if a similar thread exists for verbal then please share the link, i couldn't find one :D
@realslimshady said:Which of the following numbers can be written as the sum of three odd natural numbers?A.5021B.4445C.3339D.1233with approach please
is it three odd consecutive nos.
then its 1233
@realslimshady said:if a similar thread exists for verbal then please share the link, i couldn't find one
@maddy2807 said:is it three odd consecutive nos.then its 1233
Bhai..3339 is also possibe..1111,1113,1115
@realslimshady said:Which of the following numbers can be written as the sum of three odd natural numbers?A.5021B.4445C.3339D.1233with approach please
Did you mean consecutive odd numbers?
1669 +1671 +1673
4445
First two not possible, as not divisible by 3.
3339 =1111+1113+1115
1233
409+411+413
proper ques kya hai? :/
@soumitrabengeri said:Bhai..3339 is also possibe..1111,1113,1115
yaar wahi to... possible to sare hai... locha hai kuch question me... ya fir koi point clearly mention nai hai?
@maddy2807 said:yaar wahi to... possible to sare hai... locha hai kuch question me... ya fir koi point clearly mention nai hai?
Yep..exactly my doubt..waiting for some changes in the question
The inverse of the sum of the following series upto n terms can be written as
3/4 + 5/36 + 7/144 + ...
1) (n-1)^2/(n^2+2n)
2) (n^2+2n)/(n-1)^2
3) (n^2+2n)/(n+1)^2
4) (n+1)^2/(n^2+2n)
@techsurge said:gareeb aadmi ki baesti same question came in my AI paper in last semester of Engg and I left, AI ki book mein dhundo, Apti ki book mein nahi is ka kuch specific naam tha, Other engineer puys might remember . I to have forgotten my engineering , kuch decrypt ...pata nahi
INLETS is the answer
What is the sum of the first 10 numbers of this series :
4 11 22 37....
basically difference of the difference of any two consecutive numbers is the same.
Approach plz.
Thanks is advance.
4 11 22 37
7 11 15
4 4
tn = an^2+bn+c
4 = a+b+c
11=4a+2b+c
22=9a+3b+c
11=5a+b
18=8a+2b
9=4a+b ; 11=5a+b a=2,b=1 c=1
tn =2n^2+n+1
now take summation and solve.
@maddy2807 said:The inverse of the sum of the following series upto n terms can be written as3/4 + 5/36 + 7/144 + ...1) (n-1)^2/(n^2+2n)2) (n^2+2n)/(n-1)^23) (n^2+2n)/(n+1)^24) (n+1)^2/(n^2+2n)
4?
@mohnish_khiani said:What is the sum of the first 10 numbers of this series :4 11 22 37....basically difference of the difference of any two consecutive numbers is the same.Approach plz.Thanks is advance.
835?
@maddy2807 said:approach b do bhai... help others too
i dont have kaapy pen with me...
so all i did was...
just took the first term,
y=3/4
inverse would be 4/3
check for the options...only fourth satisfies.
@mohnish_khiani said:What is the sum of the first 10 numbers of this series :4 11 22 37....basically difference of the difference of any two consecutive numbers is the same.Approach plz.Thanks is advance.
S = 4 + 11 + 22 + 37 + ...
S = .......4 + 11 + 22 + ..... (Tn)
Subtract and get Tn
Tn = 4 + [7 + 11 + 15 + ....(n-1) terms] = 4 + S1
S1 = 7 + 11 + 15 +.....
S1 = .......7 + 11 + ....... (Gn)
Subtract and get Gn
Gn = 7 + [ 4 + 4 + ...(n-2) terms] = 7 + 4*(n-2) = 4n - 1
S1 = Sigma(4n-1) = 2n(n+1) - n = 2n^2 + n
Tn = 2n^2 + n + 4
S = Sigma(Tn) = (n*(n+1)*(2n+1)/3) + (n*(n+1)/2) + 4n
For n = 10, S = (10*11*21/3) + ( 10*11/2) + 40 = 770 + 55 + 40 = 865 ?
PS: This is a general approach, there might be some calculation mistakes, also for this specific question, you can just do it manually as 10 terms is less enough
@maddy2807 said:The inverse of the sum of the following series upto n terms can be written as3/4 + 5/36 + 7/144 + ...1) (n-1)^2/(n^2+2n)2) (n^2+2n)/(n-1)^23) (n^2+2n)/(n+1)^24) (n+1)^2/(n^2+2n)
4 option..by first term ..it shud be 4/3
@maddy2807 said:The inverse of the sum of the following series upto n terms can be written as3/4 + 5/36 + 7/144 + ...1) (n-1)^2/(n^2+2n)2) (n^2+2n)/(n-1)^23) (n^2+2n)/(n+1)^24) (n+1)^2/(n^2+2n)
S = (1/1^2 - 1/2^2) + (1/2^2 - 1/3^2) + (1/3^2 - 1/4^2) + .......
= (1 - 1/(n+1)^2)
S = (n^2+2n)/(n+1)^2
= (1 - 1/(n+1)^2)
S = (n^2+2n)/(n+1)^2
But since inverse is asked, answer will be option (D)
Please correct me if i am wrong
@maddy2807 said:The inverse of the sum of the following series upto n terms can be written as3/4 + 5/36 + 7/144 + ...1) (n-1)^2/(n^2+2n)2) (n^2+2n)/(n-1)^23) (n^2+2n)/(n+1)^24) (n+1)^2/(n^2+2n)
This series can be written as :
(1/(1^2)) -(1/(2^2))+(1/(2^2))-(1/(3^2))+(1/(3^2))+....
Therefore sum upto n terms = 1-(1/(n+1)^2)
It's inverse : 4th option.
The lengths of the sides of a triangle are integers, and its area is also an integer. One side is 21 and the perimeter is 48 . What is the shortest side of the triangle?
p.s. - Currently No OA