible by 4 or 5??


sum =16200

 

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

 

2)  If  x = 1 + 1/3 + 1/3² +.............∞

 

    y = 1 + 1/4 + 1/ 4² +............∞



xy=2

 

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

can u simply ellaburate pls

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Originally posted by : mukesh kumar singh

> 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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