find the sum of series :1/2+1/6+1/12+1/20+.........+1/156+1/182 what is sum if taken to infinite terms ??
Amit31 Saysfind the sum of series :1/2+1/6+1/12+1/20+.........+1/156+1/182 what is sum if taken to infinite terms ??
series is
1/n+1/(n+4)+1/(n+10)+...1/(n+180)
addition of 4 to first term,6 to second term...
2,6,12,20,30,42,56,72,90,110,132,156,182=>13 terms.
for 2 terms,it is 2/3
for 3 terms,it is 3/4
for 4 terms,it is 4/5
for 13 terms,it is 13/14
for n terms,it is n/(n+1)
Amit31 Saysfind the sum of series :1/2+1/6+1/12+1/20+.........+1/156+1/182 what is sum if taken to infinite terms ??
Dunno how you would calculate it for infinite terms.... 😞
1/2+1/6+1/12+1/20+.........+1/156+1/182
= 1/(1 x 2) + 1/(2 x 3) + 1/(3 x 4) +..............1/(13 x 14)
= 1/1 - 1/2 + 1/2 - 1/3 + ............1/13 - 1/14
= 1 - 1/14 = 13/14
Dunno how you would calculate it for infinite terms.... 😞
1/2+1/6+1/12+1/20+.........+1/156+1/182
= 1/(1 x 2) + 1/(2 x 3) + 1/(3 x 4) +..............1/(13 x 14)
= 1/1 - 1/2 + 1/2 - 1/3 + ............1/13 - 1/14
= 1 - 1/14 = 13/14
What u did is correct, in case of infinity the answer wld be 1, as each term wld be canceled with the succeeding term.
else as shashank found for n terms,it is n/(n+1)
now, Lt n->inf Sigma{n/n+1 }
Lt 1/n -> 0 Sigma{ 1/(1+1/n)} = 1
let bi is the number of birds and br be the number of branches
from the info in the question we can arrive at I and II
solve the two eqns to get br = 3
Can some one explain this answer clearly?? I got confused of this question.Pls. help me
1) Four Bells ring at the intervals of 6,8,12,&18 seconds.They started ringing together at 12'0 Clock.After how many seconds they will ring together again??
a) 72 b) 84 c) 60 d) 48
2)For the above problem, How many times they eill ring together durnig the next 12 mins.
a) 9 b) 10 c) 11 d) none
3)The Product of digits of two digit number is twice as large as the sum of its digits.If we subtract "27" from the number, we get a number consisting of same digits written in reverse order.Find the No.??
4)Find no. of Zero's in the Product 1 X 2^2 X 3^3 X 4^4 X 5^5...............100^100
1) Four Bells ring at the intervals of 6,8,12,&18 seconds.They started ringing together at 12'0 Clock.After how many seconds they will ring together again??
a) 72 b) 84 c) 60 d) 48
2)For the above problem, How many times they eill ring together durnig the next 12 mins.
a) 9 b) 10 c) 11 d) none
3)The Product of digits of two digit number is twice as large as the sum of its digits.If we subtract "27" from the number, we get a number consisting of same digits written in reverse order.Find the No.??
4)Find no. of Zero's in the Product 1 X 2^2 X 3^3 X 4^4 X 5^5...............100^100
1) Take Lcm(6,8,12,18 ) = 72
2)(12*60)/Lcm(6,8,12,18 ) = 720/72 = 10
3)ab = 2(a+b), 10a+b - 27 = 10b+a
9a-9b = 27, a-b = 3
easily we can see the number is 63
4)calculate number of 5's
5 + 15 + 50 + 35 + 45 + 55 + 65 + 150 + 85 + 95 + 700 = 1300
HI,
Pls answer these following qsns......
1) Three numbers are such that second is as much lesser than the third as the first is lesser than the second.If the product of two smaller numbers is 85 and product of two larger numbers is 115.Find the middle no
a) 9 b) 8 c) 10 d) 30
2) AB+XY=2XP where A!=0 all the letters signify different digits from 0 to 9 , then the value of "A" is
a) 6 b) 7 c)9 d)any number more than "6"
3) X-4 + Y-4 = 4 , then how many Integer values can (X,Y) have??
4) A Two digit number is thrice as large as the sum of its digits and square of that sum is equal to trebled of required number,then find the no.??
solve and explain....
Thanks,
HI,
Pls answer these following qsns......
1) Three numbers are such that second is as much lesser than the third as the first is lesser than the second.If the product of two smaller numbers is 85 and product of two larger numbers is 115.Find the middle no
a) 9 b) 8 c) 10 d) 30
2) AB+XY=2XP where A!=0 all the letters signify different digits from 0 to 9 , then the value of "A" is
a) 6 b) 7 c)9 d)any number more than "6"
3) X-4 + Y-4 = 4 , then how many Integer values can (X,Y) have??
4) A Two digit number is thrice as large as the sum of its digits and square of that sum is equal to trebled of required number,then find the no.??
solve and explain....
Thanks,
1)a,b,c
b-a = c-b {Implies they are in AP}
ab = 85, bc = 115
c/a = 115/85 = 23/17{ since they are prime, they can be equated to c & a}
17, b, 23
so,b = 20
2)Its not possible!
two digit numbers when added can't giv a number more than 200,{Max is 99 + 99 = 198}
If the question is AB +XY = 1XP then out of options A can take 9 since A + x = x, so A can be 9 or 0.
3)If we draw a graph, we get a right angle triangle with vertices (0,0),(0,4),(4,0)
Now the integer points which lies with in hte area of triangle are
(1,3)(3,1)
(2,2)
(0,4)(4,0)
4)(10a + b) = 3*(a+b) ..... i
(a+b)^2 = = 3*(10a+b) ... ii
solve we get a+b = 9
so number is 27
hello puy.........my doubt is
The numbers 2,4,6,8...98,100 are multiplied together. The number of zeros at the end of the product must be :
13
12
11
10
9
hello puy.........my doubt is
The numbers 2,4,6,8...98,100 are multiplied together. The number of zeros at the end of the product must be : 13
12
11
10
9
Calculate the no of 5 in that
10,20,30,40,60,70,80,90 - one 5's each
50,100 - two 5's each
So total = 8+4 =12 zeros
Calculate the no of 5 in that
10,20,30,40,60,70,80,90 - one 5's each
50,100 - two 5's each
So total = 10 zeros
from 10 to 90 total 8 zero
and for 50 & 100 total 4 zero
so ans =12
from 10 to 90 total 8 zero
and for 50 & 100 total 4 zero
so ans =12
but the ans given is 11
shona_ne Saysbut the ans given is 11
It is not given factorial to calculate number of 5's
Its just 2*4*6*8*10*.....*100
zeros are contributed by 10,20,30,...100
so will be having 11 zeros at the end
It is not given factorial to calculate number of 5's
Its just 2*4*6*8*10*.....*100
zeros are contributed by 10,20,30,...100
so will be having 11 zeros at the end
but menacebhai 50 and 100 both have 2 times 5
jigar_p_civil Saysbut menacebhai 50 and 100 both have 2 times 5
multiply them individually
50*100 = 5000
we are not going to consider number of 5's
unless it is
50! ,here number of zeros wld be + = 10 + 2 = 12
multiply them individually
50*100 = 5000
we are not going to consider number of 5's
unless it is
50! ,here number of zeros wld be + = 10 + 2 = 12
ok i got it
but if question like 2,4,6,.........1000
then how can we judge?
multiply them individually
50*100 = 5000 * some even no = 10000
we are not going to consider number of 5's
unless it is
50! ,here number of zeros wld be + = 10 + 2 = 12
But we have other even nos which on mulltiplication gives one more zero...see it in bold
sumitbce SaysBut we have other even nos which on mulltiplication gives one more zero...see it in bold
Maafi 😃
we get 12 zeros at the end
coz, extra zero is contributed by 50
ok i got it
but if question like 2,4,6,.........1000
then how can we judge?
single zeros, 10,20,...,90,110120,...190,..like wise till 990 we get 99 zeros
Double zeros contributed by 100,200 ...till 900 we get 9 more
so, 99+9 = 108
3 zeros contibuted by 1000 so,108+3= 111
and there are few more zeros contributed by
50,150,250,350...950 so 10 more
total will be 111 + 10 = 121 zeros in all